3 Optimizing Functions of Two Variables. Chapter 7 Section 3 Optimizing Functions of Two Variables 533

(b) Read about the principle of diminishing returns in an economics text. Then write a paragraph discussing the economic factors that might account for this phenomenon.

53. It is estimated that the weekly output at a certain plant is given by Q(x, y)  1,175x 483y  3.1x2y 1.2x3 2.7y2units, where x is the number of skilled workers and y is the number of unskilled workers employed at the plant. Currently the workforce consists of 37 skilled and 71 unskilled workers.

(a) Store the output function as

1,175X  483Y  3.1(X^2)*Y  1.2(X^3)  2.7(Y^2)

Store 37 as X and 71 as Y and evaluate to obtain Q(37, 71). Repeat for Q(38, 71) and Q(37, 72).

(b) Store the partial derivative Qx(x, y) in your calculator and evaluate Qx(37, 71).

Use the result to estimate the change in output resulting when the workforce is increased from 37 skilled workers to 38 and the unskilled workforce stays fixed at 71. Then compare with the actual change in output, given by the dif-ference Q(38, 71)  Q(37, 71).

(c) Use the partial derivative Qy(x, y) to estimate the change in output that results

when the number of unskilled workers is increased from 71 to 72 while the num-ber of skilled workers stays at 37. Compare with Q(37, 72)  Q(37, 71). 54. Repeat Problem 53 with the output function

Q(x, y)  1,731x  925y  x2y 2.7x2 1.3y3/2 and initial employment levels of x 43 and y  85.

Suppose a manufacturer produces two VCR models, the deluxe and the standard, and that the total cost of producing x units of the deluxe and y units of the standard is given by the function C(x, y). How would you find the level of production x a and y  b that results in minimal cost? Or perhaps the output of a certain production process is given by Q(K, L), where K and L measure capital and labor expenditure, respectively. What levels of expenditure K0and L0result in maximum output?

In Section 4 of Chapter 3, you learned how to use the derivative f(x) to find the largest and smallest values of a function of a single variable f(x), and the goal of this section is to extend those methods to functions of two variables f(x, y). We begin with a definition.

Optimizing

Functions of

Two Variables

3

In geometric terms, there is a relative maximum of f(x, y) at P(a, b) if the surface z f(x, y) has a “peak” at the point (a, b, f(a, b)); that is, if (a, b, f(a, b)) is at least as high as any nearby point on the surface. Similarly, a relative minimum of f(x, y) occurs at Q(c, d) if the point (c, d, f(c, d)) is at the bottom of a “valley,” so (c, d, f(c, d)) is at least as low as any nearby point on the surface. For example, in Figure 7.11, the func-tion f(x, y) has a relative maximum at P(a, b) and a relative minimum at Q(c, d).

FIGURE 7.11 Relative extrema of the function f(x, y).

The points (a, b) in the domain of f(x, y) for which both fx(a, b)  0 and fy(a, b) 

0 are said to be critical points of f. Like the critical points for functions of one vari-able, these critical points play an important role in the study of relative maxima and minima.

To see the connection between critical points and relative extrema, suppose f(x, y) has a relative maximum at (a, b). Then the curve formed by intersecting the surface z  f(x, y) with the vertical plane y  b has a relative maximum and hence a horizontal tangent when x  a (see Figure 7.12a). Since the partial derivative fx(a, b) is the slope of this tangent, it follows that fx(a, b)  0. Similarly, the curve

formed by intersecting the surface z f(x, y) with the plane x  a has a relative max-imum when y  b (see Figure 7.12b), and so fy(a, b)  0. This shows that a point

at which a function of two variables has a relative maximum must be a critical point. A similar argument shows that a point at which a function of two variables has a rel-ative minimum must also be a critical point.

C

P

Relative Extrema ■ The function f(x, y) is said to have a relative maximum at the point P(a, b) in the domain of f if f(a, b)  f(x, y) for all points sufficiently close to P.

