# The Mathematical Association of America Published and Distributed by

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The Mathematical Association of America Published and Distributed by
Project 1 The Mathematical Association of America Published and Distributed by Marketing Computer Drives Part 2: Calculus & Optimization Graphing Functions Trend Lines Demand, Revenue, Cost, and Profit Differentiation Using Solver Integration Marketing Example Video introduction by Prof. Christopher Lamoureux Department of Finance University of Arizona. Title Mathematics for Business Decisions Part 2 Release 2, 2009 C I © 2009 by The Arizona Board of Regents for The University of Arizona. All rights reserved.

Graphing Functions, Plotting Points
The ability to produce informative, attractively formatted plots is of importance in almost all areas of business. Graphing the functions that are used in marketing, or probability densities and cumulative distributions for continuous random variables requires some special computer skills. In particular, many points are needed for Excel to accurately graph these, and other “smooth” functions. Example 1. We will start by plotting a simple function which could be the p.d.f. of a continuous random variable that takes values in the interval [1, 3]. To graph f over the interval [0, 4], we subdivide [0, 4] with a large number of points. For this example, we will use 501 points x0, x1, , x500 that are evenly spaced from 0 to 4, with x0 = 0 and x500 = 4. T C I (material continues)

Graphing Functions. Plotting Points: page 2
Graphing, Points With this arrangement, the distance between any two adjacent points is (4 - 0)/500 = This means that x0 = 0.000, x1 = 0.008, x2 = 0.016, , x500 = In the sheet Graph 1 of the Excel file Graph Examples.xlsx we have entered 0 in Cell B9 and in Cell B10. The pattern is extended by dragging these two cells down to Cell B509. The formula for f is entered in Cell C9, using the IF function. Recall that IF is found in the Logical list under Paste Function. It takes the form IF(statement, rule 1, rule 2) If the statement is true, IF uses rule 1. When the statement is false, IF uses rule 2. We will use -0.75*B9^2+3*B for rule 1 and 0 for rule 2. With these rules, we want our statement to be 1  x  3, that is, 1  B9  3. Graph Examples.xlsx (material continues) T C I

Graphing Functions. Plotting Points: page 3
Graphing, Points Excel does not recognize a continued inequality such as 1  B9  3. We must realize that this means 1  B9 and B9  3. These are entered using the AND function, found in the Logical list of the Insert Function Dialog Box. Graph Examples.xlsx (material continues) T C I

Graphing Functions. Plotting Points: page 4
Graphing, Points The And function is used as the statement in the IF function, completing the definition of f in Cell C9. Cell C9 in now dragged down to Cell C509. Graph Examples.xlsx T C I (material continues)

We select this type of plot and enter our data range.
Graphing Functions. Plotting Points: page 5 Graphing, Points The best graph of our 501 points is made using an XY (Scatter) plot, with smoothed lines and no markers. To find this, click Insert, then on the Create Chart box We select this type of plot and enter our data range. Create Chart Graph Examples.xlsx (material continues) T C I

Graphing Functions. Plotting Points: page 6
Graphing, Points Right click in the plot region, then left click on Select Data. Once the data is entered, we can format the plot in any way that we like. Graph Examples.xlsx (material continues) T C I

Graphing Functions. Plotting Points: page 7
Graphing, Points The best effect is usually obtained by clicking on the plotted points, clicking on Format/ Selected Data Series, and then selecting a medium line width with no markers. Graph Examples.xlsx T C I (material continues)

Graphing Functions. Plotting Points: page 8
Graphing, Points Here is our final graph, copied from Graph Examples.xlsx. Graph Examples.xlsx (material continues) T C I

Graphing Functions. Plotting Points: page 9
Graphing, Points Start with a blank Excel file and use 701 evenly spaced points to plot the graph of f(t) = 2,000e0.06t over the interval [-12, 12]. Recall that Excel uses EXP(x) for ex. This graph shows you the value of \$2,000, invested at 6%, compounded continuously for t years. When t is negative, f(t) gives the past value. When t is positive, f(t) gives the future value of the \$2,000. Exercise 1 Use the plot that you made in Exercise 1 to estimate the length of time that it will take for \$2,000 to grow to \$3,000, at 6% compounded continuously. Exercise 2 Start with a blank Excel file and use 501 evenly spaced points to plot the graph of over the interval [-2, 10]. Note that this function is the c.d.f. for an exponential random variable with parameter  = 1.5. Exercise 3 Excel T C I (material continues)

Graphing Functions. Plotting Points: page 10
Graphing, Points Plot the graph of f(x) =  x over the interval [0, 2,000]. Exercise 4 Start with a blank Excel file and use 501 evenly spaced points to plot the graph of over the interval [4, 4]. To enter the constant  in Excel, type PI(). In Excel, the square root function is entered as SQRT. Exercise 5 (i) Have Excel display vertical gridlines, spaced 0.5 units apart, in the graph that you made in Exercise 5. Add horizontal gridlines, spaced 0.1 units apart, so that the area of each gridline rectangle is 0.05 square units. (ii) Use the rectangles to estimate the total area of the region between the x-axis and the graph of f, over the interval [4, 4]. Exercise 6 Excel T C I (material continues)

Graphing Functions, Several Functions
Graphing two or more functions in the same plot can often help us understand the connections between the quantities that are represented by the functions. Example 2. Here are two functions that are typical of those which we will encounter in our study of marketing. We will plot both f and g over the interval [0, 2,000], using 401 equally spaced points x0, x1, , x400, with x0 = 0 and x400 = 2,000. The distance between any two consecutive points is 2,000/400 = 5. This means that x0 = 0, x1 = 5, x2 = 10, , x400 = 2,000. In the sheet Graph 2 of Graph Examples.xlsx we have entered 0 in Cell B9 and 5 in Cell B10. The pattern is extended by dragging these two cells down to Cell B409. T C I (material continues)

Graphing Functions. Several Functions: page 2
Graphing, Several The formulas for f and g are entered in Cells C9 and D9, using the IF function for g. We select an XY (Scatter) plot, with smoothed lines and no markers. Right click in the plot region, then left click on Select Data. Enter the indicated data range. Graph Examples.xlsx (material continues) T C I

Graphing Functions. Several Functions: page 3
Graphing, Several Click on Add under Legend Entries (Series). We name Series 1 as f and enter its data ranges. Series 2 is named g, and its data is entered. Finally, the plot is formatted as desired. Graph Examples.xlsx (material continues) T C I

Graphing Functions. Several Functions: page 4
Graphing, Several Look at the graph in the sheet Graph 2 of Graph Examples.xlsx to see how the final formatting was done. The graph is copied on this page. Use the plot of f and g from Example 2 to estimate the two values of x at which f(x) = g(x). Exercise 7 Graph Examples.xlsx (material continues) T C I

Graphing Functions. Several Functions: page 5
Graphing, Several Show the graphs of both f(x) = 60x + 20,000 and g(x) = 0.2x x in the same plot, over the interval [0, 1,000]. Exercise 8 Let (i) Show the graphs of both f and g in the same plot, over the interval [0, 8]. Recall that in Excel, the square root function is entered as SQRT. (ii) Use your plot to estimate the coordinates of the point at which f(x) = g(x). (iii) How would you describe the relationship between the graphs of f and g at the point where they intersect? Exercise 9 The c.d.f. for an exponential random variable X, with parameter  is given by Show three graphs of F, with  = 2, 3, and 4, in a single plot, over the interval [5, 15]. Exercise 10 Excel T C I (material continues)

Graphing, A Graphing Utility
Graphing Functions. A Graphing Utility Graphing, A Graphing Utility Click here for information on running the macro in Graphing.xlsm. 3. A GRAPHING UTILITY It is important that you know how to create your own graphs. This will be necessary for special plotting needs. Understanding the process of computer graphing also helps you know what the computer is doing and assists you in your interpretation of the results. However, in routine situations the creation of plots can be time consuming and could distract us from our main goal of understanding the business applications of a function. For this reason, we are supplying an Excel file, Graphing.xlsm, that serves as a graphing utility. To use Graphing.xlsm, open the file and follow the directions in the blue box. Be sure to note that you are to enter a formula for f(x) in terms of the letter x, not a cell reference. x is the only letter that can be used as a variable in Graphing.xlsm. Graphing.xlsm T C I (material continues)

Graphing Functions. A Graphing Utility: page 2
Before running Graphing.xlsm, be sure that Excel is set for Automatic Calculation. When entering a function in Graphing.xlsm, you can use the letters s, t, u, v, or w as constants. If this is done, you must enter values for any letters that you use. For example, to plot f(x) = x s, where s is a number that you enter in Cell L17, enter the following function in Cell C18. = x^s Accept the function and then click on the Graph button at the right of the plot. Entering a new number in Cell L17 will automatically change the function and redraw the plot. If values are entered in Cells L17:L21; then constants s, t, u, v, or w may also be used to describe the range, [a, b], of the plot. If these constant names are used, they must be entered in Cells H18:I18 preceded with "=" signs. For example, =s. Graphing.xlsm T C I (material continues)

Graphing Functions. A Graphing Utility: page 3
Use Graphing.xlsm to plot the graph of f(x) = 2,000e0.06t over the interval [-12, 12]. Exercise 11 Use Graphing.xlsm to plot the graph of F(x) over the interval [-5, 15] Note that this function is the c.d.f. for an exponential random variable with parameter  = 3.5. Exercise 12 Use Graphing.xlsm to plot the graph of fX over the interval [-5, 15] Note that this function is the p.d.f. for an exponential random variable with parameter  = 3.5. Exercise 13 Graphing.xlsm T C I (material continues)

Graphing Functions. A Graphing Utility: page 4
Redo Exercise 13, but leave the parameter  as a constant. Changing  and recomputing the sheet should change the plot correspondingly. Hint: use the built-in constant, s, as a replacement for . Exercise 14 Let  = 1, and use Graphing.xlsm to plot the graph of over the interval [4, 8]. To enter the constant  in Excel, type PI(). In Excel, the square root function is entered as SQRT. Exercise 15 (i) Redo Exercise 15, but leave the parameter  as a constant. Changing  and recomputing the sheet should change the plot correspondingly. Hint: use the built-in constant, s, as a replacement for . (ii) Display graphs for  = 0, 1, 2, and 4. Exercise 16 (material ends) Graphing.xlsm T C I

Trend Lines, Fitting Models
Mathematical analysis of business situations often requires formulas for the functions that are used as models. Unfortunately, the real world almost never provides us with formulas. What we do find in business are data points that provide approximate values for the models. Trend lines allow us to use data points to generate formulas which can be used for business planning. Example 1. (An Exponential Model) A county Child Protective Services, CPS, agency has records showing the number of cases that it has handled during the years from 1998 through 2006. Noting that the case load is increasing rather rapidly, the agency would like to predict how many cases it is likely to handle in the year 2010. The Excel file Case Load.xlsx shows a plot of the nine data points. For convenience in later work, we have rescaled time as the number of years after 1998. Case Load.xlsx T C I (material continues)

Trend Lines. Fitting Models: page 2
Trend Lines, Models We would like to find a formula for the function that represents the underlying pattern of the case load growth. To do this, we must first use our judgment to select the type of functions that we will consider. For the CPS data we note the following. (i) The data points appear to increase and curve slightly upward. (ii) Many changing populations have patterns of exponential growth. (iii) Experience in public administration around the country suggests that social service case loads increase exponentially. For these reasons we will assume that the number of cases handled in a year that is t years after 1998, is given by an exponential function of the form f(t) = uevt, where u and v are constants. Recall that e is a constant that is the base for the natural exponential function. It can be shown that e  The exponential function is defined for all values of x, and is entered in Excel with upper case letters as EXP(). Once we have selected a form for f(t), Excel can find values for the constants u and v that produce the function of the given type, that best fits the data points. Such a function is called a trend line. Case Load.xlsx T C I (material continues)

Trend Lines. Fitting Models: page 3
Trend Lines, Models We will select an exponential trend line for the CPS data. Open Case Load.xlsx to see a plot of the nine original data points. To add a trend line: 1. Left click on any data point. 2. Right click to see the pull-down menu. 3. Click on Add Trendline. Case Load.xlsx T C I (material continues)

Trend Lines. Fitting Models: page 4
Trend Lines, Models 4. Click on Exponential. 5. Click on Trendline Options. Case Load.xlsx (material continues) T C I

Trend Lines. Fitting Models: page 5
Trend Lines, Models 6. To move to 2010, set Forward to 4. 8. Click on Close. 7. Click on Display Equation on chart. Case Load.xlsx (material continues) T C I

Trend Lines. Fitting Models: page 6
Trend Lines, Models The equation of the exponential trend line is displayed on the plot. We see that the coefficients are u = and v = To use these numbers in the Excel file, you must either copy them manually, or with copy/paste. Case Load.xlsx (material continues) T C I

Trend Lines. Fitting Models: page 7
Trend Lines, Models Let f(t) = 2, e0.1217t. Since we are using t as the number of years since 1998, the expected case load in a year y is f(y  1998). Case Load.xlsx shows that the agency can expect 10,225 cases in the year Their load has been growing at a yearly rate of approximately 12%. The extension of a trend line beyond the range of the known data is called extrapolation. This must be done with great caution. It is not unreasonable to expect the current pattern of growth to continue for the next four years. However, it would be absurd to expect the same circumstances to continue for the next 100 years. As Case Load.xlsx shows, f(2106  1998) = 1,212,295,365. Obviously, we should not assume that CPS will handle over one billion cases in the year 2106. The choice of a basic form for a trend line is of major importance. Further training in business will help you recognize the appropriate types of functions to use in common situations. Important! Case Load.xlsx T C I (material continues)

Trend Lines. Fitting Models: page 8
Trend Lines, Models Example 2. (A Linear Model) We will illustrate the effect of different types of trend lines by fitting a linear function to the CPS data points. This means looking for the best fitting curve of the form f(t) = at + b. This is done in exactly the same way that we used for the exponential model, except that we select the Linear box on Excel’s menu. Once the coefficients of the trend line are computed, we can click on the equation box and format the numbers. For the linear trend line, we have displayed a and b as integers. Case Load.xlsx T C I (material continues)

Trend Lines. Fitting Models: page 9
Trend Lines, Models We see that the coefficients are a = 487 and b = 2,112. To use these numbers in the Excel file, you must either copy them manually, or with copy/paste. Case Load.xlsx (material continues) T C I

Trend Lines. Fitting Models: page 10
Trend Lines, Models Let f(t) = 487t + 2,112. The expected case load in a year y is f(y  1998). Case Load.xlsx shows that the agency can expect 7,956 cases in the year This is approximately 22% fewer cases than were predicted with the exponential model. Will growth be exponential, linear, or neither? Only time can tell with any certainty. The difference in choice of models is particularly noticeable over long periods of extrapolation. As computed in Case Load.xlsx, the linear model predicts a load of 54,708 cases in the year This is approximately % of the 1,212,295,365 cases predicted by the exponential model! Ask a tutor How do I know what type of trend line to use? Case Load.xlsx T C I (material continues)

Trend Lines. Fitting Models: page 11
Trend Lines, Models (i) Use the exponential model in the CPS example to predict the case load in the year (ii) Repeat Part i, using the linear model. Exercise 1 (i) Use the exponential model in the CPS example to predict the case load in the year (ii) Repeat Part i, using the linear model. Exercise 2 (i) Fit a 3rd degree polynomial trend line through the data points in the CPS example. (ii) What case load does this model predict for the year 2010? Exercise 3 (i) Fit a 6th degree polynomial trend line through the data points in the CPS example. (ii) What case load does this model predict for the year 2010? (iii) Does increasing the degree of a polynomial trend line seem to improve its predictive value? Exercise 4 Case Load.xlsx T C I (material continues)

Trend Lines. Fitting Models: page 12
Trend Lines, Models During the first 6 days of a special promotion, a small business records the following sales information. The dollar amount listed for each day is the total cumulative sales from the start of the promotion. (i) Fit a Power trend line to the data and use it to estimate the total sales during the first 9 days of the promotion. (ii) Repeat Part i, using a Linear trend line. (iii) Which model do you think is more realistic? Exercise 5 Fit an exponential trend line to the following data and use the line’s formula to predict the highest closing price during July. Exercise 6 Excel T C I (material continues)

Trend Lines. Fitting Models: page 13
Trend Lines, Models (i) Fit a linear trend line to the data in Exercise 6, and use this model to predict the highest closing price during July. (ii) Fit a 3rd degree polynomial trend line to the data in Exercise 6, and use this model to predict the highest closing price during July. (iii) Do you think that trend lines can provide a reliable way to predict stock prices? Exercise 7 Your company, which manufactures and distributes beach wear, has kept sales records for the months of March, April, May, June, and July of the current year. These indicate that sales volume is increasing. Which, if any, of the types of trend lines that are available in Excel might be used to predict sales volume for the rest of the year? Explain your answer. Exercise 8 Excel T C I (material continues)

Trend Lines. Fitting Models: page 14
Trend Lines, Models The Flimsy Plastic company produces toys for children in the age group from 3 to 6 years. A regional sales office has kept records on the number of children in this range that live in its territory. (i) Experiment with several different types of trend lines to determine which model best fits the data. (ii) Use the type of trend line that you select to predict the number of 3-6 year olds that will be in the sales territory in the year 2011. Exercise 9 3-6 Year Olds in Sales Territory Year 1997 1998 1999 2000 2001 Population (K’s) 390 410 440 460 470 2002 2003 2004 2005 2006 510 540 570 580 620 Excel T C I (material ends)

Demand, Revenue, Cost, & Profit, Income
The amount of revenue that a producer receives from the sale of a good depends on the number of units that are sold and on the price per unit that is paid. In a given market, some quantity q of the product can be sold at a price p per unit. These are related by the demand function D(q). Specifically, p = D(q) is the price at which q units of the good can be sold. Example 1. A regional restaurant chain plans to introduce a new buffalo steak dinner. Managers tested various prices in their establishments and arrived at the following estimates for weekly sales. For example, they believe that D(2,800) is approximately \$ In addition, their experience suggests a quadratic demand function. In the sheet Income of the Excel file Dinners.xlsm we have plotted the four given points and fitted a 2nd degree polynomial trend line to the data. Dinners.xlsm (material continues) T C I

