Chapter 7 Functions of Several Variables

1. Chapter 7Functions of Several Variables

2. 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

3. §7.1 Examples of Functions of Several Variables

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

5. Functions of More Than One Variable

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

7. 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 Cost (per sq ft) Area (sq ft)

8. 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

9. 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.

10. 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.

11. Level Curves

12. 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.

13. §7.2 Partial Derivatives

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

15. Partial Derivatives

16. 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.

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

18. 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.

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

20. Local Approximation of f(x, y)

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 12h. That is,

22. 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).

23. 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.

24. §7.3 Maxima and Minima of Functions of Several Variables

25. 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

26. Relative Maxima & Minima

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

28. Second-Derivative Test

29. 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.

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.

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

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).

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.

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.

35. 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

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).

37. §7.4 Lagrange Multipliers and Constrained Optimization

38. Section Outline • Background and Steps for Lagrange Multipliers • Using Lagrange Multipliers • Lagrange Multipliers in Application

39. 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.

40. Steps For Lagrange Multipliers L-1 L-2 L-3

41. 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

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

43. 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

44. 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’.

45. 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.

46. 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

47. 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

48. §7.5 The Method of Least Squares

49. Section Outline • Least Squares Error • Least Squares Line (Regression Line) • Determining a Least Squares Line

50. Least Squares Error