Chapter 7 Functions of Several Variables GoldsteinSchneiderLayAsmar Calculus

Chapter Outline q Examples of Functions of Several Variables q Partial Derivatives q Maxima and Minima of Functions of Several Variables q Lagrange Multipliers and Constrained Optimization q The Method of Least Squares q Double Integrals Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 2

§ 7. 1 Examples of Functions of Several Variables Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications,

Section Outline q Functions of More Than One Variable q Cost of Material q Tax and Homeowner Exemption q Level Curves Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 4

Functions of More Than One Variable Definition Example Function of Several Variables: A function that has more than one independent variable Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 5

Functions of More Than One Variable EXAMPLE Let . Compute g(1, 1) and g(0, -1). SOLUTION Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 6

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 Cost (per sq ft) 3 5 5 5 Area (sq ft) xy yz yz xz Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 7

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 3 xy. Continuing in this way, we see that the total cost is Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 8

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 9

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 10

Level Curves Definition Level Curves: For a function f (x, y), a family of curves with equations f (x, y) = c where c is any constant Example An example immediately follows. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 11

Level Curves EXAMPLE Find a function f (x, y) that has the curve y = 2/x 2 as a level curve. SOLUTION 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 12

§ 7. 2 Partial Derivatives Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright ©

Section Outline q Partial Derivatives q Computing Partial Derivatives q Evaluating Partial Derivatives at a Point q Local Approximation of f (x, y) q Demand Equations q Second Partial Derivative Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 14

Definition: Partial Derivatives Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014,

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) = x 2 and g(x) = e 3 x. To compute , we only differentiate factors (or terms) that contain y and we interpret x to be a constant. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 16

Computing Partial Derivatives CONTINUED This is the given function. Differentiate ln y. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 17

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 18

Evaluating Partial Derivatives at a Point EXAMPLE Let Evaluate at (x, y, z) = (2, -1, 3). SOLUTION Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 19

Partial Derivative Rule Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014,

Local Approximation of f (x, y) We can generalize the interpretations of following general fact: to yield the 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 21

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 12 h. That is, Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 22

Demand Equations EXAMPLE 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 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). Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 23

Second Partial Derivative EXAMPLE Let . Find SOLUTION We first note that This means that to compute must take the partial derivative of , we with respect to x. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 24

§ 7. 3 Maxima and Minima of Functions of Several Variables Goldstein/Schneider/Lay/Asmar, Calculus and

Section Outline q Relative Maxima and Minima q First Derivative Test for Functions of Two Variables q Second Derivative Test for Functions of Two Variables q Finding Relative Maxima and Minima Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 26

Relative Maxima & Minima Definition 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 27

First-Derivative Test If one or both of the partial derivatives does not exist, then there is no relative maximum or relative minimum. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 28

Second-Derivative Test Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 30

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 31

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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 32

Finding Relative Maxima & Minima CONTINUED Since , we know, by the second-derivative test, that f (x, y) has a relative maximum at (1, 3). Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 33

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 34

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 35

Finding Relative Maxima & Minima CONTINUED We now determine the corresponding value of y by replacing x with 443 in the equation y = -0. 1 x + 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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 36

Finding Relative Maxima & Minima CONTINUED Since , we know, by the second-derivative test, that R(x, y) has a relative maximum at (443, 236). Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 37

First Derivative Test – 3 Variables Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright

§ 7. 4 Lagrange Multipliers and Constrained Optimization Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14

Section Outline q Background and Steps for Lagrange Multipliers q Using Lagrange Multipliers q Lagrange Multipliers in Application Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 40

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 41

Steps For Lagrange Multipliers L-1 L-2 L-3 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e

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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 43

Using Lagrange Multipliers CONTINUED From the first two equations we see that Therefore, Substituting this expression for x into the third equation, we derive Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 44

Using Lagrange Multipliers CONTINUED Using y = 3/5, we find that 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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 45

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’. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 46

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 47

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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 48

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 Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 49

§ 7. 5 The Method of Least Squares Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14

Section Outline q Least Squares Error q Least Squares Line (Regression Line) q Determining a Least Squares Line Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 51

Least Squares Error Definition 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 52

Least Squares Line (Regression Line) Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright ©

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 caraccident related deaths in 2012. (It is interesting that, while the number of drivers is obviously increasing with time, the number of car-accidentrelated deaths is actually decreasing, maybe because of improvements in car -manufacturing technologies and added safety measures. ) Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 54

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 55

Determining a Least Squares Line CONTINUED Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright

Determining a Least Squares Line CONTINUED Therefore, the equation of the least-squares line is y = -. 39 x + 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 car-related accidental deaths to be 39. 06 thousand in 2012. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 57

§ 7. 6 Double Integrals Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright ©

Section Outline q Double Integral of f (x, y) over a Region R q Evaluating Double Integrals q Double Integrals in “Application” Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 59

Double Integral of f (x, y) over a Region R Definition 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 60

The Double Integral Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014,

Evaluating Double Integrals EXAMPLE Calculate the iterated integral. SOLUTION Here g(x) = x and h(x) = 2 x. 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 62

Double Integrals in “Application” EXAMPLE 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. 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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 63

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. Goldstein/Schneider/Lay/Asmar, Calculus and Its Applications, 14 e Copyright © 2018, 2014, 2010 Pearson Education Inc. Slide 64