Demand, Revenue, Cost, & Profit. Income: page 2
D, R, C, & P, Income Define the demand function to be D(q) = aq2 + bq + c, where a =  , b =  , and c = Dinners.xlsm T C I (material continues)

Demand, Revenue, Cost, & Profit. Income: page 3
D, R, C, & P, Income The amount of money that a producer receives from the sale of q units of its product is called revenue, and is denoted by the revenue function R(q). If D(q) is the demand function, and revenue will come from selling q units at a price of D(q) dollars per unit, then R(q) = qD(q). In the case of the restaurants, the amount of money that diners pay for q of the buffalo steak dinners per week is weekly revenue. The following graph is copied from the sheet Income in Dinners.xlsm. Setting a high price for the dinners will result in very few orders, and a small revenue. Setting a low price for the dinners will also bring in low revenue. Dinners.xlsm (material continues) T C I

Demand, Revenue, Cost, & Profit. Income: page 4
D, R, C, & P, Income It appears from the graph that the maximum revenue would result from selling approximately 2,300 buffalo steak dinners. Looking back at the graph of the demand function, we see that a price of around \$20 can be expected to result in the sale of 2,300 dinners. Open the sheet Income in Dinners.xlsm and explore this with the Computation cells. (i) Find the price, per dinner, that will result in the sale of 2,600 buffalo steak dinners. (ii) What revenue can be expected from 2,600 dinners? Exercise 1 Dinners.xlsm (material continues) T C I

Demand, Revenue, Cost, & Profit. Income: page 5
D, R, C, & P, Income (i) Experiment with different values in Cell B19 of the sheet Income in Dinners.xlsm to find the number of dinners that would be sold at a price of \$ (ii) What revenue can be expected if dinners are priced at \$19.95? Exercise 2 Suppose that the demand function for a good is given by D(q) = 0.1q (i) Use Graphing.xlsm to plot D(q) and the revenue function, R(q). (ii) Estimate the price that will yield the greatest revenue. Exercise 3 Suppose that D(q) =  q is the demand function for a certain model of audio speaker. (i) Use Graphing.xlsm to plot D(q) and the revenue function, R(q). (ii) Estimate the price that will yield the greatest revenue. Exercise 4 Graphing.xlsm T C I (material continues)

Demand, Revenue, Cost, & Profit. Income: page 6
D, R, C, & P, Income The demand function for a new type of car alarm system is shown in the adjacent plot. Note that the quantities sold are given in thousands. Estimate revenue, in dollars, from selling 300,000 alarms. Exercise 5 (i) Use your team’s data and Trend Lines to find the formula for a quadratic demand function. (ii) Plot D(q) and R(q) for your team’s product. Use the same units as in the work on the Class Project in the sheet Functions of Marketing Focus.xlsm. Exercise 6 Excel T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit D, R, C, & P, Expenses & Profit 2. EXPENSES AND PROFIT Every producer incurs costs in the production of its product. These usually include fixed costs, which do not depend upon the amount of a good that is produced, and variable costs. Fixed costs might include such things as plant overhead, minimal labor costs, and debt service. Variable (production) costs cover such items as materials, labor, and distribution expenses. We will denote fixed cost by C0, and denote the variable cost for q units of a good by VC(q). A producer’s total cost function, C(q), for the production of q units is given by C(q) = C0 + VC(q). Example 2. We will continue Example 1, and develop a cost function for the restaurant chain’s new buffalo steak dinners. Managers estimate that the new menu item will have to support \$9,000 out of the complete weekly fixed cost of operating the chain. Hence, C0 = \$9,000. The head chef makes the following estimates for the costs of preparing various numbers of dinners. (material continues) T C I

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 2 D, R, C, & P, Expenses & Profit Notice that there is some economy of scale. The cost of preparing the first thousand dinners is \$14,000, of preparing the second thousand dinners is an additional \$8,000, and of preparing the third thousand dinners is an additional \$6,000. In order to develop a formula for the variable cost function, we will have Excel fit a trend line through the three estimated data points. The managers know that, in former menu offerings, food preparation costs have usually turned out to follow power function models. That is, VC(q) = uqv, for some constants u and v. In the sheet Profit of the file Dinners.xlsm, we have plotted the data points and added a power trend line. Excel finds that u = 177 and v = Thus, VC(q) = 177q0.663, and the total weekly cost function for the buffalo steak dinners is C(q) = C0 + VC(q) = 9, q0.633. Dinners.xlsm T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 3 D, R, C, & P, Expenses & Profit Dinners.xlsm (material continues) T C I

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 4 D, R, C, & P, Expenses & Profit The producer of any good is interested in profit. We let P(q) be the profit obtained from producing and selling q units of a good at the price D(q). Profit = Revenue  Cost P(q) = R(q)  C(q) The sheet Profit of Dinners.xlsm shows a plot of both the revenue and cost functions on a single set of axes. As the graph shows, revenue appears to exceed cost in a range of approximately 700 to 3,100 dinners per week. The sheet Profit also plots the profit function, P(q). A maximum profit of roughly \$14,000 seems to result from the preparation and sale of approximately 2,000 dinners per week. Note that this is somewhat less than our estimate of 2,300 dinners per week that would produce the maximum revenue for the chain. Open the sheet Profit and explore this with the Computation cells. Ask a tutor How can I recog-nize the graphs of demand, cost, and profit functions? Dinners.xlsm T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 5 D, R, C, & P, Expenses & Profit Dinners.xlsm T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 6 D, R, C, & P, Expenses & Profit (i) Find the cost of preparing 2,300 buffalo steak dinners. (ii) What revenue can be expected from preparing and selling 2,300 dinners? (iii) What weekly profit can be expected from preparing and selling 2,300 dinners? Exercise 7 (i) Find the cost of preparing 1,500 buffalo steak dinners. (ii) What revenue can be expected from preparing and selling 1,500 dinners? (iii) What weekly profit can be expected from preparing and selling 1,500 dinners? Exercise 8 (i) Experiment with computation in the sheet Profit of Dinners.xlsm to find the number of dinners that could be sold at a price of \$18. (ii) Find the cost of preparing these dinners. (iii) What profit can be expected from preparing and selling buffalo steak dinners at \$18? Exercise 9 Dinners.xlsm T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 7 D, R, C, & P, Expenses & Profit Returning to our plot of the demand function, D(q), we can estimate that a price of approximately \$22 would produce sales of 2,000 dinners. In summary, the restaurant chain should prepare around 2,000 buffalo steak dinners per week and price them at approximately \$22. If this is done, they can expect a weekly profit of around \$14,000 from the new menu item. The graph of P(q) shows us how sensitive the weekly profit is to deviation from the optimal price per dinner of around \$22. For example, using rough estimates from the plot of the demand function, it appears that raising the price of a dinner to \$25 would reduce the demand to around 1,600 dinners per week. Estimating from the plot of the profit function, we see that this would reduce the chain’s weekly profit to around \$13,000. That is a 7% drop from the maximum profit of approximately \$14,000. Dinners.xlsm T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 8 D, R, C, & P, Expenses & Profit Estimate the percentage drop in weekly profit that would result from dropping the price of buffalo steak dinners from \$22 to \$20. Exercise 10 Use the graphs in this section to estimate the range of dinner prices that would yield weekly profits of at least \$10,000. Exercise 11 Use the graphs in this section to estimate the range of dinner prices that would yield weekly revenues of at least \$35,000. Exercise 12 Use the graphs in this section to estimate the weekly profit that would result from spending \$30,000 to prepare buffalo steak dinners. Exercise 13 Dinners.xlsm T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 9 D, R, C, & P, Expenses & Profit Refer to the demand function given in Exercise 3. Suppose that the fixed cost for producing this good is C0 = \$12,000 and that the variable costs are given by (i) Use Graphing.xlsm to plot D(q) and P(q). (ii) Use your graphs to estimate the number of units that should be produced, and how they should be priced in order to attain the maximum profit. (iii) Approximately what maximum profit might be expected? Exercise 14 Refer to the demand function given in Exercise 4. Suppose that the fixed cost for producing the speakers is C0 = \$60,000 and that it costs \$110 to produce each speaker. (i) Use Graphing.xlsm to plot D(q) and P(q). (ii) Use your graphs to estimate the number of speakers that should be produced, and how they should be priced in order to attain the maximum profit. (iii) Approximately what maximum profit might be expected? Exercise 15 Graphing.xlsm Excel (material continues) T C I

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 10 Refer to the demand function whose graph is shown in Exercise 5. The cost function for the alarms is plotted in the adjacent graph. Note that the quantities of alarms are given in thousands and the total costs are given in millions of dollars. (i) Estimate the number of alarms that would be sold at a price of \$250 each. (ii) Estimate the revenue, in dollars, that would result from the sale of alarms priced at \$250. (iii) Estimate the cost, in dollars, of producing the alarms that would be sold at \$250. (iv) Estimate the profit, in dollars, that would result from the sale of alarms at \$250. D, R, C, & P, Expenses & Profit Exercise 16 T C I (material continues)

D, R, C, & P, Expenses & Profit
Demand, Revenue, Cost, & Profit. Expenses & Profit: page 11 D, R, C, & P, Expenses & Profit (i) Use your team’s data to find a formula for the cost function of your product. (ii) Plot C(q) and P(q) for your team’s data. Use the same units as in the work on the Class Project in the sheet Functions of Marketing Focus.xlsm. Study the Focus pages to see how to use the custom programmed function COST. Exercise 17 SALE Everything Must Go! Excel T C I (material continues)

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives How can demand, revenue, cost, and profit functions help us price 12-GB drives? Obviously, Card Tech would like to price its new drives in such a way that profit is maximized. To compute values of their profit function, they must first develop demand, revenue, and cost functions. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Our information on test markets and variable costs is summarized in the sheet Data of the Excel file Marketing Focus.xlsm. Since we are dealing with quite large numbers, we will adopt some conventions for units.  Prices for individual drives are given in dollars.  Revenues from sales in the national market are given in millions of dollars.  Quantities of drives in the test markets are actual numbers of drives.  Quantities of drives in the national market are given in thousands of drives. At the top of the sheet Functions in Marketing Focus.xlsm, these units are applied to our data, and the projected yearly sales in the national market are computed for each test market price. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Since it was assumed that the demand function is quadratic, the data points for national sales are plotted and fitted with a second degree polynomial trend line. Excel finds that D(q) =  q2  q where D(q) is the price, in dollars, at which q thousand drives can be sold. We formatted the trend line equation so that its coefficients are displayed with 8 decimal places. Further precision would result in only very minor changes in computed values. This apparent increase in accuracy would be spurious, since it exceeds the accuracy of our market data. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

R(q) = (D(q)q1,000)/1,000,000 = D(q)q/1,000
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives In our units, the revenue function R(q) is to give the revenue, in millions of dollars from selling q thousand drives. D(q) gives the price, in dollars per drive, at q thousand drives. Hence, D(q)q1,000 gives the revenue, in dollars, from selling q thousand drives. To express this revenue in millions of dollars, we divide by 1,000,000. R(q) = (D(q)q1,000)/1,000,000 = D(q)q/1,000 (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

on the project Marketing Computer Drives
Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives It appears that we might obtain the maximum revenue, if we sold approximately 1,400 thousand drives. Cells C91:E91 in the sheet Functions of Marketing Focus.xlsm allow us to compute D(q) and R(q) for any value of q. We find that D(1,400) = \$ and R(1,400) = million dollars. We want the total cost function, C(q), to give the cost, in millions of dollars, of producing q thousand drives. C(q) = fixed cost + variable cost for q thousand drives (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Let marginal cost denote the price, in dollars per drive, for production at a given number of drives. Our cost information can now be summarized in the following table. The Batch Size column displays the numbers of drives, in K’s, that will be produced in first, second, and all further production lots. C(q) depends upon 7 numbers: q (Quantity), Fixed Cost, Batch Size 1, Batch Size 2, Marginal Cost 1, Marginal Cost 2, and Marginal Cost 3. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives For example, suppose that we produce 1,400 thousand drives. The fixed cost is 135 million dollars. The first 800 thousand drives cost \$160 per drive, for a total cost of 8001,000160/1,000,000 = 800160/1,000 = 128 million dollars. The next 400 thousand drives cost \$128 per drive, for a total cost of 4001,000128/1,000,000 = 400128/1,000 = 51.2 million dollars. The remaining 200 thousand drives cost \$72 per drive, for a total cost of 2001,00072/1,000,000 = 20072/1,000 = 14.4 million dollars. Adding these, we find that C(1,400) = = million dollars. It is possible, though very tedious, to use nested IF functions in Excel to produce a formula for C(q). This is demonstrated in Cell D125 of the sheet Functions in Marketing Focus.xlsm. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives When it is difficult or very complicated to compute with built-in Excel functions, we can seek help from custom programming. In particular, the Visual Basic language can be used to construct user defined functions in any Excel file. These are accessed via pull-down menus, in the same way that the built-in functions are used. Given the complexity of our cost function, it is desirable to custom program the formula in Visual Basic and attach it to the Excel file. This has been done in Marketing Focus.xlsm, with the resulting function, COST, being placed in the list of functions under User Defined. Look in the formula bar for Cell E125 in the sheet Functions of Marketing Focus.xlsm to see how the function is used. To view the Visual Basic code for COST in Office 2007, click on the Developer tab, then on Visual Basic. In earlier versions, click on Tools/Macro/Visual Basic Editor. In the side bar, select VBAProjects (Marketing Focus.xlsm), double click on Modules, then double click on Module1. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives You are not expected to do any Visual Basic programming in Mathematics for Business Decisions. However, you will probably find it convenient to use the function COST while working on your team’s project. To transfer the Cost function from Marketing Focus.xlsm into another Excel file, open both files. Open Module1 (as directed above), click on its icon in the side bar under VBAProject (Marketing Focus.xlsm) and drag the icon into the VBAProject side bar area of the new file. You can also move modules containing macros or custom functions with export/import. To do this, open the module, click on File/Export File, then select a folder to contain a copy of the module. To attach the module to another Excel file, open the Visual Basic Editor in that file, click on File/Import File, and open the module. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives The Cost function can be added to Graphing.xlsm and plotted with that utility, or it can be plotted as in the sheet Functions of Marketing Focus.xlsm. It is most interesting to display the graphs of both R(q) and C(q) on the same set of axes. We will make a profit when R(q) > C(q) and will operate at a loss where R(q) < C(q). It appears that we can make a profit by selling between approximately 650 and 1,650 thousand drives. Verify this with numerical experimentation in Cells B170:F170 in the sheet Functions of Marketing Focus.xlsm. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Our main interest is in profit, which is given by revenue minus cost. Let P(q) be the profit, in millions of dollars, from selling q thousand drives. P(q) = R(q)  C(q) A plot of the profit function has been copied from Marketing Focus.xlsm into the next Focus page. Each of the two “humps” represent a local maximum profit. It appears that the local maximum that occurs on the right, at approximately 1,280 thousand drives, is slightly higher than the local maximum on the left. Hence, the largest possible profit, or global maximum, is approximately P(1,280). This is close to 42 million dollars. Looking at the plot of the demand function, we estimate that the 1,200 thousand drives should be priced at approximately \$280 each. Verify this with numerical experimentation in Cells B170:F170 in the sheet Functions of Marketing Focus.xlsm. What we need now is some way to replace graphical estimates with more precise computations. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives WHAT SHOULD YOU DO? Each team should now find formulas for, and make plots of, its demand, revenue, cost, and profit functions. Work with the same units and coefficient precision as in the Class Project. Use your plots to estimate the production range that would yield a profit, and to estimate the maximum possible profit. (material continues) Marketing Focus.xlsm Class Project T C I D, R, C, & P, Focus

Differentiation, Marginal Analysis
When working with the cost function for a product it is often helpful to determine the instantaneous rate of change in total cost per unit of the good. This cost per unit at a given level of production is called the marginal cost. Example 1. We consider the cost function C(q) = C0 + VC(q) = 9, q0.633 that was developed in the Expenses and Profit section of Demand, Revenue, Cost, and Profit. Recall that a restaurant chain is planning to introduce a new buffalo steak dinner. C(q) is the cost, in dollars, of preparing q dinners per week. We let MC(q) be the marginal cost at q dinners. That is, MC(q) is the cost for an additional dinner, when q dinners are being prepared. For the moment, we will think of this as the cost of preparing the next, or (q + 1)st, dinner. Since C(q + 1) is the cost of preparing q + 1 dinners, and C(q) is the cost of preparing q dinners, the (q + 1)st dinner costs C(q + 1)  C(q) dollars. This gives a First Plan formula for the marginal cost. (material continues) T C I

