© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 62 Chapter 7 Functions of Several Variables.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 2 of 62  Examples of Functions of Several Variables  Partial Derivatives  Maxima and Minima of Functions of Several Variables  Lagrange Multipliers and Constrained Optimization  The Method of Least Squares  Double Integrals Chapter Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 4 of 62  Functions of More Than One Variable  Cost of Material  Tax and Homeowner Exemption  Level Curves Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 5 of 62 Functions of More Than One Variable DefinitionExample Function of Several Variables: A function that has more than one independent variable

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 7 of 62 Cost of MaterialEXAMPLE SOLUTION (Cost) Find a formula C(x, y, z) that gives the cost of material for the rectangular enclose in the figure, with dimensions in feet, assuming that the material for the top costs $3 per square foot and the material for the back and two sides costs $5 per square foot. TOPLEFT SIDERIGHT SIDEBACK 3555 xyyz xz Area (sq ft) Cost (per sq ft)

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 8 of 62 Cost of Material The total cost is the sum of the amount of cost for each side of the enclosure, CONTINUED Similarly, the cost of the top is 3xy. Continuing in this way, we see that the total cost is

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 9 of 62 Tax & Homeowner ExemptionEXAMPLE (Tax and Homeowner Exemption) The value of residential property for tax purposes is usually much lower than its actual market value. If v is the market value, then the assessed value for real estate taxes might be only 40% of v. Suppose the property tax, T, in a community is given by the function where v is the estimated market value of a property (in dollars), x is a homeowner’s exemption (a number of dollars depending on the type of property), and r is the tax rate (stated in dollars per hundred dollars) of net assessed value. Determine the real estate tax on a property valued at $200,000 with a homeowner’s exemption of $5000, assuming a tax rate of $2.50 per hundred dollars of net assessed value.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 10 of 62 Tax & Homeowner ExemptionSOLUTION We are looking for T. We know that v = 200,000, x = 5000 and r = 2.50. Therefore, we get CONTINUED So, the real estate tax on the property with the given characteristics is $1875.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 11 of 62 Level Curves DefinitionExample Level Curves: For a function f (x, y), a family of curves with equations f (x, y) = c where c is any constant An example immediately follows.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 12 of 62 Level CurvesEXAMPLE SOLUTION Find a function f (x, y) that has the curve y = 2/x 2 as a level curve. Since level curves occur where f (x, y) = c, then we must rewrite y = 2/x 2 in that form. This is the given equation of the level curve. Subtract 2/x 2 from both sides so that the left side resembles a function of the form f (x, y). Therefore, we can say that y – 2/x 2 = 0 is of the form f (x, y) = c, where c = 0. So, f (x, y) = y – 2/x 2.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 14 of 62  Partial Derivatives  Computing Partial Derivatives  Evaluating Partial Derivatives at a Point  Local Approximation of f (x, y)  Demand Equations  Second Partial Derivative Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 15 of 62 Partial Derivatives DefinitionExample Partial Derivative of f (x, y) with respect to x: Written, the derivative of f (x, y), where y is treated as a constant and f (x, y) is considered as a function of x alone If, then

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 16 of 62 Computing Partial DerivativesEXAMPLE SOLUTION Compute for To compute, we only differentiate factors (or terms) that contain x and we interpret y to be a constant. This is the given function. Use the product rule where f (x) = x 2 and g(x) = e 3x. To compute, we only differentiate factors (or terms) that contain y and we interpret x to be a constant.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 18 of 62 Computing Partial DerivativesEXAMPLE SOLUTION Compute for To compute, we treat every variable other than L as a constant. Therefore This is the given function. Rewrite as an exponent. Bring exponent inside parentheses. Note that K is a constant. Differentiate.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 19 of 62 Evaluating Partial Derivatives at a PointEXAMPLE SOLUTION Let Evaluate at (x, y, z) = (2, -1, 3).

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 21 of 62 Local Approximation of f ( x, y )EXAMPLE SOLUTION Let Interpret the result We showed in the last example that This means that if x and z are kept constant and y is allowed to vary near -1, then f (x, y, z) changes at a rate 12 times the change in y (but in a negative direction). That is, if y increases by one small unit, then f (x, y, z) decreases by approximately 12 units. If y increases by h units (where h is small), then f (x, y, z) decreases by approximately 12h. That is,

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 22 of 62 Demand EquationsEXAMPLE SOLUTION The demand for a certain gas-guzzling car is given by f (p 1, p 2 ), where p 1 is the price of the car and p 2 is the price of gasoline. Explain why is the rate at which demand for the car changes as the price of the car changes. This partial derivative is always less than zero since, as the price of the car increases, the demand for the car will decrease (and visa versa). is the rate at which demand for the car changes as the price of gasoline changes. This partial derivative is always less than zero since, as the price of gasoline increases, the demand for the car will decrease (and visa versa).