Likewise, f(x, y) has a relative minimum at Q(c, d) if f(c, d)  f(x, y) for all points (x, y) sufficiently close to Q.

FIGURE 7.12 The partial derivatives are zero at a relative extremum. Here is a more precise statement of the situation.

Although all the relative extrema of a function must occur at critical points, not every critical point of a function is necessarily a relative extremum. Consider, for example, the function f(x, y)  y2  x2, whose graph, which resembles a saddle, is sketched in Figure 7.13. In this case, fx(0, 0)  0 because the surface has a relative

maximum (and hence a horizontal tangent) “in the x direction,” and fy(0, 0)  0

because the surface has a relative minimum (and hence a horizontal tangent) “in the y direction.” Hence (0, 0) is a critical point of f, but it is not a relative extremum. For a critical point to be a relative extremum, the nature of the extremum must be the same in all directions. A critical point that is neither a relative maximum nor a rela-tive minimum is called a saddle point.

Here is a procedure involving second-order partial derivatives that you can use to decide whether a given critical point is a relative maximum, a relative minimum, or a saddle point. This procedure is the two-variable version of the second derivative test for functions of a single variable that you saw in Chapter 3, Section 2.

T

S

P

T

Critical Points and Relative Extrema ■ A point (a, b) in the domain of f(x, y) for which the partial derivatives fxand fyboth exist is called

a critical point of f if both

fx(a, b)  0 and fy(a, b)  0

If the first-order partial derivatives of f exist at all points in some region R in the xy plane, then the relative extrema of f in R can occur only at critical points. y z x y z x (a, b) (a) (b) Horizontal tangent Horizontal tangent (a, b, f(a, b)) (a, b, f(a, b)) (a, b) x y z Saddle point

FIGURE 7.13 The surface z 

Notice that there is a saddle point at the critical point (a, b) only when the quan-tity D in the second partials test is negative. If D is positive, there is either a relative maximum or a relative minimum in all directions. To decide which, you can restrict your attention to any one direction (say, the x direction) and use the sign of the sec-ond partial derivative fxxin exactly the same way as the single variable second

deriv-ative was used in the second derivderiv-ative test given in Chapter 3; namely, a relative minimum if fxx(a, b)  0

a relative maximum if fxx(a, b) 0

You may find the following tabular summary a convenient way of remembering the conclusions of the second partials test:

The proof of the second partials test involves ideas beyond the scope of this text and is omitted. The following examples illustrate how the test can be used.

Find all critical points for the function f(x, y)  x2 y2and classify each as a rela-tive maximum, a relarela-tive minimum, or a saddle point.

Solution Since

fx 2x and fy 2y

Sign of D Sign of fxx Behavior at (a, b)

  Relative minimum   Relative maximum

 Saddle point

The Second Partials Test ■ Suppose that (a, b) is a critical point of the function f(x, y). Let

D fxx(a, b)fyy(a, b)  [ fxy(a, b)]2

If D 0, then f has a saddle point at (a, b).

If D 0 and fxx(a, b) 0, then f has a relative maximum at (a, b).

If D 0 and fxx(a, b)  0, then f has a relative minimum at (a, b).

If D 0, the test is inconclusive and f may have either a relative extremum or a saddle point at (a, b).

the only critical point of f is (0, 0). To test this point, use the second-order partial derivatives fxx  2 fyy 2 and fxy  0 to get D(x, y)  fxxfyy ( fxy) 2  2(2)  0  4 That is, D(x, y)  4 for all points (x, y) and, in particular,

D(0, 0)  4  0

Hence, f has a relative extremum at (0, 0). Moreover, since fxx(0, 0)  2  0

it follows that the relative extremum at (0, 0) is a relative minimum. For reference, the graph of f is sketched in Figure 7.14.

FIGURE 7.14 The surface z x2_{ y}2_{with a relative minimum at (0, 0).}

Find all critical points for the function f(x, y)  y2 x2and classify each as a rela-tive maximum, a relarela-tive minimum, or a saddle point.