Differentiation, Marginal
Differentiation. Marginal Analysis: page 2 Differentiation, Marginal We can use a calculator or Excel to compute values of MC(q). For example, Thinking in terms of money, the marginal cost at the level of 1,000 dinners, is approximately \$8.88 per dinner. Similar computations show that MC(2,000)  \$6.88 and MC(3,000)  \$5.93. Since the marginal cost per dinner depends upon the number of dinners currently being prepared, it is helpful to look at a plot of MC(q) against q. This is created in the sheet M Cost of the Excel file Dinners.xlsm. Excel T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 3 Differentiation, Marginal Visiting that file, we see that the First Plan marginal cost in Column D is very large, when only a small number of dinners are prepared. MC(0) = \$177.00/dinner. However, the cost per dinner drops rapidly. MC(100) is down to \$20.63/dinner, and MC(3,999) = \$5.34/dinner. When we talk about the cost per dinner, when q dinners are being prepared, it is not clear whether we should consider the cost of the qth or the (q + 1)st dinner. Rather than arbitrarily choosing to move ahead to the (q + 1)st dinner, as in the First Plan, we get a better indication of the marginal cost at q by averaging the change in both directions from q. We will call this the Second Plan for computing marginal cost. Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Marginal Analysis: page 4 Differentiation, Marginal The marginal costs, computed with the Second Plan, are shown in Column E of M Cost in the Excel file Dinners.xlsm. What can we learn from MC(q)? The restaurant managers might want to know how many dinners would need to be prepared per week in order to get the price per dinner for further dinners down to \$8 or less. Glancing down Column E in the sheet M Cost shows that MC(q)  \$8 for q  1,327. One further refinement is necessary in our computation of marginal cost. MC(q) is best defined as the instantaneous rate of change in total cost, per unit. The formula is only an average change over an interval of length 2. Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 5 Differentiation, Marginal To improve the computation, we let mathematics do something that cannot be done in the real world. We consider a non-integer change in the number of dinners that are prepared. If h is any small number then is the average change over an interval of length 2h. Since the numerator is a difference of two numbers and the entire expression is a quotient, it is usually called a difference quotient. As h is assigned smaller and smaller values, the difference quotient gives better and better approximations for MC(q). We indicate that MC(q) is an instantaneous rate of change, by writing Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 6 Differentiation, Marginal In practice, we will compute marginal cost by evaluating the difference quotient at a small value of h. Since excessively small values may cause numerical problems, we will often use h = This method is the Final Plan, that is used in Column F of M Cost in the Excel file Dinners.xlsm. Explain, in terms of real world dinners and dollars, why it is plausible that, for buffalo steak dinners, MC(q) is very large for small values of q, and that it is always positive. Exercise 1 Use Excel, with h = , to compute MC(1,000) for buffalo dinners, rounded to 2 decimal places. Exercise 2 Excel Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 7 Differentiation, Marginal Use Excel, with h = , to compute MC(2,500) for buffalo dinners, rounded to 4 decimal places. Exercise 3 Marginal analysis is also used with revenue, demand, and profit functions. For example, the marginal revenue, MR(q), at q dinners is the instantaneous rate of change in revenue per dinner, when q dinners are being prepared and ordered at the demand price per dinner. This is defined by and is computed by evaluating the difference quotient for a small value of h. Similar definitions apply to the marginal demand, MD(q), and marginal profit, MP(q), which are defined by the following limits. Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 8 Values for all of our mar- ginal functions are computed in the sheets M Cost and M Profit of the Excel file Dinners.xlsm. The graphs of MD(q), MR(q), and MP(q) are also displayed in those sheets. Differentiation, Marginal Many aspects of the demand function are reflected in properties of the difference quotients for marginal demand, and in the marginal demand function. D(q) is always decreasing. Hence, all difference quotients for marginal demand are negative, and MD(q) is always negative. The more rapidly D(q) drops, the more negative are the difference quotients, and the further negative is MD(q). Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 9 Differentiation, Marginal positive, and MR(q) is positive. For example, MR(1,300) is approximately \$20. Thus, when 1,300 dinners are prepared and sold, the restaurant chain takes in \$20 more for each extra dinner. Likewise, where R(q) is decreasing, MR(q) is negative. This shows that the maximum revenue will occur at the value of q where the marginal revenue is equal to 0. Computations in the sheet M Profit show that MR(2,309) = \$0.01 and MR(2,310) = \$ Hence, the maximum revenue occurs at either 2,309 or 2,310 dinners. Direct computation shows that the maximum revenue is R(2,310) = \$45, Where the revenue function R(q) is increasing, the difference quotients for marginal revenue are Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 10 Differentiation, Marginal Marginal analysis can tell us a great deal about the profit function. Refer back to these plots while reading the next pages. Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 11 Differentiation, Marginal Since P(q) = R(q)  C(q), profit will increase with more dinners if the increase in revenue per dinner is greater than the increase in cost per dinner. This happens where MR(q) > MC(q). Similarly, profit will decrease with more dinners if the change in revenue per dinner (positive or negative) is less than the increase in cost per dinner. This happens where MR(q) < MC(q). Profit stops increasing, and starts to decrease at its maximum value. Hence, the maximum profit must occur where MR(q) = MC(q). From the plot of MR(q) and MC(q), it appears that the two graphs cross at a point where q is slightly greater than 2,000. The computations in the sheet M Profit of Dinners.xlsm show that MR(2,025) = \$6.84 = MC(2,025). Hence, the maximum profit will occur at q = 2,025 dinners. Direct computation shows that D(2,025) = \$22.21 and that P(2,025) = \$14,052. Considering only the mathematics of our analysis, the restaurant chain should expect to sell 2,025 buffalo steak dinners per week, priced at \$22.21 per dinner, for a total profit of \$14,052. Note that all of this information is consistent with the estimates that we made in the section Demand, Revenue, Cost, and Profit. Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 12 Differentiation, Marginal This must be where MP(q) changes from positive to negative. Thus, the maximum profit occurs when the marginal profit is zero, MP(q) = 0. The computations in Column F of M Profit in Dinners.xlsm show that MP(2,025) = \$ This agrees with the value found from the marginal analysis of revenue and cost. Marginal analysis of profit offers another way to determine the maximum profit. Where profit, P(q), is increasing, marginal profit, MP(q), is positive. Where P(q) is decreasing, MP(q) is negative. The change from increasing to decreasing profit occurs at the maximum profit. Dinners.xlsm T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 13 Differentiation, Marginal Exercises 4, 5, and 6 refer to the situation that was discussed in Exercises 3 and 14 of Demand, Revenue, Cost, and Profit. The demand and cost functions for a good were given by D(q) = 0.1q and respectively. Use the methods of Marginal Analysis to plot the marginal cost function, MC(q), and the marginal revenue function, MR(q), on the same set of axes. Exercise 4 (i) Use the methods of Marginal Analysis to plot MP(q). (ii) Use your graph to estimate the value of q that maximizes profit. (iii) What is the maximum possible profit? (iv) At what price should the good be sold, in order to realize the maximum profit? Exercise 5 (i) Use your graphs from Exercise 4 to estimate the value of q that makes MR(q) = MC(q). (ii) How does this value relate to your work in Exercise 5? Exercise 6 Excel T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 14 Differentiation, Marginal Exercises 7, 8, and 9 refer to the situation that was discussed in Exercises 4 and 15 of Demand, Revenue, Cost, and Profit. The demand and cost functions for one model of audio speaker were given by D(q) =  q and C(q) = 60, q, respectively. Use the methods of Marginal Analysis to plot the marginal cost function, MC(q), and the marginal revenue function, MR(q), on the same set of axes. Exercise 7 (i) Use the methods of Marginal Analysis to plot MP(q). (ii) Use your graph to estimate the value of q that maximizes profit. (iii) What is the maximum possible profit? (iv) At what price should the speakers be sold, in order to realize the maximum profit? Exercise 8 (i) Use your graphs from Exercise 7 to estimate the value of q that makes MR(q) = MC(q). (ii) How does this value relate to your work in Exercise 8? Exercise 9 Excel T C I (material continues)

Differentiation, Marginal
Differentiation. Marginal Analysis: page 15 Differentiation, Marginal Exercises 10 and 11 refer to the demand and cost functions for car alarm systems that are plotted in Exercises 5 and 16 of Demand, Revenue, Cost, and Profit. Use your conceptual understanding of marginal demand and the plot of the demand function to sketch a rough graph of MD(q). Exercise 10 Use your conceptual understanding of marginal cost and the plot of the cost function to sketch a rough graph of MC(q). Exercise 11 Sketch what you believe might be a typical plot of a marginal revenue function. Explain the reasoning behind your sketch. Exercise 12 Excel T C I (material continues)

Differentiation, Derivatives
Difference quotients are used in many business situations, other than in marginal analysis. For this reason, it is convenient to have a standard name and common notation for limits of such quotients. Let f be a function which is defined on an open interval containing a real number x. Suppose that approaches a number m as h is taken to be smaller and smaller. We write and call m the derivative of f at x. To show that m depends on both the function, f, and the number, x, we denote the derivative by f (x). (material continues) T C I

Differentiation, Derivatives
Derivatives: page 2 This is read as “the derivative of f, at x, is f prime of x.” The process of computing f (x) is called differentiation and f (x) is interpreted as the instantaneous rate of change in f(x) with respect to x. Differentiation, Derivatives Click here for technical information about our definition. In the new terminology, marginal demand, MD(q), is the derivative of the demand function, D(q). Hence, MD(q) = D(q). Likewise, MC(q) = C(q), MR(q) = R(q), and MP(q) = P(q). In this notation, the units of the marginal functions will reflect the units of the original functions. We can compute the approximate value of a derivative, f (x), by having Excel compute for a very small value of h. Example 2. Evaluate f (2), if f(x) = x We let h = and approximate f (2) with the difference quotient (material continues) T C I

Differentiation, Derivatives
Differentiation. Derivatives: page 3 Differentiation, Derivatives The following computation can be done with a calculator or with Excel. If it is likely that we will need to evaluate f (x) for several different values of x, then Excel is a more efficient tool. Example 3. Let f(x) = x Use Excel to plot both f(x) and f (x) over the interval from 3 to 3. The work is shown in the file Example 3.xlsx and the resulting graph is shown on the following page. Example 3.xlsx (material continues) T C I

Differentiation, Derivatives
Differentiation. Derivatives: page 4 Differentiation, Derivatives Let f(x) = 1  ex/2. Use Excel and the method of Example 3 to plot both f(x) and f (x) over the interval from 0 to 10. Exercise 13 Let f(x) = 2x. (i) Use Excel and the method of Example 3 to plot the graph of f  over the interval [0, 4]. (ii) Use the interpretation of the derivative as an instantaneous rate of change to explain the form of your graph. Exercise 14 Example 3.xlsx Graphing.xlsm T C I (material continues)

Differentiation, Derivatives
Differentiation. Derivatives: page 5 Differentiation, Derivatives The evaluation of difference quotients provides a good understanding of differentiation, but it is rather tedious. Carrying out this much work, while in the middle of a business application, could easily distract us from the basic business problem. What we are doing is called numerical differentiation. Many software packages, but (unfortunately) not Excel, have built-in functions that perform numerical differentiation. To meet this need, we have created an Excel file Differentiating.xlsm, that serves as a numerical differentiation utility. To use Differentiating.xlsm, open the file and follow the instructions in the blue box. In particular, be sure that Excel is set for Automatic Calculation. The file works in much the same way as Graphing.xlsm. Once you have entered the formula for f(x), you can get numerical values for f(x) and f (x), and can see the graphs of these functions over any interval. Differentiating.xlsm (material continues) T C I

Differentiation, Derivatives
Differentiation. Derivatives: page 6 Differentiation, Derivatives From now on, unless otherwise indicated, we will assume that all differentiation is performed with Differentiating.xlsm. When entering a function in Differentiating.xlsm, you can use the letters s, t, u, v, or w as constants. If this is done, you must enter values in Cells O23:O27 for any letters that you use. For example, suppose that you want to differentiate f(x) = x s, where s is a number that is in Cell O23. You would enter = x^s in Cell C24. Accept the function and then click on the Differentiate button at the right of the plots. Since Excel is set for automatic calculation, entering a new number in Cell O23 will automatically change the function, redraw the plots, and recompute the derivative. Click here for information on running the macro in Differentiating.xlsm. Differentiating.xlsm (material continues) T C I

Differentiation, Derivatives
Differentiation. Derivatives: page 7 Differentiation, Derivatives If values are entered in Cells O23:O27; then constants s, t, u, v, or w may also be used to describe the range, [a, b], of the plot. If these constant names are used, they must be entered in Cells I24:J24 preceded with "=" signs. For example, =s. If f '(x) is constant, the displayed plot in Differentiating.xlsm will be distorted. To correct this, format the y-axis to have realistic fixed minimum and maximum values. Caution! Example 4. Use Differentiating.xlsm to redo Examples 1 and 2. Open Differentiating.xlsm and enter =x^3+6 in Cell C24. Click on  to accept the function, and then click on the Enter button on the Differentiating toolbar. Enter 2, 3, and 3 in Cells E24, I24, and J24, respectively. The resulting computation and plot of both the function and its derivative are shown on the following page. Differentiating.xlsm (material continues) T C I

Differentiation, Derivatives
Derivatives: page 8 Differentiation, Derivatives Use Differentiating.xlsm to redo Part i of Exercise 5. Exercise 15 Use Differentiating.xlsm to redo Part i of Exercise 8. Exercise 16 Differentiating.xlsm (material continues) T C I

Differentiation, Derivatives
Differentiation. Derivatives: page 9 Differentiation, Derivatives Exercises 17, 18, and 19 refer to the situation that was discussed in Exercises 3 and 14 of Demand, Revenue, Cost, and Profit. The demand and cost functions for a good were given by D(q) = 0.1q and respectively. Use Differentiating.xlsm to plot marginal cost, C, over the interval from 0 to 1,500. Exercise 17 (i) Use Differentiating.xlsm to plot marginal profit, P over the interval from 0 to 1,500. (ii) Experiment with the Computation boxes to find a value for q that is greater than 100, and at which P(q) = 0. (Hint: This value of q might not be an integer.) Exercise 18 (i) Use Differentiating.xlsm to plot marginal revenue, R over the interval from 0 to 1,500. (ii) Experiment with the Computation boxes to find a value for q at which R(q) = 0. Exercise 19 Differentiating.xlsm T C I (material continues)

Differentiation, Derivatives
Differentiation. Derivatives: page 10 Differentiation, Derivatives Exercises 20, 21, and 22 refer to the situation that was discussed in Exercises 4 and 15 of Demand, Revenue, Cost, and Profit. The demand and cost functions for audio speakers were given by D(q) =  q and C(q) = 60, q, respectively. Use Differentiating.xlsm to plot marginal revenue, R, over the interval from 0 to 2,100. Exercise 20 (i) Use Differentiating.xlsm to plot marginal profit, P, over the interval from 0 to 2,100. (ii) Experiment with the Computation boxes to find a value for q at which P(q) = 0. (Hint: This value of q might not be an integer.) Exercise 21 Use Differentiating.xlsm to plot marginal cost, C, over the interval from 0 to 2,100. Hint: remember the caution about using Differentiating.xlsm, when the graph of f  is distorted. Exercise 22 Differentiating.xlsm T C I (material continues)

Differentiation, Properties & Applications
Differentiation. Properties and Applications Differentiation, Properties & Applications 3. PROPERTIES AND APPLICATIONS A few basic properties of derivatives can save us time while working on business problems. Since a derivative represents the instantaneous rate of change in a function, it is not surprising that multiplying a function, f, by a constant, a, also multiplies its derivative by the same constant. If g(x) = af(x), then g(x) = af (x). Thus, for example, doubling the production costs for a product will double the marginal cost. The derivative of the sum or difference of two functions is the sum or difference of their derivatives. If h(x) = f(x)  g(x), then h (x) = f (x)  g(x). This property is often used with marginal profit. If R, C, and P, are the revenue, cost, and profit functions for a good, then P(q) = R(q)  C(q). Thus, P(q) = R(q)  C(q). (material continues) T C I

Differentiation. Properties & Applications: page 2
Differentiation, P. & A. Since the marginal profit is equal to the marginal revenue minus the marginal cost, the marginal profit is equal to zero exactly when the marginal revenue is equal to the marginal cost. Some useful properties of differentiation are clear from the definition of a derivative. A function, f, is constant on an interval, if f(x) is the same number for all values of x in the interval. For example, we might have f(x) = 100 for all numbers, x, with 0  x  10. Since the function does not change, the derivative of a constant function is zero. Approximately constant demand functions are encountered in the case of some commodities, such as wheat or corn, where a change in production level by a single farmer has no effect on the price per bushel for his or her crop. In such a case the marginal demand is zero. A function of the form f(x) = mx + b, where m and b are constants, is called linear. We know that the derivative of f is the sum of the derivatives of mx and b. Since b is a constant, its derivative is zero. The graph of y = mx is a straight line, with a slope of m. Thus, the instantaneous rate of change for mx is always equal to m. (material continues) T C I

Differentiation. Properties & Applications: page 3
Differentiation, P. & A. Putting these facts together, we see that, if f(x) = mx + b, then f (x) = m, for all values of x. Linear functions are often used as models for parts of demand and cost functions. In this case, there is no need for numerical differentiation. For example, if the cost function for a good is given by C(q) = 120q + 5,000, then its marginal cost is C(q) = 120, for all quantities, q. If f(x) = 0.75x , find a formula for f (x). Exercise 23 Let f(x) = 3g(x) and suppose that g(5) = 2. What is f (5)? Exercise 24 When 12,000 mountain bicycles are being produced and sold the marginal revenue is \$895 and the marginal cost is \$787. (i) What is the marginal profit at this production level? (ii) What does this number mean in terms of bicycles and dollars? Exercise 25 (material continues) T C I

Differentiation. Properties & Applications: page 4
Differentiation, P. & A. A demand function, D(q) = 0.8q + 200, gives the price, in dollars, at which q items can be sold. (i) Find a formula for the marginal demand. (ii) Relate your formula to dollars and items. Exercise 26 The demand function D(q) =  q for audio speakers has been considered in previous exercises. The corresponding revenue function is given by R(q) = qD(q) =  q q. (i) Use Differentiating.xlsm to plot R(q) and use Graphing.xlsm to plot the function g(q) =  q (ii) What do your plots suggest about the form of the marginal revenue function? Exercise 27 Use the formulas for differentiation to explain what happened in the solution of Exercise 23. Exercise 28 Since this stuff takes a lot of effort, you might want to have a backup plan. Have you considered OMPHALOSKEPSIS? (material continues) Differentiating.xlsm Graphing.xlsm T C I

