Chapter 7 Functions of Several Variables

Chapter Outline 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

§ 7.1 Examples of Functions of Several Variables

Section Outline Functions of More Than One Variable Cost of Material Tax and Homeowner Exemption Level Curves

Functions of More Than One Variable Definition Example Function of Several Variables: A function that has more than one independent variable

Functions of More Than One Variable EXAMPLE Let . Compute g(1, 1) and g(0, -1). SOLUTION

Cost of Material EXAMPLE (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. SOLUTION TOP LEFT SIDE RIGHT SIDE BACK 3 5 xy yz xz Cost (per sq ft) Area (sq ft)

Cost of Material CONTINUED The total cost is the sum of the amount of cost for each side of the enclosure, Similarly, the cost of the top is 3xy. Continuing in this way, we see that the total cost is

Tax & Homeowner Exemption EXAMPLE (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.

Tax & Homeowner Exemption CONTINUED SOLUTION We are looking for T. We know that v = 200,000, x = 5000 and r = 2.50. Therefore, we get So, the real estate tax on the property with the given characteristics is $1875.

Level Curves Definition Example 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.

Level Curves EXAMPLE Find a function f (x, y) that has the curve y = 2/x2 as a level curve. SOLUTION Since level curves occur where f (x, y) = c, then we must rewrite y = 2/x2 in that form. This is the given equation of the level curve. Subtract 2/x2 from both sides so that the left side resembles a function of the form f (x, y). Therefore, we can say that y – 2/x2 = 0 is of the form f (x, y) = c, where c = 0. So, f (x, y) = y – 2/x2.

§ 7.2 Partial Derivatives

Section Outline Partial Derivatives Computing Partial Derivatives Evaluating Partial Derivatives at a Point Local Approximation of f (x, y) Demand Equations Second Partial Derivative

Definition: Partial Derivatives

Computing Partial Derivatives EXAMPLE Compute for SOLUTION 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) = x2 and g(x) = e3x. To compute , we only differentiate factors (or terms) that contain y and we interpret x to be a constant.

Computing Partial Derivatives CONTINUED This is the given function. Differentiate ln y.

Computing Partial Derivatives EXAMPLE Compute for SOLUTION 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.

Evaluating Partial Derivatives at a Point EXAMPLE Let Evaluate at (x, y, z) = (2, -1, 3). SOLUTION

Partial Derivative Rule

Local Approximation of f (x, y) We can generalize the interpretations of to yield the following general fact: Partial derivatives can be computed for functions of any number of variables. When taking the partial derivative with respect to one variable, we treat the other variables as constant.

Local Approximation of f (x, y) EXAMPLE Let Interpret the result SOLUTION 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,

Demand Equations EXAMPLE The demand for a certain gas-guzzling car is given by f (p1, p2), where p1 is the price of the car and p2 is the price of gasoline. Explain why SOLUTION 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).

Second Partial Derivative EXAMPLE Let . Find SOLUTION We first note that This means that to compute , we must take the partial derivative of with respect to x.

§ 7.3 Maxima and Minima of Functions of Several Variables

Section Outline 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

Relative Maxima & Minima Definition Example 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 near b. Examples are forthcoming. Definition Example 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 near b. Examples are forthcoming.

First-Derivative Test If one or both of the partial derivatives does not exist, then there is no relative maximum or relative minimum.

Second-Derivative Test

Finding Relative Maxima & Minima EXAMPLE 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. SOLUTION We first use the first-derivative test.

Finding Relative Maxima & Minima CONTINUED Now we set both partial derivatives equal to 0 and then solve each for y. Now we may set the equations equal to each other and solve for x.

Finding Relative Maxima & Minima CONTINUED We now determine the corresponding value of y by replacing x with 1 in the equation y = x + 2. 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

Finding Relative Maxima & Minima CONTINUED Since , we know, by the second-derivative test, that f (x, y) has a relative maximum at (1, 3).