Solution Since fx 2x and fy 2y y z x Relative minimum

EXAMPLE 3.2

EXAMPLE 3.2

E x p l o r e !

E x p l o r e !

Refer to Example 3.2. Store

f(x, y)  y2 x2as Y1  L12 X2_{, where L1  {2, 1, 0,}

0.8, 1.5}. Graph these curves using the decimal window [4.7, 4.7]1 by [3.1, 3.1]1, paying close attention to the or-der in which the shapes appear. Describe what you observe.

the only critical point of f is (0, 0). To test this point, compute the second-order par-tial derivatives fxx 2 fyy  2 and fxy 0 to get D(x, y)  fxxfyy ( fxy) 2  2(2)  0  4 That is, D(x, y)  4 for all points (x, y) and, in particular,

D(0, 0)  4 0

It follows that f must have a saddle point at (0, 0). The graph of f is shown in Fig-ure 7.15.

Solving the equations fx 0 and fy 0 simultaneously to find the critical points

of a function of two variables is rarely as simple as in Examples 3.1 and 3.2. The algebra in the next example is more typical. Before proceeding, you may wish to refer to the Algebra Review at the back of the book, in which techniques for solving sys-tems of two equations in two unknowns are discussed.

Find all critical points for the function f(x, y)  x3 y3 6xy and classify each as a relative maximum, a relative minimum, or a saddle point.

Solution Since fx 3x 2  6y and fy 3y 2  6x

you find the critical points of f by solving simultaneously the two equations 3x2 6y  0 and 3y2 6x  0

From the first equation, you get y   which you can substitute into the second equation to find

  6x  0 or x(x3 8)  0

The solutions of this equation are x  0 and x  2. These are the x coordinates of the critical points of f. To get the corresponding y coordinates, substitute these values of x into the equation y   (or into either one of the two original equations). You will find that y 0 when x  0 and y  2 when x  2. It follows that the critical points of f are (0, 0) and (2, 2).

x2 2 3x4 4 x2 2

EXAMPLE 3.3

EXAMPLE 3.3

FIGURE 7.15 The surface z 

y2 _{ x}2_{with a saddle point at} (0, 0).

The second-order partial derivatives of f are fxx  6x fyy 6y and fxy 6 Hence, D(x, y)  fxxfyy ( fxy) 2  36xy  36  36(xy  1) Since D(0, 0)  36[0(0)  1]  36 0

it follows that f has a saddle point at (0, 0). Since

D(2, 2)  36[2(2)  1]  108  0

and fxx(2, 2)  6(2)  12  0

you see that f has a relative minimum at (2, 2). These results are summarized in the following table:

In the next example, you will see how to apply the theory of relative extrema to solve an optimization problem from economics. Actually, you will be trying to find the absolute maximum of a certain function. It turns out, however, that the absolute and relative maxima of this function coincide. In fact, in the majority of two-variable opti-mization problems in the social sciences, the relative extrema and absolute extrema coincide. For this reason, the theory of absolute extrema for functions of two vari-ables will not be developed in this text, and you may assume that the relative extremum you find as the solution to a practical optimization problem is actually the absolute extremum.

The only grocery store in a small rural community carries two brands of frozen apple juice, a local brand that it obtains at the cost of 30 cents per can and a well-known national brand that it obtains at the cost of 40 cents per can. The grocer estimates that if the local brand is sold for x cents per can and the national brand for y cents per can, approximately 70  5x  4y cans of the local brand and 80  6x  7y cans of the national brand will be sold each day. How should the grocer price each brand to maximize the profit from the sale of the juice?