Differentiation. Properties & Applications: page 5
Differentiation, P. & A. So far we have used differentiation for simple numerical problems and as new terminology for marginal analysis. Many other types of business problems can be solved with differentiation. As an illustration, we will apply differentiation to a situation in inventory control. Example 5. A large computer manufacturer uses 50,000 CD drives at a uniform rate throughout the year. These are outsourced, and must be ordered in batches from the supplier. Drives cost \$20 each, and there is an additional \$1,200 expense per order for processing, shipping, receiving and handling the drives. Thus, it is cheaper to order in large batches. The drives must be kept in inventory from the time that they arrive until they are used in computers. The estimated cost of keeping each drive stored is one cent per day. The cost of maintaining the inventory suggests placing smaller orders, that will be used up more quickly. Assuming that orders can be placed so that the drives arrive just in time to be used, how often should the company order a batch of drives, how many should be ordered, and how much will it cost them, per year, to order and store the CD drives? T C I (material continues)

Differentiation. Properties & Applications: page 6
Differentiation, P. & A. Suppose that C(t) is the total yearly cost if drives are ordered every t days, starting from the first of the year. Our goal is to find the value of t that minimizes C(t). If orders are placed every t days, then there will be 365/t orders per year, and each order will be for drives. In order to understand the problem, let It(x) be the number of drives in the company’s inventory at x days after the arrival of an order. As an example, we will plot It(x), assuming that the company orders every 30 days, that is for t = 30. In this case, each order will be for drives. Note that, on average, there will be one-half the number of parts that are ordered in storage. T C I (material continues)

Differentiation. Properties & Applications: page 7
Differentiation, P. & A. In general, if the company orders every t days, the mean number of drives in the inventory will be T C I (material continues)

Differentiation. Properties & Applications: page 8
Differentiation, P. & A. At one cent per day per drive for storage, it will cost \$0.01365 = \$3.65 per year to store a single drive. We can now set up a formula for C(t). Differentiating.xlsm T C I (material continues)

Differentiation. Properties & Applications: page 9
Differentiation, P. & A. This expression for C(t) is entered into Differentiating.xlsm as f(x). The resulting plots are shown below. Large scale plots show that f(x) has only one possible minimum value, and that the minimum seems to occur when x is close to 40. Differentiating.xlsm (material continues) T C I

Differentiation. Properties & Applications: page 10
Differentiation, P. & A. Where f(x) is decreasing, its difference quotients are negative, and it derivative is negative. Where f(x) is increasing, f (x) is positive. Hence, the minimum value of f(x) occurs where f (x) = 0. This is exactly what is shown in the graphs. By experimenting with the values of x in Cell E24 of Differentiating .xlsm, we find that f ( ) = and that f( ) = 1,020,928. Computation shows that days is equal to 41 days, 20 hours, 33 minutes, and 56 seconds. While this is very precise, it is not a highly practical time interval at which to place orders! The natural plan would be to round this time to 42 days. Putting 42 into Cell E24 of Differentiating.xlsm, shows that f(42) = 1,020,929. Since this is only \$1 more than the minimum possible value of f(x), it does no harm to order every 42 days. Returning to the language of our problem, we have found that the most economical plan is to order (50,000/365)t  5,753 drives every 42 days. This will result in a total cost of approximately \$1,021,000 per year for procuring and storing the drives. Differentiating.xlsm (material continues) T C I

Differentiation. Properties & Applications: page 11
Differentiation, P. & A. Redo Example 5, assuming that the fixed cost of an order is \$1,000 and that it costs 1.5 cents per day to store a CD drive. Exercise 29 Redo Example 5, assuming that the fixed cost of an order is \$1,500 and that it costs 0.8 cents per day to store a CD drive. Exercise 30 (i) Use your team’s data to plot MR, MC, and MP for its product. (ii) Prepare computational cells, as in the sheet Marginal of Marketing Focus.xlsm and use these to answer your team’s Questions 1-4 for its product. Use the same units as in the work on the Class Project in Marketing Focus.xlsm. Study of the Focus pages will help with this work. Exercise 31 Differentiating.xlsm T C I (material continues)

Differentiation, Tangents & Slopes
The subject of differentiation is much more sophisticated than we have indicated in our presentation. However, the numerical methods that we have introduced provide good intuition, and are adequate for a wide range of simple business problems. Differentiation is one of the main components of the mathematical subject, calculus. There is a geometric interpretation of the derivative that often helps in the understanding of business graphs. Recall that the slope of a straight line is defined to be the change in height divided by the horizontal change between any two points on the line. We would like to extend this notion to define “slope” for the graph of a function f at a single point (x, f(x)). Example 6. Let and consider the point (1, f(1)) = (1, 1). Moving an increment h units to the left and to the right of (1, 1), locates the points (1  h, f(1  h)) and (1 + h, f(1 + h)) on the graph of f. The straight line connecting these points is the secant line determined by the increment h. * This section is not needed for the study of other material in Mathematics for Business Decisions. (material continues) T C I

Differentiation. Tangents & Slopes: page 2
Differentiation, T. & S. Notice that the slope, mh, of the blue secant line is given by the difference quotient 2h f(1 + h)  f(1  h) (1, f(1)) y The red line that we will call tangent to the graph of f at the point (1, f(1)) has a slope which is the limit of the slopes of the secant lines, as the increment, h, approaches 0. That is, (material continues) T C I

Differentiation. Tangents & Slopes: page 3
Differentiation, T. & S. Since we see that the derivative, f (1), is the slope of the tangent line at (1, f(1)). The Excel file Slope.xlsx plots both the tangent and secant lines for selected values of h. Open that file and experiment with different increments. For relatively small values of h, the blue secant line appears to coincide with the red tangent line. What we have just learned about the tangent line for at the point (1, f(1)) is also true for any differentiable function. If f(a) exists, then the slope of the tangent line to the graph of f at the point (a, f(a)) is defined to be f(a). For this reason, f(a) is called the slope of the graph of f at the point (a, f(a)). This is often shortened to “the slope of f at a.” In practical situations, the slope, f (a), of f at (a, f(a)) is interpreted as the instantaneous rate of change of f(x) at a. This follows from the fact that the slope mh of a secant line is the average rate of change in f(x) over the interval from a  h to a + h. The limit of this average rate, as the increment, h, approaches 0, is the instantaneous rate of change. Slope.xlsx T C I (material continues)

Differentiation. Tangents & Slopes: page 4
Differentiation, T. & S. Since we know the slope of the tangent line and have one point on that line, we can determine its equation. If f (a) exists, then the tangent line to the graph of f at the point (a, f(a)) has the equation y = f (a)(x  a) + f(a). When using the equation for a tangent line it is important to distinguish between a and x. a is the first coordinate of the point at which the line is tangent. x is simply the independent variable in the equation of the tangent line. Caution Example 7. Find the slope of the graph of f(x) = x2  4x + 4 at the point (3, f(3)), and find an equation for the tangent line at that point. Differentiating.xlsm shows that f (3) = 2. Hence, the slope of f is 2 at x = 3. This means that f(x) is increasing at an instantaneous rate of 2 units per unit increase in x, when x = 3. In the tangent line equation, we have a = 3, and f(a) = f(3) = 32  43 + 4 = 1. Differentiating.xlsm T C I (material continues)

Differentiation. Tangents & Slopes: page 5
y Differentiation, T. & S. Tangent Line (i) Find the slope of the graph of f(x) = 2x at the point (1, f(1)), and (ii) find an equation for the tangent line at that point. (iii) Use Excel to show both the graph of f and the graph of the tangent line in a single plot. Plot both functions over the interval [0, 3]. Exercise 32 Excel Differentiating.xlsm T C I (material continues)

Differentiation. Tangents & Slopes: page 6
Differentiation, T. & S. (i) Find the slope of the graph of the natural logarithm function, f(x) = ln(x) at the point (2, f(2)), and (ii) find an equation for the tangent line at that point. (iii) Use Excel to show both the graph of f and the graph of the tangent line in a single plot. Plot both functions over the interval [1, 5]. Exercise 33 Let f(x) = 1 + x/2. (i) Find the slope of the graph of f at the point (4, f(4)), and (ii) find an equation for the tangent line at that point. (iii) Use Excel to show both the graph of f and the graph of the tangent line in a single plot over the interval [0, 6]. (iv) Use the formulas for differentiation to explain what happened in Part iii. Exercise 34 Many formulas have been developed which allow us to find the derivative of a function symbolically, without the need for numerical approximation. These will not be needed in our work with differentiation, but they might be of interest to you. To see a few examples, click on symbolic. Information Excel Differentiating.xlsm T C I (material continues)

Differentiation, Focus
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives How can differentiation help us price 12-GB drives? We must consider the effect of the units that we are using. Recall; from the sheet Functions in Marketing Focus.xlsm and from the Focus pages of the section Demand, Revenue, Cost, and Profit; that C(q) gives the cost, in millions of dollars for producing q thousand drives. C'(q) gives the rate of change in C(q), with respect to q. (material continues) Marketing Focus.xlsm Class Project T C I Differentiation, Focus

Differentiation, Focus
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Hence, C'(q) gives marginal cost in millions of dollars per thousand drives, at a production level of q thousand drives. From our variable cost data, we see that the marginal cost for q thousand drives is \$162 per drive if q is less than or equal to 800; \$128 per drive if q is greater than 800, but less than or equal to 1,200; and \$72 per drive for q greater than 1,200. These values give marginal cost in dollars per drive at a production level of q thousand drives. How does this relate to the derivative? 1,000,000C'(q) is in dollars per thousand drives, and (1,000,000C'(q))/1,000 is in dollars per drive. The last quantity simplifies to 1,000C'(q). Hence, where C is differentiable, 1,000C'(q) gives the marginal cost, MC(q), in dollars per drive, at a production level of q thousand drives. (material continues) Marketing Focus.xlsm Differentiation, Focus Class Project T C I

Differentiation, Focus
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives For 12-GB drives, R(q) gives the revenue, in millions of dollars, from the sale of q thousand drives, and R'(q) gives the rate of change in R(q), with respect to q. Hence, R'(q) gives marginal revenue in millions of dollars per thousand drives, at a production level of q thousand drives. Since our data for marginal profit occurs naturally in dollars per drive, we will us these units for all 12-GB drive marginal functions. (1,000,000R'(q))/1,000 is in dollars per drive. This simplifies to 1,000R'(q). MR(q) = 1,000R'(q) is the marginal revenue, MR(q), in dollars per drive, when q thousand drives are being sold. The functions MR(q), and MC(q) are plotted in the sheet Marginal of Marketing Focus.xlsm. These graphs are also shown on the next page. Note that the marginal revenue is positive where the revenue function is increasing and negative where revenue is decreasing. Since cost increases with quantity, marginal cost is always positive. (material continues) Marketing Focus.xlsm Differentiation, Focus Class Project T C I

Differentiation, Focus
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives P(q) = R(q)  C(q) gives the profit, in millions of dollars, from the sale of q thousand drives. Since P(q) = R(q)  C(q), we know that, in dollars per drive, MP(q) = MR(q)  MC(q) = 1,000R'(q)  1,000C'(q) = 1,000(R'(q)  C'(q)) = 1,000P'(q). (material continues) Marketing Focus.xlsm Differentiation, Focus Class Project T C I

Differentiation, Focus
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Since, in dollars per drive, MP(q) = 1,000P'(q), we see that MP(q) and P'(q) always have the same sign, and are equal to zero at exactly the same values of q. Hence, MP(q) = 0 if and only if R (q) = C (q). The above graphs are copied from the sheet Marginal in Marketing Focus.xlsm. The changes in cost per drive at q = 800 and q = 1,200 thousand drives cause drops in the plot of MC(q) and produce corresponding jumps upward in MP(q). Due to the second gap, marginal profit crosses the axis in two places. These reflect the two local maxima, or “humps”, in the profit function. (material continues) Marketing Focus.xlsm Differentiation, Focus Class Project T C I

Differentiation, Focus
From the plot of marginal profit, it appears that MP(q) = 0 when q is approximately equal to 1,150 or 1,250. Looking at the plot of profit, it seems likely that profit has local maxima at the zeros of MP. Cells B107:I107 of the sheet Marginal in Marketing Focus.xlsm compute values for all of our functions, when a value of q is entered in B107. Experimentation with different values of q produces the following results. As we have guessed from the graphs, the local maximum at the larger value of q is greater than at the smaller value of q. According to our model, Card Tech would earn a maximum profit of \$42,176,000 from selling 1,262, GB drives, priced at \$ This answers Questions 1, 2, and 3 from the original discussion of our Class Project. on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives (material continues) Marketing Focus.xlsm Differentiation, Focus Class Project T C I

Differentiation, Focus
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives We will answer Question 4 for our Class Project, by investigating the sensitivity of the maximum profit to changes from the optimal price. Numerical experimentation in Cells B107:I107 of the sheet Marginal in Marketing Focus.xlsm shows that raising the price per drive by \$2 to \$ would lower the number of drives sold to 1,250,431, and decrease the profit to million dollars. This is a drop of only 0.08% from the maximum possible profit. Similar work shows a 0.08% drop in profit if the price per drive is lowered to \$ Clearly, profit is not very sensitive to \$2 changes in price. On the other hand, a \$10 change in price would cause a noticeable effect on profits. Raising the price per drive to \$295.88, drops the profit by 2.01% and lowering the price to \$275.88, drops the profit by 1.91%. (material continues) Marketing Focus.xlsm Differentiation, Focus Class Project T C I

Differentiation, Focus
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Since profit is not very sensitive to small changes in price, Card Tech would probably adjust the “optimum” price of \$ to a more common value, such as \$ or \$ Experimentation in the sheet Marginal of Marketing Focus.xlsm, shows that these could be expected to reduce the profit by only 0.02%. WHAT SHOULD YOU DO? Each team should now plot MR, MC, and MP for its product and prepare computational cells, as in the sheet Marketing. Use these to answer your team’s Questions 1-4 for its product. Use the same units as we have used in the Class Project. (material continues) Marketing Focus.xlsm Differentiation, Focus Class Project T C I

Using Solver, Excel’s Solver
The utility Solver is one of Excel’s most useful tools for business analysis. This allows us to maximize, minimize, or find a predetermined value for the contents of a given cell by changing the values in other cells. Moreover, this can be done in such a way that it satisfies extra constraints that we might wish to impose. Example 1. The size limitations on boxes shipped by your plant are as follows. (i) Their circumference is at most 100 inches. (ii) The sum of their dimensions is at most 120 inches. You would like to know the dimensions of such a box that has the largest possible volume. Let H, W, and L be the height, width, and length of a box; respectively; measured in inches. We wish to maximize the volume of the box, V = HWL, subject to the limitations that the circumference C = 2H + 2W  100 and the sum S = H + W + L  120. This problem is set up in the Excel file Shipping.xlsx. We will outline its solution with screen captures and directions. First, enter any reasonable values for the dimensions of the box in Cells B7:D7. Shipping.xlsx (material continues) T C I

Using Solver. Excel’s Solver: page 2
W H L FRAGILE Crush slowly Using Solver. Excel’s Solver: page 2 Using Solver, Solver To use Solver, click on Data, then Solver in the Analysis box. In older versions of Excel select Tools in the main Excel menu, then click on Solver. To use Solver, click on Data, then Solver in the Analysis box. In older versions of Excel select Tools in the main Excel menu, then click on Solver. Enter cell that computes volume. Select Max. Enter cells that contain dimensions Click on Add. Computer Problem? (material continues) Shipping.xlsx T C I

Using Solver. Excel’s Solver: page 3
Using Solver, Solver The requirement that the circumference be at most 100 inches is called a constraint. We want to have the contents of Cell E7 be at most 100. Enter cell that computes circumference. Select <=. Click on OK. Enter the limiting number. Repeat the above process to add the constraint F7 <= 120, then click on Solve. Shipping.xlsx T C I (material continues)

Using Solver. Excel’s Solver: page 4
Using Solver, Solver Click on Solve. Click on Keep Solver Solution. Click on OK. Shipping.xlsx T C I (material continues)

Using Solver. Excel’s Solver: page 5
The dimensions that maximize volume are now shown in Cells B8:D8. The maximum volume, the value of the circumference and the sum of the dimensions are now displayed. For a maximum volume of 43,750 cubic inches, the box should be 25 inches high, 25 inches wide, and 70 inches long. Using Solver, Solver In rare cases; such as very large or small initial values of H, W, or L; you may need to add the constraints B7 >= 0, C7 >= 0 and D7 >= 0. Suppose that the shipping company in Example 1 requires that each dimension of a box must be at least 3. Note that the longest item which can be shipped in a box has a length of (i) Modify Shipping.xlsx and use Solver to find the dimensions of an allowable box that will accept the longest possible item. (ii) What is the maximum length of such an item? Exercise 1 Shipping.xlsx T C I (material continues)

Using Solver. Excel’s Solver: page 6
Big Box shipping company limits the circumference of its boxes to at most 80 inches with the sum of their dimensions to be at most 150 inches. (i) Modify Shipping.xlsx and use Solver to find the dimensions of the box with maximum volume that will be accepted by Big Box. (ii) What is the maximum volume of a box which Big Box will ship? Exercise 2 Using Solver, Solver Rush! shipping company limits the size of the boxes that it accepts by limiting their volume to at most 16 cubic feet (27,648 cubic inches). For it to ship a box, each dimension must be between 3 and 54 inches. (i) Modify Shipping.xlsx and use Solver to find the dimensions of a Rush! box which will accept the longest possible item. Hint: Use different initial values for each dimension. (ii) What is the maximum length of such an item? Note that the longest item which can be shipped in a box has a length of Exercise 3 Shipping.xlsx T C I (material continues)