Finding Relative Maxima & Minima EXAMPLE 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. SOLUTION We first use the first-derivative test.

Finding Relative Maxima & Minima CONTINUED Now we set both partial derivatives equal to 0 and then solve each for y. Now we may set the equations equal to each other and solve for x.

Finding Relative Maxima & Minima CONTINUED We now determine the corresponding value of y by replacing x with 443 in the equation y = -0.1x + 280. 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

Finding Relative Maxima & Minima CONTINUED Since , we know, by the second-derivative test, that R(x, y) has a relative maximum at (443, 236).

First Derivative Test – 3 Variables

§ 7.4 Lagrange Multipliers and Constrained Optimization

Section Outline Background and Steps for Lagrange Multipliers Using Lagrange Multipliers Lagrange Multipliers in Application

Optimization Method of Lagrange multipliers: 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.

Steps For Lagrange Multipliers

Using Lagrange Multipliers EXAMPLE Maximize the function , subject to the constraint SOLUTION We have and The equations L-1 to L-3, in this case, are

Using Lagrange Multipliers CONTINUED From the first two equations we see that Therefore, Substituting this expression for x into the third equation, we derive

Using Lagrange Multipliers CONTINUED Using y = 3/5, we find that So the maximum value of x2 + y2 with x and y subject to the constraint occurs when x = 6/5, y = 3/5, and That maximum value is

Lagrange Multipliers in Application EXAMPLE 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. SOLUTION 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’.

Lagrange Multipliers in Application CONTINUED The area of the rectangular garden is xy. Therefore, our objective equation is Therefore, Now we calculate L-1, L-2, and L-3.

Lagrange Multipliers in Application CONTINUED From the first two equations we see that Therefore, Substituting this expression for y into the third equation, we derive

Lagrange Multipliers in Application CONTINUED Using x = 12, we find that 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

§ 7.5 The Method of Least Squares

Section Outline Least Squares Error Least Squares Line (Regression Line) Determining a Least Squares Line

Example is forthcoming. Least Squares Error Definition Example Least Squares Error: The total error in approximating the data points (x1, y1),...., (xN, yN) 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.

Least Squares Line (Regression Line)

Determining a Least Squares Line EXAMPLE The following table gives the number in thousands of car-accident-related deaths in the U.S. for certain years. (a) Find the straight line that provides the best least-squares fit to these data. (b) Use the straight line found in part (a) to estimate the number of car-accident related deaths in 2012. (It is interesting that, while the number of drivers is obviously increasing with time, the number of car-accident-related deaths is actually decreasing, maybe because of improvements in car-manufacturing technologies and added safety measures.)

Determining a Least Squares Line CONTINUED SOLUTION (a) The points are plotted in the figure below, where x denotes the number of years since 1990. The sums are calculated in the table in the following slide and then used to determine the values of A and B.

Determining a Least Squares Line CONTINUED

Determining a Least Squares Line CONTINUED Therefore, the equation of the least-squares line is y = -.39x + 47.64. (b) We use the straight line to estimate the number of car-accident-related deaths in 2012 by setting x = 22. Then we get y = -.39(22) + 47.64 = 39.06. Therefore, we estimate the number of number of car-related accidental deaths to be 39.06 thousand in 2012.

§ 7.6 Double Integrals

Section Outline Double Integral of f (x, y) over a Region R Evaluating Double Integrals Double Integrals in “Application”

Example is forthcoming. Double Integral of f (x, y) over a Region R Definition Example 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.

The Double Integral

Evaluating Double Integrals EXAMPLE Calculate the iterated integral. SOLUTION 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.

Double Integrals in “Application” EXAMPLE Calculate the volume over the following region R bounded above by the graph of f (x, y) = x2 + y2. R is the rectangle bounded by the lines x = 1, x = 3, y = 0, and y = 1. SOLUTION 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.

Double Integrals in “Application” CONTINUED Now we carry out the integration with respect to y. 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.