P

O

P

Critical Point (a, b) D(a, b) fxx(a, b) Behavior at (a, b)

(0, 0)  Saddle point

(2, 2)   Relative minimum

Solution Since

 

it follows that the total daily profit from the sale of the juice is given by the function f(x, y)  (x  30)(70  5x  4y)  (y  40)(80  6x  7y)

 5x2_{ 10xy  20x  7y}2_{ 240y  5,300}

Compute the partial derivatives

fx 10x  10y  20 and fy  10x  14y  240

and set them equal to zero to get

10x  10y  20  0 and 10x 14y  240  0

or x  y  2 and 5x 7y  120

Then solve these equations simultaneously to get

x 53 and y 55

It follows that (53, 55) is the only critical point of f. Next apply the second partials test. Since

fxx 10 fyy 14 and fxy 10

you get D(x, y)  fxxfyy ( fxy)2 10(14) (10)2 40

and since D(53, 55)  40  0 and fxx(53, 55)  10 0

it follows that f has a (relative) maximum when x 53 and y  55. That is, the gro-cer can maximize profit by selling the local brand of juice for 53 cents per can and the national brand for 55 cents per can.

A planner for Acme Corporation plots a grid on a map and determines that Acme’s three most important customers are located at A(1, 5), B(0, 0), and C(8, 0), where units are in miles. At what point W(x, y) should a warehouse be located in order to minimize the sum of the distances from P to A, B, and C (see Figure 7.16).







of the local brand







EXAMPLE 3.5

EXAMPLE 3.5

FIGURE 7.16 Locations of busi-nesses A, B, and C and ware-house W.

Solution

The point W(x, y) where the sum of the distances is minimized is the same point that minimizes the sum of the squares of the distances; namely,

S(x, y)  [(x  1)2 (y  5)2]  (x2 y2)  [(x  8)2 y2]

W to A W to B W to C

(By working with the squares of distances, we eliminate the square roots and make the calculations easier.) To minimize S, begin by computing the partial derivatives

Sx 2(x  1)  2x  2(x  8)  6x  18

Sy 2(y  5)  2y  2y  6y  10

Then Sx 0 and Sy 0 when

6x 18  0 6y 10  0

or x  3 and y  . Since Sxx 6, Sxy 0, and Syy 6, you get

D SxxSyy Sxy 2  (6)(6)  02  36  0 and Sxx  6  0

Thus, the sum of squares is minimized at the map point W .

Suppose in the process of analyzing a particular phenomenon we gather the set of data plotted in Figure 7.17a. The data points seem to lie roughly in a straight line, but what line? In other words, given a collection of points (x1, y1), (x2, y2), . . . ,

(xn, yn), what line y mx  b “best fits” the data?

T

M

L

S









aedddgbeddgdc

afdbfddc

agfdddefbgddddffc

FIGURE 7.17 Least-squares approximation of a set of data: (a) a collection of data points and (b) the least-squares criterion.

One of the most frequently used procedures for determining the best-fitting line is to compute the sum of squares S of the vertical distances from the data points to the line y mx  b (Figure 7.17b). The sum S will be a function of the two vari-ables m (slope) and b (y intercept), and we obtain the best-fitting line by using the optimization procedures of this section to minimize the function S(m, b). Here is an example.

Use the least-squares criterion to find the equation of the line that is closest to the three points (1, 1), (2, 3), and (4, 3).

Solution

As indicated in Figure 7.18, the sum of the squares of the vertical distances from the three given points to the line y mx  b is

d12 d22 d32 (m  b  1)2 (2m  b  3)2 (4m  b  3)2

This sum depends on the coefficients m and b that define the line, so the sum can be thought of as a function S(m, b) of the two variables m and b. The goal, therefore, is to find the values of m and b that minimize the function

S(m, b)  (m  b  1)2 (2m  b  3)2 (4m  b  3)2

You do this by setting the partial derivatives and equal to zero to get

 2(m  b  1)  4(2m  b  3)  8(4m  b  3)  42m  14b  38  0 ∂S ∂m ∂S ∂b ∂S ∂m x y (x1, y1) d1 _d 2 d3 x1 x2 x3 (x2, y2) (x3, y3)

(b) The least-squares criterion

y  mx  b

x y

(a) A collection of data points

EXAMPLE 3.6

EXAMPLE 3.6

E x p l o r e !