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 23 of 62 Second Partial DerivativeEXAMPLE SOLUTION Let. Find We first note that This means that to compute, we must take the partial derivative of with respect to x.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 25 of 62  Relative Maxima and Minima  First Derivative Test for Functions of Two Variables  Second Derivative Test for Functions of Two Variables  Finding Relative Maxima and Minima Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 26 of 62 Relative Maxima & Minima DefinitionExample Relative Maximum of f (x, y): f (x, y) has a relative maximum when x = a, y = b if f (x, y) is at most equal to f (a, b) whenever x is near a and y is is near b. Examples are forthcoming. DefinitionExample Relative Minimum of f (x, y): f (x, y) has a relative minimum when x = a, y = b if f (x, y) is at least equal to f (a, b) whenever x is near a and y is is near b. Examples are forthcoming.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 27 of 62 First-Derivative Test If one or both of the partial derivatives does not exist, then there is no relative maximum or relative minimum.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 29 of 62 Finding Relative Maxima & MinimaEXAMPLE SOLUTION Find all points (x, y) where f (x, y) has a possible relative maximum or minimum. Then use the second-derivative test to determine, if possible, the nature of f (x, y) at each of these points. If the second-derivative test is inconclusive, so state. We first use the first-derivative test.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 30 of 62 Finding Relative Maxima & Minima Now we set both partial derivatives equal to 0 and then solve each for y. CONTINUED Now we may set the equations equal to each other and solve for x.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 31 of 62 Finding Relative Maxima & Minima We now determine the corresponding value of y by replacing x with 1 in the equation y = x + 2. CONTINUED So we now know that if there is a relative maximum or minimum for the function, it occurs at (1, 3). To determine more about this point, we employ the second-derivative test. To do so, we must first calculate

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 32 of 62 Finding Relative Maxima & Minima Since, we know, by the second-derivative test, that f (x, y) has a relative maximum at (1, 3). CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 33 of 62 Finding Relative Maxima & MinimaEXAMPLE SOLUTION A monopolist manufactures and sells two competing products, call them I and II, that cost $30 and $20 per unit, respectively, to produce. The revenue from marketing x units of product I and y units of product II is Find the values of x and y that maximize the monopolist’s profits. We first use the first-derivative test.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 34 of 62 Finding Relative Maxima & Minima Now we set both partial derivatives equal to 0 and then solve each for y. CONTINUED Now we may set the equations equal to each other and solve for x.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 35 of 62 Finding Relative Maxima & Minima We now determine the corresponding value of y by replacing x with 443 in the equation y = -0.1x + 280. CONTINUED So we now know that revenue is maximized at the point (443, 236). Let’s verify this using the second-derivative test. To do so, we must first calculate

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 36 of 62 Finding Relative Maxima & Minima Since, we know, by the second-derivative test, that R(x, y) has a relative maximum at (443, 236). CONTINUED

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 38 of 62  Background and Steps for Lagrange Multipliers  Using Lagrange Multipliers  Lagrange Multipliers in Application Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 39 of 62 Optimization In this section, we will optimize an objective equation f (x, y) given a constraint equation g(x, y). However, the methods of chapter 2 will not work, so we must do something different. Therefore we must use the following equation and theorem.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 41 of 62 Using Lagrange MultipliersEXAMPLE SOLUTION Maximize the function, subject to the constraint We have and The equations L-1 to L-3, in this case, are

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 42 of 62 Using Lagrange Multipliers From the first two equations we see that CONTINUED Therefore, Substituting this expression for x into the third equation, we derive

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 43 of 62 Using Lagrange Multipliers Using y = 3/5, we find that CONTINUED So the maximum value of x 2 + y 2 with x and y subject to the constraint occurs when x = 6/5, y = 3/5, and That maximum value is