Using Solver. Excel’s Solver: page 7
Using Solver, Solver Example 2. Return to the restaurant management example that has been studied in Demand, Revenue, Cost and Profit; and Differentiation. We will consider the demand, profit, and marginal profit functions; D(q), P(q), and MP(q); that were developed in those sections. Recall that a restaurant chain is planning to introduce a new buffalo steak dinner. D(q) is the price, in dollars, at which q dinners per week would be ordered. P(q) is the profit from selling q dinners priced at D(q) dollars and MP(q) =P(q) is the marginal profit. What is the minimum number of dinners that the restaurants would have to sell in order to make a profit? In the sheet Solver of Dinners.xlsm we have plotted D(q), P(q), and P(q); and have built computation boxes that evaluate the functions at q. Referring to the graph in that sheet we see that P(q) is first positive when q is approximately equal to Solver is now used to find a value close to 700, in Cell G11 that makes Cell I11 equal to 0. Dinners.xlsm T C I (material continues)

Using Solver. Excel’s Solver: page 8
Using Solver, Solver Dinners.xlsm T C I (material continues)

Using Solver. Excel’s Solver: page 9
Using Solver, Solver Solver finds that dinners would be needed to produce a non-negative profit. Since this makes no sense in terms of actual meals, the restaurants would need to serve at least 690 buffalo steak dinners in order to make a profit. Recomputing the Excel sheet shows that this would still require a price of at least \$29.13 per dinner. Different initial values for q may cause Solver to produce slightly different values in the target cell. Warning! Three questions are of great interest to the restaurant managers. (i) How should the buffalo steak dinners be priced in order to obtain the maximum profit? (ii) How many dinners per week could they expect to sell at that price? (iii) What maximum profit can they expect? We can use Solver and the computation boxes in the sheet Solver of Dinners.xlsm to answer all of these questions. We start with a guess value of 2,100 for q, and then maximize P(q). Dinners.xlsm T C I (material continues)

Using Solver. Excel’s Solver: page 10
Using Solver, Solver Dinners.xlsm T C I (material continues)

Using Solver. Excel’s Solver: page 11
Using Solver, Solver P(q) has a maximum value when q = As expected, we note that the marginal profit, MP(q), is equal to zero when P(q) is at a maximum. Customers do not order dinners! Hence, q = is not a practical answer. If we add the constraint that Cell G11 is to be an integer, then Solver can find the whole number of dinners which will maximize profit. Dinners.xlsm T C I (material continues)

Using Solver. Excel’s Solver: page 12
Using Solver, Solver Ask a tutor This completes our work with the restaurant example. When we first considered the problem in Demand, Revenue, Cost, and Profit; we used graphs to estimate that the restaurant chain should prepare around 2,000 buffalo steak dinners per week and price them at approximately \$22, expecting a weekly profit of around \$14,000. In Differentiation, we used marginal analysis to refine these estimates, concluding that they should expect to sell 2,025 buffalo steak dinners per week, priced at \$22.21, for a total profit of \$14,052. Finally, Solver has given us a If Solver can maximize or minimize any function, why should I use derivatives for optimization? much easier way to obtain the same information. Although such precision is clearly unnecessary, we could use Solver to find the expected profit from more realistic pricing. 2,059 dinners, priced at \$21.95 would yield an expected weekly profit of \$14,040. Entrée Buffalo Steak Dinner only \$21.95 New! Dinners.xlsm T C I (material continues)

Using Solver. Excel’s Solver: page 13
Ha, ha, ha!!! Using Solver. Excel’s Solver: page 13 Using Solver, Solver Use Solver to find the maximum number of dinners that the restaurant chain in our example could sell and still make a non-negative profit. Exercise 4 Oh, Oh. Use Solver to find the expected maximum profit if the buffalo dinners are priced at (i) \$19.95 or (ii) at \$24.95. Exercise 5 (i) Use Solver to determine the optimum number of buffalo dinners for the restaurant chain, by finding where MP(q) = 0. (ii) Why do you think that Solver returns a non-integer value for q, even if a constraint is added, requesting only integer values? Exercise 6 Dinners.xlsm T C I (material continues)

Using Solver. Excel’s Solver: page 14
Using Solver, Solver (i) Use Solver to find the number qrev of buffalo dinners that will produce the maximum revenue in the restaurant example. (ii) What price should be put on the dinners in order to sell qrev of them? (iii) What maximum revenue can be expected? Exercise 7 Consider the situation that was discussed in Demand, Revenue, Cost, and Profit. The demand and cost functions for a good were given by D(q) = 0.1q and respectively. (i) Use Differentiating.xlsm to plot profit, P, and marginal profit, P, over the interval from 0 to 1,500. (ii) Use Solver to find a value for q that is greater than 100, and maximizes profit. (iii) Use Solver to find a value for q that is greater than 100, and at which P(q) = 0. Exercise 8 Refer to the good whose demand function is given in Exercise 8. (i) Use Differentiating.xlsm to plot revenue, R, and marginal revenue, R, over the interval from 0 to 1,500. (ii) Use Solver to find a value for q that maximizes revenue. (iii) Use Solver to find a value for q at which R(q) = 0. Exercise 9 (material continues) Dinners.xlsm Differentiating.xlsm Excel T C I

Using Solver. Excel’s Solver: page 15
Using Solver, Solver Consider the situation that was discussed in Demand, Revenue, Cost, and Profit. The demand and cost functions for a certain model of audio speaker are D(q) =  q and C(q) = 60, q, respectively. (i) Use Differentiating.xlsm to plot profit, P, and marginal profit, P. (ii) Use Solver to find a value for q that maximizes profit. (iii) Use Solver to find a value for q at which P(q) = 0. Exercise 10 Refer to the audio speakers whose demand function is given in Exercise 10. (i) Use Differentiating.xlsm to plot revenue and marginal revenue. (ii) Use Solver to find a value for q that maximizes revenue. (iii) Use Solver to find a value for q at which R(q) = 0. Exercise 11 In Differentiation, we considered the function C(t) that gives the total yearly expense of purchasing and storing CD drives, if drives are ordered every t days, starting from the first of the year. Use Solver to find the value of t that minimizes C(t). Exercise 12 Differentiating.xlsm Excel T C I (material continues)

Using Solver. Excel’s Solver: page 16
Using Solver, Solver Consider the Class Project on pricing 12-GB computer drives. (i) Use Solver to find the number q0 of drives, in thousands, such that MR(q0) = 0. (ii) What price should be put on the drives in order to sell q0 drives? (ii) What profit would result from selling q0 drives? Exercise 13 (i) Use Solver and your team’s data to check your earlier numerical results. (ii) Modify your project files and use Solver to answer your team’s Questions Study of the Focus pages will help with this work. Exercise 14 Marketing Focus.xlsm T C I (material continues)

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives How can Solver help us price 12-GB drives? We will apply Solver to the computation table in Rows 106 and 107 of the sheet Marginal in the Excel file Marketing Focus.xlsm. Starting with initial values of q = 600 and 1,700, and finding values of q that make P(q) = 0, we see that a positive profit will result from selling between thousand and 1, thousand drives. (material continues) Marketing Focus.xlsm Using Solver, Focus Class Project T C I

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives With an initial value of 1,300 for q, Solver finds a maximum value of \$42,176,000 for P(q). Changing the cell formatting to display three decimal places shows that maximum profit results from the production and sale of 1,262,120 drives. Since D(1, ) = \$285.91, the theoretical optimum price for the drives is \$ It is interesting to note that the lack of sensitivity of profit to price results in rather unstable results when Solver maximizes. The prices and quantities, but not the maximum profit, often depend upon the initial value of q. Marginal analysis provides more accurate results. The computation table in the sheet Marginal shows that MP(1, ) = \$0.07 per drive. Since there is a positive marginal profit at 1, thousand drives, the maximum profit will actually occur at a slightly higher production level. (material continues) Marketing Focus.xlsm Using Solver, Focus Class Project T C I

Calculus, Mathematics, Tests, Homework, Computers
Taking an initial value of 1,300 for q and using Solver to find q such that MP(q) = 0, we find a sales level of 1, thousand drives. These should be priced at \$285.88, and their sale will produce a profit of million dollars. We will use 1,262,274 drives as our optimum marketing level. In order to duplicate these results, you may need to click on Options in the Solver window, and reset Precision to The values obtained with Solver are in very close agreement with the results that we found by numerical experimentation in the Focus section of Differentiation. The use of Solver can increase accuracy and save a tremendous amount of time by eliminating the need for graphical estimation and numerical experimentation. Differentiation and Solver do have limitations, and cannot be used to replace thought and understanding. For example, the change in marginal cost at a production level of 1,200 thousand drives gives P(q) a local minimum at exactly 1,200. However, the graph of P(q) is not smooth at 1,200. Since it has an angled ”corner,” P(1,200) does not exist. If Solver is constrained to look near 1,200, it cannot find a value of q such that P(q) =0. on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives (material continues) Marketing Focus.xlsm Using Solver, Focus Class Project T C I

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Due to the lack of sensitivity of profit to production level, Solver can only approximate the level that produces the local minimum 1,200. If it is constrained to look only between 1,190 and 1,210, given an initial value of 1,201, and asked to find a value of q that minimizes P(q); Solver returns 1, Question 6 in the description of the Class Project asks us to find the profit that Card Tech could expect, if they price the drives at \$ We use Solver in the computation table at the bottom of the sheet Marginal. This shows that D(q) = \$299.99, when q = 1, , and that the profit from selling 1,176,693 drives will be \$41,779,000. Questions 7 and 8 from the Class Project ask how much Card Tech should pay for an advertising campaign that would increase demand for the 12-GB drives by 10% at all price levels. We go to the sheet Functions in Marketing Focus.xlsm. Setting the Demand Factor in Cell F11 at 1.10 increases the sales in all of the test markets by 10%. When the sheet is recalculated, new coefficients for the trend line are displayed in the plot of the demand data. These must by copied into Cells H28:H30. (material continues) Marketing Focus.xlsm Using Solver, Focus Class Project T C I

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives We now return to the computation table at the bottom of the sheet Marginal and use Solver to find a value for q such that MP(q) = 0. Starting with an initial value of q = 1,390, the optimum strategy is to sell 1,388,590 drives, priced at \$ per drive. This will result in a maximum profit of \$\$69,192,000. The increase in profit due to the 10% increase in demand is \$69,192,000  \$42,176,000. These values are rounded to the nearest thousands of dollars. Computation in Excel with unrounded values shows that the difference is \$27,015,000, rounded to thousands. Hence, Card Tech should pay no more than \$27,015,000 for the advertising campaign. It is interesting to note that the optimum selling price per drive remained unchanged by the change in demand. Question 9 in the description of the Class Project asks if it would be wise for Card Tech to put \$15,000,000 into training and streamlining which would reduce its non-fixed, variable costs by 7%. To answer this, we open the sheet Functions of Marketing Focus.xlsm and enter 0.93 as the Marginal Cost Factor in Cell G102. When the sheet is recalculated, this reduces all of the marginal costs by 7%. (material continues) Marketing Focus.xlsm Using Solver, Focus Class Project T C I

on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Jump start your business career! Click on EXECUTIVE LEVEL THINKING. Solver can now be used with MP in the computation table of the sheet Marginal. With q = 1,280 as an initial value, we find a maximum profit of \$\$55,061,00. Since this is up \$12,885,000 from the present maximum profit of \$42,176,000, Card Tech would not expect to recover the \$15,000,000 cost for training and streamlining. This makes the investment look unwise. WHAT SHOULD YOU DO? Study the file Marketing Focus.xlsm and all of the earlier Focus sections. When you have a clear understanding of the marketing process, perform similar computations with your team’s data. In particular, each team should now use Solver to answer Questions and in its team project. (material continues) Marketing Focus.xlsm Using Solver, Focus Class Project T C I

Integration, Area Under A Curve
There is a very convenient way to picture the revenue that results from selling q items with a demand function D for a good. Since R(q) is the product of q and D(q), it is the area of a rectangle that is q units wide and D(q) dollars high. It is easy to picture such a rectangle under the graph of the demand function. q D(q) Demand Function Revenue T C I (material continues)

Integration. Area Under A Curve: page 2
Since q is a number of items and D(q) is the price in dollars per item, the area of any region under the graph of the demand function represents income from sales. The unit of such an area is (items)(dollars/items) = dollars. In particular, the area of the entire region under the graph of the demand function and over the horizontal axis represents the total possible revenue for the good. The total possible revenue is the money that the producer would receive if everyone who wanted the good, bought it at the maximum price that he or she was willing to pay. This is the greatest possible revenue that a seller or producer could obtain when operating with a given demand function. Demand Function Total Possible Revenue T C I (material continues)

Integration. Area Under A Curve: page 3
We have seen that our revenue from selling a good at a single price that must be paid by all buyers is represented by the area of a rectangle that is part of the region under the graph of the demand function. The remaining space under the graph of the demand function is partitioned into two other regions. These represent income that is lost from selling at a single price. Some buyers would have been willing pay a higher price for the good than we charged. The total extra amount of money that people who bought the good would have paid is called the consumer surplus. It is represented by the area of the region under the graph of the demand function and above the revenue rectangle. q D(q) Revenue Consumer Surplus Not Sold Demand Function T C I (material continues)

Integration. Area Under A Curve: page 4
Fixing a single price means that some people can buy the good for less money that they would have been willing to pay. It also means that some potential customers do not buy the good, because they feel that the price is too high. The total amount of this lost income, which we will call not sold, is represented by the area of the region under the graph of the demand function to the right of the revenue rectangle. We learned how to find the single price that maximizes the revenue that we receive from selling a good. This is an important tool. However, the skilled marketer is always looking for creative ways to claim some of the lost income that is represented by the other regions under the graph of the demand function. The first step in this direction is computing the dollar values of the total possible revenue, the consumer surplus, and the not sold lost income. Example 1. We revisit the problem of pricing buffalo steak dinners that was considered in several earlier sections. The restaurants’ demand function is given by D(q) =  q2  q Taking q = 2,300 dinners as an example, we find that D(2,300) = \$19.99. T C I (material continues)

Integration. Area Under A Curve: page 5
If the restaurants priced buffalo steak dinners at \$19.99, they could expect to sell 2,300 dinners per week, and take in revenue of approximately (\$19.99)(2,300) = \$45,977. 1000 2000 3000 4000 5000 \$8 \$16 \$24 \$32 Not Sold Consumer Surplus q = 2,300 D(2,300) = \$19.99 Revenue \$45,977 2,300 \$19.99 Demand Function D(q) =  q2  q \$0 It appears that we will realize only a little more than half of the total possible revenue, if we sell at the single price of \$19.99. T C I (material continues)

Integration. Area Under A Curve: page 6
Watch the effect of different prices on the various regions. Click anywhere in the plot to run the animation. See Using The Computer Text for information on controlling a media animation. It appears that the maximum revenue occurs when approximately 2,328 buffalo steak dinners are sold, priced at \$ This gives rough graphical confirmation of our earlier computation. Computer Problem? T C I (material continues)

Integration. Area Under A Curve: page 7
Experimentation with the animation on the preceding page finds that the approximate maximum revenue from buffalo steak dinners is \$45,972. This comes from the sale of 2,328 dinners. At that quantity, the consumer surplus is shown as \$15, 945 and the lost revenue from dinners not sold is \$18,082. Adding the areas of these three regions, we find that the total possible revenue is \$45,972 + \$15, \$18,082 = \$79,999. Notice that, with single price selling, we can claim only \$45,972/\$79,999  57% of the total possible revenue. Integration, Area In Differentiation we found that the maximum profit from the sale of buffalo steak dinners resulted from the sale of 2,025 dinners, priced at \$22.21 per dinner. (i) Experiment with the animation on the preceding page to approximate the consumer surplus, revenue, and lost income from not sold dinners. Hint: the quantity 2,007 is closest to 2,025 in the animation. (ii) Use your results from Part i to approximate the total possible revenue. (iii) What percentages of the total possible revenue are accounted for by consumer surplus, revenue, and lost income from not sold dinners? Exercise 1 T C I (material continues)

Integration. Area Under A Curve: page 8
In order to deal effectively with marketing problems we will have to learn how to compute the areas of regions under the graphs of demand functions. This is only one of many practical business applications that involve the computation of areas. As a second example, recall from your work in Part 1 of Mathematics for Business Decisions that probability information about a continuous random variable, X, is carried by its probability density function, fX. Specifically, P(a  X  b) is given by the area under the graph of fX and over the interval [a, b]. A a b fX Area of region A = P(a  X  b) T C I (material continues)

Integration. Area Under A Curve: page 9
Computing probabilities for continuous random variables (such as time and money), defining and computing the mean of a continuous random variable, computing the value of an income stream, and computing consumer surpluses; all involve finding the areas of regions that are bounded by the graphs of functions. Integration is the mathematical tool that helps us make these computations. The development of integration is a major undertaking, which will require time and effort. However, there is no way to acquire the quantitative information necessary for many business decisions without this important tool. We will start with a very simple problem. What is the area of the region R that is enclosed between the x-axis and the graph of f(x) = 2x  x2/2, for x between 1 and 4? The first step is to graph the function. Try this with Graphing.xlsm and compare your plot with the following. Graphing.xlsm T C I (material continues)

Integration. Area Under A Curve: page 10
3 2 Note that the region R is completely contained in a rectangle of height 2 and width 3. Hence, the area of R is less than 6. 1.5 2 R completely contains a rectangle of height 1.5 and width 2. Hence, the area of R is greater than 3. f(4) = 24  42/2 = 8  8 = 0 f(1) = 21  12/2 = 2  1/2 = 1.5 We should never start a computation without having some idea of the expected results. The most common of all area formulas states that the area of a rectangle is the product of its height times its width. T C I (material continues)

Integration. Area Under A Curve: page 11
We can improve on our crude estimates by using rectangles in a different way. The interval from 1 to 4 can be divided into 6 subintervals, each of width Next, we form rectangles over the subintervals. Since we want the rectangles to approximate the area of R, it seems reasonable to cut each one off at the place where the graph of f crosses the middle of its width. As is always the case, the introduction of mathematical notation simplifies our discussion. The points 1, 1.5, 2, 2.5, 3, 3.5, and 4 will be named x0, x1, x2, x3, x4, x5, and x6. Let x = 0.5, which is the width of each subinterval. Denote the midpoints of the 6 subintervals by m1, m2, m3, m4, m5, and m6. Since m1 is the midpoint of [1, 1.5], we have m1 = ( )/2 = Likewise, m2 = 1.75, m3 = 2.25, m4 = 2.75, m5 = 3.25, and m6 = 3.75. T C I (material continues)