E x p l o r e !

Refer to Example 3.6. Store the points (1, 1), (2, 3), and (4, 3) into the data list feature of your graphing calculator, with x co-ordinates in L1 and y in L2. Find the equation of the least-squares regression line and graph it along with the data points. Now place the line y 0.6x 1 in Y2 of the equation editor. Which line appears to fit the data better and why?

FIGURE 7.18 Minimize the sum d2

1 d22 d23.

and  2(m  b  1)  2(2m  b  3)  2(4m  b  3)  14m  6b  14  0

and solving the resulting simplified equations 21m 7b  19

7m 3b  7 simultaneously for m and b to conclude that

m and b 1 Since Smm  42 Smb 14 Sbb  6 we have D SmmSbb Smb 2  (42)(6)  (14)2  56

So D  0 and Smm  0, and the second partials test tells us that the critical point

corresponds to a relative minimum. Thus, the line that best fits the given three points has the equation y  x 1.

The procedure illustrated in Example 3.6 can be generalized to find the best-fit-ting line y mx  b for an arbitrary set of data points (x1, y1), (x2, y2), . . . , (xn, yn).

4 7





Specifically, you would minimize the sum of squares function S(m, b)  (mx1 b  y1)

2

 . . .  (mxn b  yn)

2

It can be shown that the minimum occurs when

and

where, for simplicity, we have dropped the indices in the sums. For instance,

The derivation of these formulas involves several complicated algebraic steps and is omitted. For practice, apply the formulas to the data in Example 3.6 to assure your-self that they do indeed yield the result you found directly. We close with an applied example that illustrates how data can be efficiently organized for substitution into the formulas to obtain a least-squares line.

A college admissions officer has compiled the following data relating students’ high-school and college grade-point averages:

High-school GPA 2.0 2.5 3.0 3.0 3.5 3.5 4.0 4.0

College GPA 1.5 2.0 2.5 3.5 2.5 3.0 3.0 3.5

Find the equation of the least-squares line for these data and use it to predict the col-lege GPA of a student whose high-school GPA is 3.75.

Solution

Let x denote the high-school GPA and y the college GPA and arrange the calculations as follows: x2_



EXAMPLE 3.7

EXAMPLE 3.7

Use the least-squares formula with n 8 to get

and

The equation of the least-squares line is therefore y 0.78x  0.19

To predict the college GPA of a student whose high-school GPA is 3.75, substi-tute x  3.75 into the equation of the least-squares line. This gives

y 0.78(3.75)  0.19  3.12

which suggests that the student’s college GPA might be about 3.1.

FIGURE 7.19 The least-squares line for high-school and college GPAs.

1 2 3 4 Least-squares line: y = 0.78x + 0.19 y (college GPA) x (high-school GPA) 1 2 3 4 b 84.75(21.5) 25.5(71.25) 8(84.75) (25.5)2  0.19 m 8(71.25) 25.5(21.5) 8(84.75) (25.5)2  0.78 x y xy x2 3.0 3.5 10.5 9.0 3.5 2.5 8.75 12.25 3.5 3.0 10.5 12.25 4.0 3.0 12.0 16.0 4.0 3.5 14.0 16.0 x  25.5 y  21.5 xy  71.25 x2  84.75

A graph of the original data and of the corresponding least-squares line is shown in Figure 7.19. Actually, in practice, it is a good idea to plot the data before pro-ceeding with the calculations. By looking at the graph you will usually be able to tell whether approximation by a straight line is appropriate or whether a curve of some other shape should be used instead.

In Problems 1 through 20, find the critical points of the given function and classify each as a relative maximum, a relative minimum, or a saddle point.