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 44 of 62 Lagrange Multipliers in ApplicationEXAMPLE SOLUTION Four hundred eighty dollars are available to fence in a rectangular garden. The fencing for the north and south sides of the garden costs $10 per foot and the fencing for the east and west sides costs $15 per foot. Find the dimensions of the largest possible garden. Let x represent the length of the garden on the north and south sides and y represent the east and west sides. Since we want to use all $480, we know that We can simplify this constraint equation as follows. We must now determine the objective function. Since we wish to maximize area, our objective function should be about the quantity ‘area’.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 45 of 62 Lagrange Multipliers in Application The area of the rectangular garden is xy. Therefore, our objective equation is Therefore, CONTINUED Now we calculate L-1, L-2, and L-3.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 46 of 62 Lagrange Multipliers in Application From the first two equations we see that CONTINUED Therefore, Substituting this expression for y into the third equation, we derive

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 47 of 62 Lagrange Multipliers in Application Using x = 12, we find that CONTINUED So the maximum value of xy with x and y subject to the constraint occurs when x = 12, y = 8, and That maximum value is

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 49 of 62  Least Squares Error  Least Squares Line (Regression Line)  Determining a Least Squares Line Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 50 of 62 Least Squares Error DefinitionExample Least Squares Error: The total error in approximating the data points (x 1, y 1 ),...., (x N, y N ) by a line y = Ax + B, measured by the sum E of the squares of the vertical distances from the points to the line, Example is forthcoming.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 51 of 62 Least Squares Line (Regression Line) DefinitionExample Least Squares Line (Regression Line): A straight line y = Ax + B for which the error E is as small as possible. Example is forthcoming.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 52 of 62 Determining a Least Squares LineEXAMPLE Table 5 shows the 1994 price of a gallon (in U.S. dollars) of fuel and the average miles driven per automobile for several countries. (a) Find the straight line that provides the best least-squares fit to these data. (b) In 1994, the price of gas in Japan was $4.14 per gallon. Use the straight line of part (a) to estimate the average number of miles automobiles were driven in Japan.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 53 of 62 Determining a Least Squares Line (a) The points are plotted in the figure below. The sums are calculated in the table below and then used to determine the values of A and B. CONTINUED SOLUTION

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 54 of 62 Determining a Least Squares LineCONTINUED xyxyx2x2 1.5710,37116,282.472.4649 2.8610,18629,131.968.1796 3.31874028,929.410.9561 3.34767425,631.1611.1556 3.44745625,648.6411.8336 1.2411,09913,762.761.5376 ∑ x = 15.76∑ y = 55,526∑ xy = 139,386.4∑ x 2 = 46.1274

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 55 of 62 Determining a Least Squares LineCONTINUED Therefore, the equation of the least-squares line is y = -1365.824x + 12,841.898. (b) We use the straight line to estimate the average number of miles automobiles were driven in Japan in 1994 by setting x = 4.14. Then we get y = -1365.824(4.14) + 12,841.898 ≈ 7187. Therefore, we estimate the average number of miles per auto in Japan in 1994 to be 7187.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 57 of 62  Double Integral of f (x, y) over a Region R  Evaluating Double Integrals  Double Integrals in “Application” Section Outline

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 58 of 62 Double Integral of f ( x, y ) over a Region R DefinitionExample Double Integral of f (x, y) over a Region R: For a given function f (x, y) and a region R in the xy-plane, the volume of the solid above the region (given by the graph of f (x, y)) minus the volume of the solid below the region (given by the graph of f (x, y)) Example is forthcoming.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 60 of 62 Evaluating Double IntegralsEXAMPLE SOLUTION Calculate the iterated integral. Here g(x) = x and h(x) = 2x. We evaluate the inner integral first. The variable in this integral is y (because of the dy). Now we carry out the integration with respect to x. So the value of the iterated integral is 27/2.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 61 of 62 Double Integrals in “Application”EXAMPLE SOLUTION Calculate the volume over the following region R bounded above by the graph of f (x, y) = x 2 + y 2. R is the rectangle bounded by the lines x = 1, x = 3, y = 0, and y = 1. The desired volume is given by the double integral. By the result just cited, this double integral is equal to the iterated integral We first evaluate the inner integral.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 62 of 62 Double Integrals in “Application” Now we carry out the integration with respect to y. CONTINUED So the value of the iterated integral is 28/3. Notice that we could have set up the initial double integral as follows. This would have given us the same answer.

© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 62 Chapter 7 Functions of Several Variables.

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© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 62 Chapter 7 Functions of Several Variables.

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Presentation on theme: "© 2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 62 Chapter 7 Functions of Several Variables."— Presentation transcript:

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