Integration. Area Under A Curve: page 12
Similar calculations allow us to compute f(mi) for i = 2, 3, , 6. (material continues) T C I

Integration. Area Under A Curve: page 13
The height of the ith rectangle is f(mi) and its width is x = Hence, the area of the ith rectangle is f(mi)x. is our approximation for the area of R. The 6 rectangles do not completely fill R and, in some places, they extend outside of R. Thus, it is not clear whether our estimate is too big or too small. If we used more than 6 rectangles, it is likely that our approximation for the area of R would improve. For example, consider 20 rectangles, each with a height that is the value of f at the midpoint of its base. These are shown on the next page. The shaded region inside the rectangles appears to be quite similar to the area under the graph of f. T C I (material continues)

Integration. Area Under A Curve: page 14
We need notation that will allow us to compute the sums of areas for an arbitrary number, n, of rectangles. Let x = (4  1)/n, and let xi = 1 + ix for i = 0, 1, , n. The midpoint of [xi-1, xi] is mi = (xi-1 + xi)/2 for i = 1, 2, , n. We will let Sn denote the midpoint sum of the areas of the n rectangles. Excel can help us do the arithmetic that is needed to compute midpoint sums for large values of n. The sheet n = 20 of the file Area Example.xlsx shows that S20 = T C I (material continues)

Integration. Area Under A Curve: page 15
The sheet Any n in Area Example.xlsx allows us to compute Sn for any value of n between 1 and The sums are shown rounded to 6 decimal places. Experiment with different values of n in this sheet. We discover that Sn gets closer and closer to 4.5 as n is increased. When n is up to 1501, S1501 = 4.5, rounded to 6 decimal places. Further increases in n do not change this value. Based on this evidence, we believe that the limit of Sn, as n increases without bound, is This is denoted by We have now found that the area of the region R is 4.5 square units. Experiment with Area Example.xlsx, to find the smallest value of n for which Sn differs from 4.5 by at most 0.01. Exercise 2 Experiment with Area Example.xlsx, to find the smallest value of n for which Sn differs from 4.5 by at most Exercise 3 Area Example.xlsx T C I (material continues)

Integration. Area Under A Curve: page 16
Experiment with Area Example.xlsx, to find the smallest value of n for which Sn differs from 4.5 by at most Exercise 4 The same plan that we have just used in our example can be applied to the problem of computing the area between the horizontal axis and the graph of any function, over an arbitrary interval [a, b]. We let x = (b  a)/n, and let xi = a + ix for i = 0, 1, , n. The midpoint of [xi-1, xi] is mi = (xi-1 + xi)/2 for i = 1, 2, , n. We let denote the midpoint sum of the areas of the n rectangles. If these midpoint sums approach a number A, as n is made larger and larger, we write and consider A to be the area between the axis and the graph of f, over [a, b]. Area Example.xlsx T C I (material continues)

Integration. Area Under A Curve: page 17
Click anywhere in the plot to see the effect of n on the midpoint sums for a typical probability density function. See Using The Computer Text for information on controlling a media animation. Integration, Area Computer Problem? T C I (material continues)

Integration. Area Under A Curve: page 18
Modify sheet n = 20 in Area Example.xlsx, so that it computes S20(f, [0, 5]) for f(x) = 2x  x2/2. Exercise 5 Area Example.xlsx (i) Use Graphing.xlsm to plot f(x) = e-x over the interval [0, 10] and use this plot to estimate the area under the graph of f, above the x-axis, and over [0, 10]. (ii) Create a new Excel file that computes S1000(f, [0, 10]). (iii) Based upon this computation, what do you think is the area under the graph of f, above the x-axis, and over [0, 10]? (iv) Since f gives the non-zero part of an exponential probability density function, what would you expect to be true about this area? Exercise 6 Excel Graphing.xlsm T C I (material continues)

Integration. Area Under A Curve: page 19
Modify sheet n = 20 in Area Example.xlsx, so that it computes the sum S100(f, [0, 4]), with 100 subintervals, for f(x) = 2x  x2/2. Exercise 7 Area Example.xlsx (i) Create a new Excel file that computes S1000(f, [5, 5]), where the function f is given by the following. (ii) Based upon this computation, what do you think is the area under the graph of f, above the x-axis, and over [5, 5]? (iii) Do you think that it is likely, or unlikely, that f is a probability density function? Justify your answer. Exercise 8 Excel T C I (material continues)

Integration. Area Under A Curve: page 20
(i) Use Graphing.xlsm to plot f(x) = (1/6)x3  x2 + (3/2)x over the interval [0, 3] and use this plot to estimate the area under the graph of f, above the x-axis, and over [0, 3]. (ii) Create a new Excel file that computes S1000(f, [0, 3]). (iii) Based upon this computation, what do you think is the area under the graph of f, above the x-axis, and over [0, 3]? (iv) Do you think that it is likely, or unlikely, that f is the non-zero portion of a probability density function? Justify your answer. Exercise 9 Excel Graphing.xlsm T C I (material continues)

Integration, Integrals
What would happen if we computed midpoint sums for a function which might assume negative values in the interval [a, b]? a b + Where f(mi) < 0, the product f(mi)Dx is also negative. Thus, the midpoint sums will approximate the “signed area” of the region between the x-axis and the graph of f, over [a, b]. This is the algebraic sum of the area above the axis, minus the area below the axis. T C I (material continues)

Integration, Integrals
Integration. Integrals: page 2 Integration, Integrals For most of the functions, f, that occur in simple business applications, as n increases, the midpoint sums Sn(f, [a, b]) over any reasonable interval will approach a limiting value. This value is called the integral of f over [a, b] and is denoted by the expression The function, f, is called the integrand and the symbol  is called an integral sign. Historically, this was derived from a stretched out letter S, to remind people that an integral is a limit of sums. The dx at the end of the integral expression is placed there to tell us that the variable of integration is x, and to indicate the end of the integrand. To summarize our main result, the integral of f over [a, b] is and it represents the algebraic sum of the signed areas of the regions between the horizontal axis and the graph of f, over [a, b]. T C I (material continues)

integration, Integrals
Integration. Integrals: page 3 The Excel file Mid- point Sums.xlsm lets you experiment with midpoint sums. You can enter any function, any interval, and any number of subdivisions, n (with n  500). The program will then plot the function and the approximating rectangles. It will also display Dx and the value of the midpoint sum. Experiment integration, Integrals Click here for information on running the macro in Midpoint Sums.xlsm. (i) Use Midpoint Sums.xlsm to plot f(x) = 3x2 over the interval [0, 2], and compute Sn(f, [0, 2]) for n = 5, 10, 100, and 500. (ii) Use your results from Part i. to guess the exact value of (iii) Experiment with the interval [0, v] for several values of v, and use the method of Parts i and ii to guess the exact value of Exercise 10 Midpoint Sums.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 4 Integration, Integrals (i) Use Midpoint Sums.xlsm to plot f(x) = x3  2x over the interval [2, 2], and compute Sn(f, [2, 2]) for n = 3, 10, 40, and 99. (ii) Use the idea of signed areas to explain what you found in Part i. Exercise 11 (i) Use Midpoint Sums.xlsm to plot over the interval [1, 1], and compute Sn(f, [1, 1]) for n = 10 and (ii) Noting that the graph of f is the upper half of a circle with radius r = 1, explain the sum that you found for n = Hint: The area of a disk with radius one is p square units. Exercise 12 Excel interprets x2 as (x)2. You must enter the expression as (x2). However, Excel interprets 1  x2 correctly. No parentheses are needed. Useful Information (i) Use Midpoint Sums.xlsm to plot f(x) = x2 + 1 over the interval [2, 2], and compute S200(f, [2, 2]). (ii) Repeat Part i for the interval [ 0, 2]. (iii) Use the form of f(x) to explain the relationship between your answers to Parts i and ii. Exercise 13 Midpoint Sums.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 5 Integration, Integrals Example 2. Evaluate , rounded to 4 decimal places. We first use Graphing.xlsm to plot f over [1.3, 4.7]. From the plot, it appears that the area of the region between the x-axis and the graph of f, over the given interval might be roughly 1.4. Graphing.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 6 Integration, Integrals Next, we create an Excel file Integration Example.xlsx. Read the following description and then go to that file. On the first sheet, n = 50, we have entered values for a, b, and n. Cell E6 is defined by the formula =(C6-B6)/D6. The numbers 0, 1, …, 50 are entered in Column B. =\$B\$6+B9*\$E\$6 is entered in Cell C9, and then dragged down to Cell C59. Integration Example.xlsx T C I (material continues)

Integration, Integrals
Integration. Integrals: page 7 Integration, Integrals We entered =(C9+C10)/2 in Cell D10 and then dragged this down to Cell D59. =D10^2*EXP(-D10) was entered in Cell E10 and then dragged down to Cell E59. =E10*\$E\$6 was entered in Cell F10 and then dragged down to Cell F59. Finally, =SUM(F10:F59) was entered in Cell F6. S50(f, [1.3, 4.7]) = , when rounded to 4 decimal places. Integration Example.xlsx T C I (material continues)

Integration, Integrals
Integration. Integrals: page 8 Integration, Integrals Sheets n =100 and n = 200 in Integration Example.xlsx are made in a similar way. As they show, S100(f, [1.3, 4.7]) = and S200(f, [1.3, 4.7]) = , when rounded to 4 decimal places. Since the values of the midpoint sums do not appear to be changing once n is over 100, we can assume that rounded to 4 decimal places. (i) Use Graphing.xlsm to plot f(x) = x2 + 6x +22 over the interval [2.5, 3.1]. (ii) Based on this graph, what is your estimate for Exercise 14 Excel interprets x2 as (x)2. You must enter the expression as (x2). Remember (material continues) Integration Example.xlsx Graphing.xlsm T C I

Integration, Integrals
Integration. Integrals: page 9 Integration, Integrals Use Excel and the method of Example 2 to evaluate rounded to 2 decimal places. Exercise 15 (i) Use Graphing.xlsm to plot f(x) = x2ex over the interval [0, 6.5]. (ii) Based on this graph, what is your estimate for Exercise 16 Use Excel and the method of Example 2 to evaluate rounded to 4 decimal places. Exercise 17 Graphing.xlsm Excel T C I (material continues)

Integration, Integrals
Integration. Integrals: page 10 Integration, Integrals (i) Modify the method of Example 2 to create an Excel file that allows you to enter a positive value for b, and then evaluate rounded to 4 decimal places. (ii) Use your work from Part i to evaluate the integral for b = 1, 1.5, 2, 2.5, and 3. (iii) Based upon your work in Part ii, can you guess a formula for the exact value of in terms of b? Exercise 18 Graphing.xlsm Excel T C I (material continues)

Integration, Integrals
Integration. Integrals: page 11 Integration, Integrals Click here for information on running the macro in Integrating.xlsm. consuming. Carrying out this much work, while in the middle of a business application, could easily distract us from the basic business problem. What we are doing is called numerical integration. Many software packages, but not Excel, have built-in functions that perform numerical integration. To meet this need, we have created an Excel file, Integrating.xlsm, that serves as a numerical integration utility. To use Integrating.xlsm, open the file and follow the instructions in the blue box. Once you have entered the formula for f(x), you can see the graph of f over any interval and can get a numerical value for the integral of f over any other interval. From now on, unless otherwise indicated, we will assume that all integrations are performed with Integrating.xlsm. Creating Excel files to evaluate sums provides a good understanding of integration, but it is rather tedious and time Integrating.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 12 Excel interprets x2 as (x)2. You must enter the expression as (x2). Remember Integration, Integrals Use Integrating.xlsm to evaluate the following. Exercise 19 Use Integrating.xlsm to evaluate the following. Exercise 20 Use Integrating.xlsm to evaluate the following. Exercise 21 Integrating.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 13 Integration, Integrals Use Integrating.xlsm to evaluate the following. Exercise 22 Use Integrating.xlsm to evaluate the following. Exercise 23 Use Integrating.xlsm to evaluate the following. Exercise 24 Integrating.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 14 Integration, Integrals When entering a function in Integrating.xlsm, you can use the letters s, t, u, v, or w as constants. If this is done, you must enter values for any letters that you use. For example, suppose that you want to plot and integrate f(x) = x s, where s is a number that you enter in Cell O23. You would enter the following function in Cell C20. = x^s Accept the function and then click on the Integrate button at the right of the plot. Since Excel is set for automatic calculation, entering a new number in Cell O23 will automatically change the function, redraw the plot, and recompute the integral. If values are entered in Cells O23:O27; then constants s, t, u, v, or w may also be used to describe the range, [A, B], of the plot; and the limits, a and b, of integration. If these constant names are used, they must be entered in Cells H20:I20 and Cells K20:L20, preceded with "=" signs. For example, =s. Integrating.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 15 Integration, Integrals Use Integrating.xlsm to evaluate for s = 1, 2, 5, 10, 20, and Leave the parameter s as a constant. Changing s and recomputing the sheet should change the plot and integral correspondingly. What happens to the value of the integral as s increases? Exercise 25 (i) Use Integrating.xlsm to evaluate for u = 0.1, 0.5, 1, 2, 5, and 10. Leave the parameter u as a constant. Changing u and recomputing the sheet should change the plot correspondingly. (ii) What happens to the value of the integral as u increases? (iii) Use the plots of f(x) = (2/u2)x + 2/u and the values of f(0) to explain your answer to Part ii. Exercise 26 Integrating.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 16 Integration, Integrals (i) Use Integrating.xlsm to evaluate for  = 0.5, 1, 1.5, 2, 3, 5, 8, and 10. Leave the parameter  as a constant. Changing  and recomputing the sheet should change the plot correspond-ingly. (ii) What happens to the value of the integral as  changes? Hint: use the built-in constant, s, as a replacement for . Exercise 27 (i) Leaving b, , and  as positive constants that you can set later, use Integrating.xlsm to evaluate (ii) Display the graph and integral approximation for b = 2,  = 1, and  = 2. (iii) Display the graph and integral approximation for b = 3,  = 2, and  = (iv) Given any values for  > 1 and  > 1, what happens to the value of the integral as b increases? Hint: use the built-in constants to replace b, , and . The integrand in the above integral is the probability density function for a class of distributions named after the Swedish physicist Waloddi Weibull. Exercise 28 Integrating.xlsm T C I (material continues)

Integration, Integrals
Integration. Integrals: page 17 Integration, Integrals It is time for disclaimers and warnings. The mathematical theory of integration is much more sophisticated than the simple plan that we have presented. In many cases it is possible to evaluate integrals exactly, rather than using only numerical approximation. When approximation is used, there are much better routines than our plan of midpoint sums. One of these, called Simpson’s Rule, is used in Inte- grating.xlsm. Our goal is to have integration as a tool that can be used to solve business problems. For this purpose, the numerical understanding that we have developed and our integrating utility will serve quite well. Information Ask a tutor If there is a way to evaluate integrals exactly, why don’t we use that, and not bother with approximation? Integrating.xlsm T C I (material continues)

Integration, Applications
All of our work in defining and evaluating integrals can now pay off by providing information about business problems. We will start by revisiting Example 1 from the section Area Under A Curve. Example 3. Recall that the demand function for buffalo steak dinners is given by D(q) =  q2  q We can use the computation cells in the sheet Income of Dinners.xlsm to find that the graph of D reaches the q-axis at approximately q = 4,014. If every person who would buy a buffalo steak dinner in a given week were to pay the maximum price that he or she would be willing to pay, then the restaurants’ total possible revenue would be given by the area of the region that lies between the graph of D(q) and the q-axis. Using the animation in Area Under A Curve, we saw that the total possible revenue was \$79,999. We can now see how the computations were done in the animation. Dinners.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 2 Integration, Applications 1000 2000 3000 4000 5000 \$8 \$16 \$24 \$32 Demand Function D(q) =  q2  q \$0 Total Possible Revenue Integrating. xlsm allows us to confirm that the total possible revenue is Recall that, if dinners are sold at a single price, then the restaurants’ expected revenue is represented by the area of a rectangular region under the graph of the demand function. Integrating.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 3 Integration, Applications For example, to sell 2,300 dinners they would have to be priced at D(2,300) = \$19.99 each. In this case the expected revenue would be (\$19.99)(2,300) = \$45,977. D(q) =  q2  q 1000 2000 3000 4000 5000 \$8 \$16 \$24 \$32 Not Sold Consumer Surplus q = 2,300 D(2,300) = \$19.99 Revenue \$45,977 2,300 \$19.99 Demand Function \$0 T C I (material continues)

Integration, Applications
Integration. Applications: page 4 Integration, Applications Consumer surplus is represented by the area of the region above the revenue rectangle and under the graph of the demand function. This lost revenue is the total extra amount of money that people who bought the dinners would have been willing to pay. Note that the amount of consumer surplus depends upon the number of dinners that are sold. Looking at the plot on the preceding page, and using Integrating.xlsm, we can compute the consumer surplus for the sale of 2,300 buffalo steak dinners. For completeness we will also compute the amount of revenue that is lost from dinners which were not sold, because people were not willing to pay the fixed price of \$19.99. Integrating.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 5 Integration, Applications As a check on our work, note that the sum of revenue, consumer surplus, and money from dinners not sold is equal to the total possible revenue. \$45,977 + \$15,379 + \$18,643 = \$79,999 Our revenue animation is no longer “magic.” We now know how to do all of its calculations. Assuming that 1,800 buffalo steak dinners are sold, use the restaurants’ demand function to compute (i) the total possible revenue, (ii) the revenue, (iii) the consumer surplus, and (iv) the lost revenue from dinners that were not sold. Exercise 29 We have seen that the maximum profit from the sale of buffalo steak dinners would result from selling 2,025 dinners. At this sales level, use the restaurants’ demand function to compute (i) the total possible revenue, (ii) the revenue, (iii) the consumer surplus, and (iv) the lost revenue from dinners that were not sold. Exercise 30 Integrating.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 6 Integration, Applications Revenue computations for an arbitrary demand function work in the same way as those for the buffalo steak dinners. Let D(q) give the price per unit for a good, that would result in the sale of q units, and let qmax be the maximum number of units that could be sold at any price. That is, D(qmax) = 0. The total possible revenue is given by If qsold units are sold, then the revenue will be qsoldD(qsold). The following formulas give consumer surplus and lost revenue from units not sold. It is clear that revenue + consumer surplus + not sold = total possible revenue. T C I (material continues)