1. f(x, y)  5  x2 y2 2. f(x, y)  2x2 3y2 3. f(x, y)  xy 4. f(x, y)  x2 2y2 xy  14y 5. f(x, y)  6. f(x, y)  xy   7. f(x, y)  2x3 y3 3x2 3y  12x  4 8. f(x, y)  (x  1)2 y3 3y2 9y  5 9. f(x, y)  x3 y2 6xy  9x  5y  2 10. f(x, y)  x4 32x  y3 12y  7 11. f(x, y)  (x2 2y2) 12. f(x, y)  (x  4) ln (xy) 13. f(x, y)  x3 4xy  y3 14. f(x, y)  15. f(x, y)  16. f(x, y)  2x4 x2 2xy  3x  y2 2y  5 17. f(x, y)  18. f(x, y)  xye  19. f(x, y)  x ln  3x  xy2 20. f(x, y)  4xy  2x4 y2 4x  2y 21. A T-shirt shop carries two competing shirts, one endorsed by Michael Jordan and the

other by Shaq O’Neal. The owner of the store can obtain both types at a cost of $2 per shirt and estimates that if Jordan shirts are sold for x dollars apiece and O’Neal shirts for y dollars apiece, consumers will buy approximately 40  50x  40y Jordan shirts and 20  60x  70y O’Neal shirts each day. How should the owner price the shirts in order to generate the largest possible profit?

RETAIL SALES









P . R . O . B . L . E . M . S

7.3

P . R . O . B . L . E . M . S

7.3

22. The telephone company is planning to introduce two new types of executive com-munications systems that it hopes to sell to its largest commercial customers. It is estimated that if the first type of system is priced at x hundred dollars per system and the second type at y hundred dollars per system, approximately 40  8x  5y consumers will buy the first type and 50  9x  7y will buy the second type. If the cost of manufacturing the first type is $1,000 per system and the cost of manu-facturing the second type is $3,000 per system, how should the telephone company price the systems to generate the largest possible profit?

23. Suppose you wish to construct a rectangular box with a volume of 32 ft3. Three different materials will be used in the construction. The material for the sides costs $1 per square foot, the material for the bottom costs $3 per square foot, and the material for the top costs $5 per square foot. What are the dimensions of the least expensive such box?

24. A farmer wishes to fence off a rectangular pasture along the bank of a river. The area of the pasture is to be 6,400 yd2, and no fencing is needed along the river bank. Find the dimensions of the pasture that will require the least amount of fencing.

25. A dairy produces whole milk and skim milk in quantities x and y gallons, respec-tively. Suppose that the price of whole milk is p(x)  100  x and that of skim milk is q(y)  100  y and assume that C(x, y)  x2 xy  y2is the joint-cost function of the commodities. What should x and y be to maximize profit?

26. Repeat Problem 25 for the case where p(x)  20  5x, q(y)  4  2y, and C(x, y)  2xy  4.

27. Consider an experiment in which a subject performs a task while being exposed to two different stimuli (for example, sound and light). For low levels of the stimuli, the subject’s performance might actually improve, but as the stimuli increase, they eventually become a distraction and the performance begins to deteriorate. Suppose in a certain experiment in which x units of stimulus A and y units of stimulus B are applied, the performance of a subject is measured by the function

f(x, y)  C  xy

where C is a positive constant. How many units of each stimulus result in maxi-mum performance?

28. The social desirability of an enterprise often involves making a choice between the commercial advantage of the enterprise and the social or ecological loss that may result. For instance, the lumber industry provides paper products to society and income to many workers and entrepreneurs, but the gain may be offset by the destruction of habitable territory for spotted owls and other endangered species. Suppose the social desirability of a particular enterprise is measured by the function

D(x, y)  (16  6x)x  (y2 4xy  40) x  0, y  0

where x measures commercial advantage (profit and jobs) and y measures

SOCIAL CHOICES e1x2y2 RESPONSE TO STIMULI RETAIL SALES RETAIL SALES CONSTRUCTION CONSTRUCTION PRICING

ecological disadvantage (species displacement, as a percentage). The enterprise is deemed desirable if D 0 and undesirable if D 0.