Integration, Applications
Integration. Applications: page 7 Integration, Applications Suppose that the demand function for a good is given by D(q) = 0.1q (i) Use Graphing.xlsm to plot D(q) and estimate the total possible revenue. Explain how you arrived at your estimate. (ii) Use Integrating.xlsm to compute the total possible revenue. Exercise 31 Refer to the information in Exercise 31 and suppose that qsold = 400 units are sold. Compute (i) the revenue, (ii) the consumer surplus, and (iii) the lost revenue from units not sold. Exercise 32 Suppose that D(q) =  q is the demand function for a certain model of audio speaker. (i) Use Graphing.xlsm to plot D(q) and estimate the total possible revenue. Explain how you arrived at your estimate. (ii) Use Integrating.xlsm to compute the total possible revenue. Exercise 33 Refer to the information in Exercise 33 and suppose that qsold = 950 units are sold. Compute (i) the revenue, (ii) the consumer surplus, and (iii) the lost revenue from units not sold. Exercise 34 Graphing.xlsm Integrating.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 8 Integration, Applications The demand function for a new type of car alarm system is shown in the adjacent plot. Note that the quantities sold are given in thousands. (i) Estimate the total possible revenue, in dollars. (ii) Estimate the consumer surplus, in dollars, that would result from selling 300,000 alarms. Exercise 35 (i) Use your team’s Marketing Project data to compute the total possible revenue. (ii) Compute the consumer surplus for the quantity of goods that maximizes profit. (This is Question 5 in your team’s project.) Use the same units as in the work on the Class Project in the sheet Functions of Marketing Focus.xlsm. Exercise 36 Excel T C I (material continues)

Integration, Applications
Integration. Applications: page 9 Integration, Applications In addition to the computation of consumer surplus and total possible revenue, there are a great many applications of integration to business problems. We will conclude this section with one such application. The important uses of integration in probability will be considered in our work on Project 2. Example 4. The Plastic-Is-Us Toy Company considers its incoming revenue as an income stream, rather than as a huge collection of discrete payments. At a time t years from the start of its fiscal year on July 1, the company expects to receive revenue at the rate of A(t) million dollars per year. Records from past years indicate that Plastic-Is-Us can model its revenue rate with A(t) = 110t t4  330t t million dollars per year. Using Graphing.xlsm to plot A(t), we see that the company expects its largest rate of revenue in late November, at the height of the Christmas shopping season. The slowest rate of revenue flow appears to be during the summer months. Graphing.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 10 Integration, Applications 2 4 6 8 0.25 0.5 0.75 1 t A(t) Oct. 1 Jan. 1 April 1 July 1 The chief financial officer would like to compute the total amount of revenue that Plastic-Is-Us will receive in one year. The income stream, A(t), is a rate of change in money, given in dollars per year. Since the units along the t-axis are years, the area of a region under the graph of A(t) is given in (millions of dollars/year)(years) = millions of dollars. Since gives the area between the t-axis and the graph of A(t), over the interval [0, T], it can be shown that the integral gives the total amount of money, in millions of dollars, that will be received from the income stream in the first T years. Graphing.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 11 Integration, Applications Looking at the graph of the preceding page, we note that the average value of the income stream A(t) appears to be roughly 5 million dollars per year. Hence in one year, the company can expect a total income of approximately 5 million dollars. More precisely, we can use Integrating.xlsm to compute the total income received by Plastic-Is-Us during the period from 0 to 1 year. (Remember that we must use x, not t, as the variable of integration in Integrating.xlsm.) This is very close to our estimate of 5 million dollars for the company’s total yearly income. The same type of reasoning allows us to compute the total revenue from any income stream. Graphing.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 12 Integration, Applications The total revenue, in dollars, received from an income stream of A(t) dollars per year, starting now and continuing for the next T years is given by In addition to the total revenue, a company would often like to know the present value of its income stream during the next T years (0  t  T), assuming that money earns interest at some annual rate r, compounded continuously. We will develop a structure for this computation. Recall that A dollars, invested for t years at an annual rate r, compounded continuously, has a present value of Ae-rt dollars. Consider an income stream of A(t) dollars per year and let t  be a fixed time, in years. If we get A(t ) dollars per year for a very short period of time, say Dt years, we will get A(t )Dt dollars at approximately time t . The present value of this amount is A(t )Dt e-rt  = A(t )e-rt Dt dollars. Graphing.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 13 Integration, Applications Suppose that we receive the income stream for the next T years. Integration can be used to “sum” the amounts A(t )e-rt Dt over the interval 0  t  T. Suppose that money earns at an annual rate, r, compounded continuously. The present dollar value of an income stream of A(t) dollars per year, starting now and continuing for the next T years is given by Example 5. We return to the Plastic-Is-Us Toy Company that we considered in Example 4. Recall that they have an income stream of A(t) = 110t t4  330t t million dollars per year. The management of Plastic-Is-Us would like to know the present value of its income stream during the next year (0  t  1), assuming that money earns interest at an annual rate of 5.5%, compounded continuously. Graphing.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 14 Integration, Applications Applying the integral formula for present value to Plastic-Is-Us, we use Integrating.xlsm to find that the present value of their income stream for one year, starting on July 1, is million dollars. How can we tell if this is a sensible value? We have seen that the total revenue from the income stream during a single year is million dollars. Since this money will be received in the future, its present value is somewhat less than million dollars. Our computed present value of million dollars seems to be quite reasonable. Integrating.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 15 Integration, Applications (i) What total revenue will the Plastic-Is-Us Toy Company receive from its income stream during the period from July 1 to January 1? (ii) Assuming that money earns 5.5% compounded continuously, what is the present value, on July 1, of that income stream? Exercise 37 (i) What total revenue will the Plastic-Is-Us Toy Company receive from its income stream during the period from July 1 to October 1? (ii) Assuming that money earns 5.5% compounded continuously, what is the present value, on July 1, of that income stream? Exercise 38 (i) What is the present value of a constant income stream of A(t) = \$950,000 per year for the next three years at 6%, compounded continuously? (ii) How much total revenue will be received during the three year period. (You should be able to answer Part ii without using integration!) Exercise 39 Integrating.xlsm T C I (material continues)

Integration, Applications
Integration. Applications: page 16 Integration, Applications What is the present value of a constant income stream of A(t) = \$5,450,000 per year for the next five and one-half years at 5.7%, compounded continuously? (ii) How much total revenue will be received during the five and one-half year period. (You should be able to answer Part ii without using integration!) Exercise 40 (i) What is the present value of an income stream of A(t) = 348,000,000e0.07t dollars per year for the next two years at 5%, compounded continuously? (ii) How much total revenue will be received during the two year period. Exercise 41 Integrating.xlsm T C I (material continues)

Integration. Calculus 4. CALCULUS*
The topics of differentiation and integration are the main parts of the mathematical subject called calculus. Income streams allow us to explore a very useful and important relationship between integration and differentiation. Example 6. Suppose that we invest \$250,000 at an annual rate of 6.25%, compounded continuously. What is the instantaneous rate of change in the value of our investment per year after three and one-half years? That is, at what rate, in dollars per year, will we earn interest after 3.5 years? Recall from Part 1 of Mathematics for Business Decisions that the future value, F, of P dollars, invested for t years at a continuous annual rate r is given by F = Pert. Thus, F(t) = \$250,000e0.0625t represents the value of our investment after t years. * This section, and the corresponding section Calculus of Normal Distributions in Project 2, are not needed for the study of other material in Mathematics for Business Decisions. T C I (material continues)

Integration. Calculus: page 2
Since the derivative of a function gives its instantaneous rate of change, F(3.5) gives the rate, in dollars per year, at which we will earn interest after 3.5 years. Differentiating.xlsm shows that F(3.5)  \$311,130 and F(3.5)  \$19, After three and one-half years, the value of our investment will have grown to \$311,130, and we will be earning interest at a rate of \$19, per year. Instead of considering the change in value at just one time, we can use Differentiating.xlsm to plot F(t) over a period of ten years. Differentiating.xlsm T C I (material continues)

Integration. Calculus: page 3
Since F(t) is in dollars per year, we can think of this derivative as an income stream, which we will call A(t) = F(t). As we saw in the section Applications, the total income from this stream, during the first T years, is given by the integral of A(t) from 0 to T. Thus, gives the total amount of interest that the investment will earn in the first T years. At the end of T years the full value of the investment will be the original principal of \$250,000 plus the total amount of interest that has accumulated, This same full value of the investment after T years, can also be computed with the formula, F(t), for future value. Thus, T C I (material continues)

Integration. Calculus: page 4
We now subtract \$250,000 from both sides of the equation. Both sides of the last equation are functions of T. Since, the derivative of a difference is the difference of the derivatives, the derivative of the left side of the equation, with respect to T, is F(T), plus the derivative of \$250,000. But F(T) = A(T) and the derivative of a constant is equal to 0, so the derivative of F(T) + \$250,000, with respect to T, is equal to A(T) + 0 = A(T). Since both F(T)  \$250,000 and must have the same derivatives. T C I (material continues)

Integration. Calculus: page 5
We have just found that, if A(T) is the income stream resulting from interest on an investment with future value F(T) after T years, then the derivative of with respect to T, is equal to A(T). This exhibits an inverse relationship between integration and differentiation. If we first integrate A(t), and then differentiate the integral, we return to the original function, A(t). It is an amazing fact, discovered independently by Isaac Newton ( ) and Gottfried Wilhelm Leibniz ( ), that this relationship holds for a large class of functions. There is nothing special about using the letter A as the name for a function, and it would make no difference if we replaced the variable of integration, t, with any other letter. The important part of our result is that we are differentiating with respect to the letter that is the upper limit of integration. Thus, for many functions, f, the derivative of with respect to x, is equal to f(x). T C I (material continues)

Integration. Calculus: page 6
In fact, it can be shown that the same result holds if 0 is replaced by any other number, a. When stated formally, the inverse connection between integration and differentiation is called the Fundamental Theorem of Calculus. Fundamental Theorem of Calculus. For many of the functions, f, which occur in business applications, the derivative of with respect to x, is f(x). This holds for any number a and any x, such that the closed interval between a and x is in the domain of f. We will return to this theorem when working with probability in Project 2. For now, we will consider two examples. The first of these uses a simple function, but requires a good understanding of integrals as areas. Example 7. Let f(u) = 2 for all values of u. If x  1, then integral of f from 1 to x is the area of the region over the interval [1, x], between the u-axis and the graph of f. T C I (material continues)

Integration. Calculus: page 7
(1, 2) (x, 2) x 2 x  1 Integration. Calculus: page 7 Integration, Calculus The region whose area is represented by the integral is rectangular, with height 2 and width x  1. Hence, its area is 2(x  1) = 2x  2, and In the section Properties and Applications of Differentiation, we saw that the derivative of f(x) = mx + b is equal to m, for all values of x. Thus, the derivative of with respect to x, is equal to 2. As predicted by the Fundamental Theorem of Calculus, this is also the value of f(x). The next example uses the definition of a derivative as the limit of difference quotients. T C I (material continues)

Integration. Calculus: page 8
Example 8. Recall the income stream of A(t) = 110t t4  330t t million dollars per year that was expected by the Plastic-Is-Us toy company in Example 4 of Applications. Let G(T) be the total income that is expected during the first T years, for 0  T  1. Picking a time T = 0.5 years, we will check that the instantaneous rate of change of G(T), with respect to T, is the same as A(T). Note that We now wish to compute G(0.5). Recall that G(T) is approximated by the difference quotient for small values of h. We will let h = , and use Integrating.xlsm to evaluate G( ) and G(0.5  ). Integrating.xlsm rounds the numerical values of integrals to four decimal places. For the present calculation, we gain extra precision by copying the values from Cell N20 and keeping all of their decimal places. G( ) = G(0.5001)  G(0.5  ) = G(0.4999)  Integrating.xlsm T C I (material continues)

Integration. Calculus: page 9
These give a value of for the difference quotient rounded to four decimal places. This is the instantaneous rate of change in total income after 0.5 years. Integrating.xlsm shows the same value for A(0.5). Noting that we have verified the Fundamental Theorem of Calculus. At T = 0.5, the derivative of with respect to T, is equal to A(T). Let f(u) = 1/2 for all values of u. (i) If x  2, use the method of Example 7 to find a formula for (ii) Differ- entiate your result from Part i, with respect to x, and verify that the derivative is equal to f(x) for all values of x. Exercise 42 Graphing.xlsm T C I (material continues)

Integration. Calculus: page 10
Consider the income stream A(t) for the Plastic-Is-Us toy company that was last discussed in Example 8, and let (i) Use the method of Example 8 to evaluate the derivative of G(T), with respect to T, for T = 0.75 years. (ii) Show that your result from Part i is equal to A(0.75). Exercise 43 A company’s expected income stream is modeled by A(t) = 742,000,000e0.06t dollars per year, for the next three years. Let G(T) be the total income that they will receive during the first T years of this period.. (i) Use the method of Example 8 to evaluate the derivative of G(T), with respect to T, for T = 0.75 years. (ii) Show that your result form Part i is equal to A(0.75). Exercise 44 Integrating.xlsm T C I (material continues)

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives What can integration tell us about marketing 12-GB drives? The graph of D(q) shows that demand is positive for values of q between 0 and approximately 2,500 thousand drives. Using Solver in Rows 64 and 65 in the sheet Functions of Marketing Focus.xlsm, we find that D(2, ) = 0. Integration can be used to compute the total possible revenue that could result from sales of 12-GB drives. (material continues) Marketing Focus.xlsm Integration, Focus Class Project T C I

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives Integrating.xlsm shows that Since D(q) is in dollars per drive and q is in thousands of drives, the value of the integral is in thousands of dollars. We are considering revenue in millions of dollars, so we divide by 1,000. The total possible revenue is (650, thousand dollars)/1,000  million dollars. Question 5 asks for the consumer surplus, at the production level that maximizes profit. We have seen that this involves selling 1, thousand drives, priced at \$ each. We first use Integrating.xlsm to evaluate the integral of D(q) over the interval from 0 to 1, (material continues) Marketing Focus.xlsm Integration, Focus Class Project T C I

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives As with the total possible revenue, the value of the integral is in thousands of dollars. To express the possible revenue from selling 1, thousand drives in millions of dollars, we must divide by 1,000. (459, thousand dollars)/1,000  million dollars. Our expected revenue from selling 1, thousand drives, priced at \$ each, is 1,   360, thousand dollars. Thus, the predicted revenue is 360, /1,000  million dollars. The consumer surplus when 1, thousand drives are sold is the possible revenue from sales minus the revenue that we expect, at a price of \$ per drive. This gives a consumer surplus of ( million dollars)  ( million dollars)  million dollars. (material continues) Marketing Focus.xlsm Integration, Focus Class Project T C I

Calculus, Mathematics, Tests, Homework, Computers
on the project Calculus, Mathematics, Tests, Homework, Computers Marketing Computer Drives WHAT SHOULD YOU DO? Basic Computation. Each team should now answer Question 5 by computing the consumer surplus for its team project. Use the same units as we have used in the Class Project. Further Exploration. What more useful information can you extract from your team’s marketing data? What graphical and numerical displays can you produce to help make business decisions? Teams might want to explore the sensitivity of their conclusions to small changes in the marketing data and assumptions. Other areas of investigation could include the relationship of maximum profit to changes in demand and costs. Prepare Reports. When all work is completed, prepare files for your team’s written and oral reports. (material continues) Marketing Focus.xlsm Integration, Focus Class Project T C I

Marketing Example, Double Pricing
In the buffalo steak dinner example that we have been following, Solver can be used in the computation cells of the sheet Solver in Dinners.xlsm. We find that D(4,014) = 0. Hence, the total possible revenue is We have seen that the restaurant chain would obtain a maximum weekly profit of \$14,052 by selling 2,025 dinners, priced at \$22.21 each. Under this plan, the mangers note that 2,025 people are buying the dinners for less than they would have been willing to pay and that 4,014  2,025 = 1,989 people who would have been willing to buy dinners at a lower price are not paying money to the restaurants. Looking at the situation from another point of view, \$79,999  2,025\$22.21  \$35,024 of potential revenue is being lost. How might the restaurants collect some of this money and increase their profit? T C I (material continues)

Marketing Example. Double Pricing: page 2
Marketing, Double After consultation with the managers and chefs at the various restaurants, the management team decided to up-grade the buffalo dinners in 1/4 of the establishments in the chain. The fancier entree will be promoted as more desirable and will be sold at a higher price than the regular buffalo steak dinner. It is believed that the improved image of the up-scale dinners will make people who can afford the higher prices choose to pay more and eat at places which sell the up-graded dinners. We will assume that the chain has enough restaurants to allow a choice of dining level for all of its potential customers and that both fixed and variable costs are now distributed uniformly among all of the establishments. To spread out its clientele evenly, the chain decides to price the dinners in such a fashion that up-graded meals will account for 1/4 of the total number of buffalo steak dinners which will be purchased. In making their plans, the managers assume that the demand function for a total of q buffalo steak dinners will remain unchanged under the new plan. Hence, we still have (material continues) T C I