(a) What values of x and y will maximize social desirability? Interpret your result. Is it possible for this enterprise to be desirable?

(b) The function given in part (a) is artificial, but the ideas are not. Research the topic of ethics in industry and write a paragraph on how you feel these choices should be made.*

29. A particle of mass m in a rectangular box with dimensions x, y, z has ground state energy

where k is a physical constant. If the volume of the box satisfies xyz  V0 for

constant V0, find the values of x, y, and z that minimize the ground state energy.

30. A manufacturer is planning to sell a new product at the price of $150 per unit and estimates that if x thousand dollars is spent on development and y thousand dollars is spent on promotion, consumers will buy approximately units of the product. If manufacturing costs for this product are $50 per unit, how much should the manufacturer spend on development and how much on promotion to generate the largest possible profit from the sale of this product? [Hint: Profit  (number of units)(price per unit  cost per unit)  total amount spent on devel-opment and promotion.]

31. A manufacturer with exclusive rights to a sophisticated new industrial machine is planning to sell a limited number of the machines to both foreign and domestic firms. The price the manufacturer can expect to receive for the machines will depend on the number of machines made available. (For example, if only a few of the machines are placed on the market, competitive bidding among prospective purchasers will tend to drive the price up.) It is estimated that if the manufacturer supplies x machines to the domestic market and y machines to the foreign market, the machines will sell for thousand dollars apiece at home and for thousand dollars apiece abroad. If the manufacturer can produce the machines at the cost of $10,000 apiece, how many should be supplied to each market to generate the largest possible profit?

32. A manufacturer with exclusive rights to a new industrial machine is planning to sell a limited number of them and estimates that if x machines are supplied to the domestic market and y to the foreign market, the machines will sell for 150 x 6

PROFIT UNDER MONOPOLY

50 y 10 x 20 60x 5 y 20

PROFIT UNDER MONOPOLY

320y y 2 160x x 4 ALLOCATION OF FUNDS E(x, y, z) k 2 8m





* A good place to start is the article by K. R. Stollery, “Environmental Controls in Extractive Industries,” Land Economics, Vol. 61, 1985, page 169.

thousand dollars apiece at home and for thousand dollars apiece abroad. (a) How many machines should the manufacturer supply to the domestic market to

generate the largest possible profit at home?

(b) How many machines should the manufacturer supply to the foreign market to generate the largest possible profit abroad?

(c) How many machines should the manufacturer supply to each market to gen-erate the largest possible total profit?

(d) Is the relationship between the answers in parts (a), (b), and (c) accidental? Explain. Does a similar relationship hold in Problem 31? What accounts for the difference between these two problems in this respect?

33. Four small towns in a rural area wish to pool their resources to build a television station. If the towns are located at the points (5, 0), (1, 7), (9, 0), and (0, 8) on a rectangular map grid, where units are in miles, at what point S(a, b) should the station be located to minimize the sum of the distances from the towns?

34. In relation to a rectangular map grid, four oil rigs are located at the points (300, 0), (100, 500), (0, 0), and (400, 300) where units are in feet. Where should a maintenance shed M(a, b) be located to minimize the sum of the distances from the rigs?

In Problems 35 through 38 plot the given points and use the method of Example 3.6 to find the corresponding least-squares line.

35. (0, 1), (2, 3), (4, 2) 36. (1, 1), (2, 2), (6, 0) 37. (1, 2), (2, 4), (4, 4), (5, 2) 38. (1, 5), (2, 4), (3, 2), (6, 0)

In Problems 39 through 42, find the indicated least-squares line y  mx  b. You may use the formulas given just before Example 3.7.

39. A company’s annual sales (in units of 1 billion dollars) for its first 5 years of operation are shown in the following table:

Year 1 2 3 4 5

Sales 0.9 1.5 1.9 2.4 3.0

(a) Plot these data on a graph.