Marketing Example. Double Pricing: page 3
\$ Marketing Example. Double Pricing: page 3 Marketing, Double Some increase in costs will be necessary if the up-scale entree is to have an image that will support higher prices. Managers and chefs estimate that up-grading an entree will raise its share of the original fixed costs by 50% and its share of the original total variable costs by 10%. Since 25% of the dinners will be up-graded, fixed costs will now be (0.75)\$9,000 for the remaining regular meals and (1.5)(0.25)(\$9,000) for the up-scale meals. Recall that the variable cost for preparing a total of q dinners is 177q Regular meals will still account for 3/4 of this cost, 0.75177q Up-scale meals accounted for the other 1/4 of the original variable costs. Since production costs are 10% higher for up-graded entrees, it will cost (1.1)(0.25)177q0.633 to prepare their share of the q dinners. This information can be combined to form a new cost function. Before we can analyze profits, we must construct the new revenue function. (material continues) T C I

Marketing Example. Double Pricing: page 4
Marketing, Double Since it is planned to sell a total of q dinners, up-scale dinners will be priced so that q/4 dinners will be sold at these places. This means a price of D(q/4) dollars per meal, generating (q/4)D(q/4) dollars of revenue. The remaining (3/4)q regular dinners will have to be priced at a level that will allow the total sales of all q dinners. These will generate an additional (3/4)qD(q) dollars of revenue. These two components of revenue are shown under a plot of the demand function at q = 2,000. For example, if 2,000 dinners are produced, 500 will be up-scale entrees, priced at D(500) = \$29.59, each. The remaining 1,500 regular dinners will be priced at D(2,000) = \$22.40, each. Regular revenue Up-scale (material continues) T C I

Marketing Example. Double Pricing: page 5
Marketing, Double The double pricing plan is analyzed in the Up-Scale sheet of the Excel file Dinners.xlsm. To see the regions of regular and up-scale revenue for q dinners, enter q in Cell H24 of that sheet. Experiment with Up-Scale to confirm the following fact. Since no more that 4,014 dinners can be sold, a value of q which is greater than or equal to 4,014 yields the same revenue as selling q/4 dinners. The total revenue, R(q), that the chain will receive from producing q dinners is the sum of the up-scale and regular income. A plot of the revenue and cost functions for the double pricing plan is shown on the next page. As expected, revenue recovers some after q = 4,014. However, such production levels are absurd, since no more than 4,014 dinners can be sold. Obviously, the cost of these unsold meals continues to increase. Dinners.xlsm (material continues) T C I

Marketing Example. Double Pricing: page 6 Dinners.xlsm  T C I 
Marketing, Double It is clear that a positive profit will occur only when q is between approximately 700 and 3,600. Using Solver in the Computation box of the sheet Up-Scale, we find that a maximum weekly profit of \$16,923 can be expected to result from selling 582 up-scale dinners at \$29.41 each and 2,328  582 = 1746 regular dinners at \$19.75 each. Dinners.xlsm (material continues) T C I

Marketing Example. Double Pricing: page 7 Dinners.xlsm  T C I 
Marketing, Double The plan for up-grading the entree in some of the restaurants in the chain and then selling the buffalo steak dinners at two different prices gives an improvement over the single price plan. The maximum weekly profit of \$16,923 is up \$2,871 from the maximum weekly profit of \$14,052 under the single price plan. Dinners.xlsm (material continues) T C I

Marketing Example. Double Pricing: page 8
Marketing, Double (i) Suppose that the restaurant chain wants to sell a total of 1,000 dinners. Use Solver in the sheets Solver and Up-Scale of Dinners.xlsm to decide whether the single or double pricing plan would yield a greater weekly profit. (ii) Repeat Part i for 3,000 dinners. Exercise 1 (i) Use Solver in the sheet Up-Scale of Dinners.xlsm to find the total number of buffalo dinners that the restaurants should sell in order to price the up-scale dinners at \$ (ii) What price should be put on the regular dinners to match the up-scale price of \$29.95? Exercise 2 (i) Use Solver in the sheet Up-Scale of Dinners.xlsm to find the total number of dinners that should be sold in order to maximize the weekly revenue. (ii) How many of these dinners should be of each type, how should they be priced, and what maximum revenue could be expected? Exercise 3 Dinners.xlsm T C I (material continues)

Marketing, Optimizing Conversion
Marketing Example. Optimizing Conversion Marketing, Optimizing Conversion 2. OPTIMIZING CONVERSION Having seen that up-grading 1/4 of the buffalo dinners, and instituting a two-price system could increase profits, the managers wondered if 1/4 is the optimum fraction of establishments to convert to up-scale. Suppose that a fraction, r, of the dinners are up-graded, and that dinners are priced so that rq up-scale dinners out of a total of q dinners are sold. The remaining (1  r)q dinners will be regular meals. The same type of reasoning that we used with r = 0.25 can be used to derive the following formulas for the new revenue and cost functions. The sheet Up-Scale of Dinners.xlsm is set up to compute demand, revenue, cost, and profit for any value of r that is entered in Cell H23. Dinners.xlsm (material continues) T C I

Marketing Example. Optimizing Conversion: page 2
Marketing, Optimizing In Double Pricing, we only used the sheet Up-Scale with r = As a check on our more general analysis of the situation, we can enter r = 0 in Cell H23. This means that none of the entrees would be up-graded. Since no meals will be sold at higher prices, and there has been no increase in costs, we should obtain the same results that we found under the single pricing plan that was discussed in earlier work. Setting r = 0 and using Solver to find the maximum profit for an integer number of dinners, yields the following. These are exactly the same results that we found for single pricing. At the other extreme, we can see what would happen if we converted all of the meals to up-scale entrees. To do this, we let r = 1 in Up-Scale. Dinners.xlsm (material continues) T C I

Marketing Example. Optimizing Conversion: page 3
Marketing, Optimizing After maximizing profit with Solver, we see that the weekly profit from the up-graded meals would be only \$7,371. This is roughly half of the profit that was expected from the original single pricing. The ability to sell buffalo steak dinners at a slightly higher price does not off-set the increased costs of preparing the dinners. Of the plans investigated so far, up-grading 25% of the dinners has predicted the largest maximum weekly profit for the chain. Could they expect to do any better? Suppose that 50% of the meals were up-graded. We can enter 0.5 for r in Up-Scale, and then use Solver to find the maximum profit. The resulting \$19,010 is the highest value that we have yet found. Dinners.xlsm (material continues) T C I

Marketing Example. Optimizing Conversion: page 4
Marketing, Optimizing It is now clear that, for each fraction, r, of meals that will be up-graded, there are two new functions. Let Pmax(r) be the maximum weekly profit, and let qmax(r) be the total number of dinners that should be prepared and sold in order to realize that profit. The sheet Fraction in the Excel file Dinners.xlsm shows values of qmax(r) and Pmax(r) for values of r ranging from 0.00 to 1.00, in hundredths. The graphs and significant data have been copied from that sheet onto the following two pages. Dinners.xlsm (material continues) T C I

Optimizing Conversion: page 5
Marketing Example. Optimizing Conversion: page 5 Marketing, Optimizing Dinners.xlsm T C I (material continues)

Optimizing Conversion: page 6
We see that the highest weekly profit from buffalo steak dinners, \$19,012, will result from converting 51% of the restaurant chain’s establishments to up-scale units. To find the pricing structure for this plan, we return to the sheet Up-Scale, enter r = 0.51, and use Solver to find the maximum profit. Marketing Example. Optimizing Conversion: page 6 Marketing, Optimizing The chain should plan on selling 1,318 up-scale dinners, priced at \$26.67, and it can expect to sell 2,584  1,318 = 1,266 regular dinners, priced at \$17.41. Dinners.xlsm (material continues) T C I

Marketing Example. Optimizing Conversion: page 7
Marketing, Optimizing (i) Use Solver in the sheet Up-Scale of Dinners.xlsm to find the total number of dinners that should be sold in order to maximize weekly profit, if 60% of the chain’s establishments are up-graded. (ii) How many of these dinners should be of each type, how should they be priced, and what maximum profit could be expected? Exercise 4 What reduction in maximum profit would the restaurant chain experience in they up-graded 40% of the dinners, rather than the optimal 51%? Exercise 5 (i) What fraction of the buffalo dinners should be up-graded, in order to obtain the maximum weekly revenue from the entree? (ii) How many dinners of each type should be sold, how should they be priced, and what maximum revenue could be expected? Exercise 6 Dinners.xlsm T C I (material ends)

Back up your studying! omphaloskepsis: “The act of contemplating
Business Mathematics Students: Back up your studying! Backup Plan You never can tell. It might really help. omphaloskepsis: “The act of contemplating one’s navel, as an exercise for mystics.” (Greek, omphalos, the navel + Greek, skepsis, a viewing.) Webster’s New World Dictionary Second College Edition, 1980 Back to Differentiation

Back to Differentiation
The Derivative For further mathematical study of differentia- tion, it is common to define the derivative in terms of one-sided difference quotients, where h is allowed to assume either positive or negative values. For the purposes of numerical approximation, our symmetric definition of the derivative is preferable to the above theoretical definition. Although it will not be of concern in our work, it should be noted that the theoretical difference quotient for a function, f, at a given value of x, may not approach any fixed number as h takes on smaller and smaller values. This happens if the graph of f has a corner at the point (x, f(x)). In this case f (x) does not exist and we say that f is not differentiable at x. For example, the absolute value function, f(x) = |x|, is not differentiable at x = 0. The Derivative Back to Differentiation (material ends)

Symbolic Differentiation
The definition of a derivative may be used to establish the following general formulas for differentiation.  If r  0 is a real number and f(x) = x r, then f (x) = rx r1.  If a > 0 is a real number, and f(x) = ax, then f (x) = axln(a). In particular, the derivative of ex is the same function, ex.  If b  1 is a positive real number, and f(x) = logbx, then f (x) = (1/x)logbe. In particular, the derivative of ln(x) is 1/x. These formulas may be used in conjunction with the rules for differentiation which we discussed earlier. Example 1. Let f(x) = 5.7x x x Now f (x) = 5.7(3x31) + 1.2(2x21) + 5.3(1x11) + 0 = 17.1x x Thus, for example, f (6) = 17.1  = Back to Differentiation (material continues)

Symbolic Differentiation
Symbolic Differentiation: page 2 Example 2. Return to the buffalo steak dinner example that was considered earlier. The demand and cost functions for the dinners were given by D(q) =  q2  q and C(q) = 9, q0.663, respectively. This yields the following profit function. P(q) = R(q)  C(q) = qD(q)  C(q) =  q3  q q  9,000  177q0.633 The marginal profit, P, can be computed with our rules and formulas for differentiation. P(q) =  (3q31)  (2q21)  0  177(0.633q0.6331) =  q2  q  q0.367 Unfortunately, after all of this work, there is no closed form way to solve the equation P(q) = 0, and find an exact value for q. The only way that we could use the formula for marginal profit would be to use some type of numerical approximation to solve P(q) = 0. This type of dilemma is one of the reasons numerical differentiation is used in Mathematics for Business Decisions. Symbolic Differentiation Back to Differentiation (material ends)

Index E-L M-P Q-T U-Z County administrator example 2-123, 2-171, 2-179
Course files, needed 28 running 13 Cumulative distribution function 2-2, D Data Analysis, running 81 Demand function 1-35 marginal 1-70 Derivative 1-80, 1-219, 1-220 properties 1-89 Difference quotient 1-68 Differentiation 1-80, 1-219, 1-220 A-B Bicycle shorts example 2-138 Binomial random variable 2-3, 2-13, 2-71 Buffalo steak dinner example 1-35, 1-41, 1-64, 1-120 1-139, 1-173 Bug in random number generator 2-176 C Central limit theorem 2-117 Confidence interval 2-128 Consumer surplus 1-138 Control buttons 15 Cost function 1-41 marginal 1-64, 1-68 T C

Focus pages (continued), Distributions 2-58 Integration 1-199
Index: page 2 A-D M-P Q-T U-Z Index Focus pages (continued), Distributions 2-58 Integration 1-199 Normal Distributions 2-159 Simulating Normal Random Variables 2-184 Using Solver 1-130 Variance 2-94 Fundamental Theorem of Calculus 1-194, 2-156 I-J-K Income stream 1-181, 1-189 Integral 1-157 L Loading Data Analysis 81 E Expected value, continuous random variable 2-38 finite random variable 2-12 Exponential random variable 2-27, 2-42, 2-77 F-G-H Files needed 28 Final reports, Oral 35 Written 40 Focus pages, Demand, Revenue, Cost, and Profit 1-52 Differentiation 1-106 T C

Normal random variable 2-107, 2-119, 2-143 O Omphaloskepsis 1-218 P
Index: page 3 A-D E-L Q-T U-Z Index N Normal random variable 2-107, 2-119, 2-143 O Omphaloskepsis 1-218 P Parameter 2-47, 2-81 Preliminary reports 33 Present value of income stream 1-185 Probability density function, definition 2-16 Probability mass function, definition 2-2 (P continues) M Macros 83 Marginal, cost 1-64, 1-68 demand 1-70 profit 1-70 revenue 1-70 Mean, continuous random variable 2-38 finite random variable 2-12 sample 2-46 Midpoint sums, defined 1-149 animation 1-152 T C

finite random variable 2-70 sample 2-86 sample mean 2-92
Index: page 4 A-D E-L M-P U-Z Index Running animations 23 Reports, final oral 35 final written 40 preliminary 33 Revenue function 1-37 marginal 1-70 S-T Sample mean 2-46 Standard deviation, continuous 2-75 finite random variable 2-70 sample 2-86 sample mean 2-92 Standard normal random variable 2-107, 2-119 (S-T continues) P (continued) Profit function 1-44 marginal 1-70 Project 1, class project 54 team projects 58 Project 2, class project 62 team projects 68 Q Quotation 2-227 R Random variable, binomial 2-3, 2-13, 2-71 exponential 2-27, 2-42, 2-77 T C

Index: page 5 U V-W-X-Y-Z T  C
A-D E-L M-P Q-T Index S-T (continued) Standardized random variable 2-82 Symbol font 77 U Uniform random variable 2-27 V-W-X-Y-Z Variance, continuous random variable 2-75 finite random variable 2-69 sample 2-85 sample mean 2-92 T C

Executive Level Thinking
Marketing Computer Drives. Executive Level Thinking Executive Level Thinking The mathematical predictions for marketing models are dependent on the assumptions that we put into them. In the real world, our estimates for things such as the size of the potential national market and the future costs of production are almost never exactly correct. The goal is to build mathematical models that are robust. This means that their predications do not fluctuate widely from small errors in our data estimates. Executive level thinking involves understanding the role which our assumptions play in the predictions of our mathematical models. One of the more subtle assumptions in the project on Marketing Computer Drives is our choice of a 2nd degree polynomial as a model for the demand function. In order for us to study revenue and profit, we must have a formula for our demand function. Unfortunately, business settings never provide formulas. We must construct these mathematically, using sound business judgment. Excel does this by fitting a trend line to our test market data points. There is no way to mathematically determine the “best fitting” trend line. Close (material continues)

Marketing Computer Drives. Executive Level Thinking: page 2
We must select a type of function from the list of those that are available in Excel. Once the computer is given a specific type of function, it can find the “best fitting” trend line of that type. If we select a type of function that does not reflect the underlying demand, then Excel’s answer will give us false information about the marketing problem. As is the case with most test market data, our six points do not strongly suggest any type of function. We used a 2nd degree polynomial for our trend line, since it appears to fit the data quite well and its graph is a common shape for demand functions. How crucial is this choice in the outcome of the model? High level managers in marketing departments consider such questions very carefully. The best way to see the effect of our trend line choice is to investigate the predictions that would be made by a different type of function. One simple and useful such choice would be a linear model. For this we ask Excel to find the best fitting trend line of the form D(q) = ax + b. The results are shown in the worksheet Linear Trend Line in Marketing Focus.xlsm. The graph on the following slide is copied from that worksheet. Close (material continues)

Marketing Computer Drives. Executive Level Thinking: page 3
Looking at the two types of demand functions, we see that both models have very similar values in the middle range where they are fitted to the test market data. They differ for both quite small and quite large sales levels. The business interpretation of this is that the linear model predicts higher prices than the quadratic model at extreme sales levels. Which, if either, is correct? This is a business decision that a company must make. Close (material continues)

Marketing Computer Drives Executive Level Thinking: page 4
To see how robust a marketing model we have created, we will check the differences in the predictions of the two types of demand functions. A plot of the two revenue functions is helpful. Close (material continues)

Marketing Computer Drives. Executive Level Thinking: page 5
Using Solver in the worksheet Linear Trend Line, we find a maximum revenue of \$369,705,000 from selling 1,503,905 drives at a price of \$ per drive. This is up less than 1% from the maximum revenue of \$366,155,000 that was predicted by the quadratic trend line for demand. The wider spread of the linear model’s revenue graph on the right side reflects the model’s predictions of more sales at very low prices. This is probably irrelevant for any practical business decisions. In general, our greatest interest is in profit. Graphs of the profit functions determined by the two forms of trend lines for demand are shown in a single plot on the next slide. In the critical region near their maximum values they both have very similar “two humped” shapes. Using Solver and the computational cells in the worksheet Linear Trend Line, we can compute the price and quantity to sell that would maximize profit. These are compared with the corresponding values for the quadratic trend line model for demand. Close (material continues)

Marketing Computer Drives. Executive Level Thinking: page 6
Close (material continues)

Marketing Computer Drives. Executive Level Thinking: page 7
This is good news. Given the uncertain nature of our estimates for production costs and the randomness of the test marketing process, a variation of less than 2% in the predicted outcomes is of no practical business consequence. We have developed a robust model, whose results do not fluctuate widely depending upon our choice of a trend line form. Close (material ends)

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