(b) Find the equation of the least-squares line.

(c) Use the least-squares line to predict the company’s sixth-year sales.

40. The following table gives the percentage of high-school seniors in four different years who had tried cocaine at least once in their lives.*

DRUG ABUSE SALES MAINTENANCE CITY PLANNING 100 y 20

* L. Hoffman, S. Paris, and E. Hall, Developmental Psychology Today, McGraw-Hill, Inc., New York, 1994, page 405.

Year 1975 1980 1985 1990

Percentage

Using Cocaine 9.0 15.7 17.3 9.4

(a) Plot these data on a graph.

(b) Find the equation of the least-squares line.

(c) Use the least-squares line to predict the percentage of high school seniors in the year 2000 who used cocaine at least once.

41. On election day, the polls in a certain state open at 8:00 A.M. Every 2 hours after that, an election official determines what percentage of the registered voters have already cast their ballots. The data through 6:00 P.M. are shown below:

Time 10:00 12:00 2:00 4:00 6:00

Percentage

Turnout 12 19 24 30 37

(a) Plot these data on a graph.

(b) Find the equation of the least-squares line. (Let x denote the number of hours after 8:00 A.M.)

(c) Use the least-squares line to predict what percentage of the registered voters will have cast their ballots by the time the polls close at 8:00 P.M.

42. Recall that in Problem 40 of Section 1, Chapter 4, we gave the following table for the number of reported cases of AIDS for the period 1984–1991:

Year 1984 1985 1986 1987 1988 1989 1990 1991

Cases 4,445 8,249 12,932 12,070 31,001 33,722 41,595 43,672

(a) Plot these data on a graph with time t (years after 1984) on the x axis. Find the equation for the least-squares line for the given data.

(b) Use the least-squares line to predict the number of cases of AIDS reported in the year 2000.

43. Let f(x, y)  x2 y2 4xy. Show that f does not have a relative minimum at its critical point (0, 0), even though it does have a relative minimum at (0, 0) in both the x and y directions. [Hint: Consider the direction defined by the line y  x. That is, substitute x for y in the formula for f and analyze the resulting function of x.]

In Problems 44 through 47 find the partial derivatives fx and fy, and then use your

graphing utility to determine the critical points of each function. 44. f(x, y)  (x2 3y  5)ex22y2

SPREAD OF AIDS VOTER TURNOUT

45. f(x, y) 

46. f(x, y)  6x2 12xy  y4 x  16y  3 47. f(x, y)  2x4 y4 x2(11y  18)

48. Sometimes you can classify the critical points of a function by inspecting its level curves. In each case shown in the figure, determine the nature of the critical point of f at (0, 0).

In many applied problems, a function of two variables is to be optimized subject to a restriction or constraint on the variables. For example, an editor, constrained to stay within a fixed budget of $60,000, may wish to decide how to divide this money between development and promotion in order to maximize the future sales of a new book. If x denotes the amount of money allocated to development, y the amount allo-cated to promotion, and f(x, y) the corresponding number of books that will be sold, the editor would like to maximize the sales function f(x, y) subject to the budgetary constraint that x y  60,000.

For a geometric interpretation of the process of optimizing a function of two vari-ables subject to a constraint, think of the function itself as a surface in three-dimen-sional space and of the constraint (which is an equation involving x and y) as a curve in the xy plane. When you find the maximum or minimum of the function subject to the given constraint, you are restricting your attention to the portion of the surface that lies directly above the constraint curve. The highest point on this portion of the surface is the constrained maximum, and the lowest point is the constrained mini-mum. The situation is illustrated in Figure 7.20.

y y x x (a) (b) f = 1 f = 2 f = 3 f = 1 f = –3 f = –2 f = –1 f = 3 f = 2 f = 1 f = –3 f = –2 f = –1 f =1 3 f =1 2 LEVEL CURVES x2 xy  7y2 x ln y

Constrained

Optimization:

The Method

of Lagrange

Multipliers

4