4 Constrained Optimization: The Method of Lagrange Multipliers. Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 551

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1 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 551 LEVEL CURVES f(, ) ln 46. f(, ) f(, ) (11 18) 48. Sometimes ou can classif the critical points of a function b inspecting its level curves. In each case shown in the figure, determine the nature of the critical point of f at (0, 0). f = 1 f = 2 f = 3 f = 3 f = 2 f = 1 2 f = 1 f = 1 f = 1 3 f = 3 f = 2 f = 3 f = 2 (a) (b) f = 1 f = 1 4 Constrained Optimization: The Method of Lagrange Multipliers In man applied problems, a function of two variables is to be optimized subject to a restriction or constraint on the variables. For eample, an editor, constrained to sta within a fied budget of $60,000, ma wish to decide how to divide this mone between development and promotion in order to maimize the future sales of a new book. If denotes the amount of mone allocated to development, the amount allocated to promotion, and f(, ) the corresponding number of books that will be sold, the editor would like to maimize the sales function f(, ) subject to the budgetar constraint that 60,000. For a geometric interpretation of the process of optimizing a function of two variables subject to a constraint, think of the function itself as a surface in three-dimensional space and of the constraint (which is an equation involving and ) as a curve in the plane. When ou find the maimum or minimum of the function subject to the given constraint, ou are restricting our attention to the portion of the surface that lies directl above the constraint curve. The highest point on this portion of the surface is the constrained maimum, and the lowest point is the constrained minimum. The situation is illustrated in Figure 7.20.

2 552 Chapter 7 Calculus of Several Variables z Unconstrained maimum Constrained maimum Constraint curve FIGURE 7.20 Constrained and unconstrained etrema. You have alread seen some constrained optimization problems in Chapter 3. (For instance, recall Eample 5.1 of Section 3.5.) The technique ou used in Chapter 3 to solve such a problem involved reducing it to a problem of a single variable b solving the constraint equation for one of the variables and then substituting the resulting epression into the function to be optimized. The success of this technique depended on solving the constraint equation for one of the variables, which is often difficult or even impossible to do in practice. In this section, ou will see a more versatile technique called the method of Lagrange multipliers, in which the introduction of a third variable (the multiplier) enables ou to solve constrained optimization problems without first solving the constraint equation for one of the variables. More specificall, the method of Lagrange multipliers uses the fact that an relative etremum of the function f(, ) subject to the constraint g(, ) k must occur at a critical point (a, b) of the function F(, ) f(, ) [g(, ) k] where is a new variable (the Lagrange multiplier). To find the critical points of F, compute its partial derivatives F f g F f g F (g k) and solve the equations F 0, F 0, and F 0 simultaneousl, as follows: F f g 0 or f g F f g 0 or f g F (g k) 0 or g k Finall, evaluate f(a, b) at each critical point (a, b) of F.

3 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 553 There is a version of the second partials test that can be used to determine what kind of constrained relative etremum corresponds to each critical point (a, b) of F. The techniques required to carr out such an analsis are discussed in more advanced tets, but in this tet, we will assume that if f has a constrained maimum (minimum), it will be given b the largest (smallest) of the critical values f(a, b). Here is a summar of the procedure used in the method of Lagrange multipliers. A Procedure for Appling the Method of Lagrange Multipliers Step 1. Write the problem in the form: Maimize (minimize) f(, ) subject to g(, ) k Step 2. Simultaneousl solve the equations f (, ) g (, ) f (, ) g (, ) g(, ) k Step 3. Evaluate f at all points found in step 2. If the required maimum (minimum) eists, it will be the largest (smallest) of these values. A geometric justification of the multiplier method is given at the end of this section. In the following eample, the method is used to solve the problem from Eample 5.1 in Section 5 of Chapter 3. EXAMPLE 4.1 The highwa department is planning to build a picnic area for motorists along a major highwa. It is to be rectangular with an area of 5,000 square ards and is to be fenced off on the three sides not adjacent to the highwa. What is the least amount of fencing that will be needed to complete the job? Picnic area Solution Label the sides of the picnic area as indicated in Figure 7.21 and let f denote the amount of fencing required. Then, Highwa FIGURE 7.21 Rectangular picnic area. f(, ) 2 The goal is to minimize f subject to the constraint that the area must be 5,000 square ards; that is, subject to the constraint 5,000

4 554 Chapter 7 Calculus of Several Variables Let g(, ) and use the partial derivatives f 1 f 2 g and g to obtain the three Lagrange equations 1 2 and 5,000 From the first and second equations ou get 1 and 2 (since 0 and 0), which implies that 1 or 2 2 Now substitute 2 into the third Lagrange equation to get 2 2 5,000 or 50 and then use 50 in the equation 2 to get 100. It follows that 100 and 50 are the values that minimize the function f(, ) 2 subject to the constraint that 5,000. That is, the optimal picnic area is 100 ards wide (along the highwa), etends 50 ards back from the road, and requires ards of fencing. EXAMPLE 4.2 Find the maimum and minimum values of the function f(, ) subject to the constraint Solution Let g(, ) 2 2 and use the partial derivatives f f g 2 and g 2 to get the three Lagrange equations 2 2 and Neither nor can be zero if all three of these equations are to hold (do ou see wh?), and so ou can rewrite the first two equations as 2 and 2 which implies that or 2 2

5 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 555 Now substitute 2 2 into the third equation to get or 2 If 2, it follows from the equation 2 2 that 2 or 2. Similarl, if 2, it follows that 2 or 2. Hence, the four points at which the constrained etrema can occur are (2, 2), (2, 2), ( 2, 2), and ( 2, 2). Since f(2, 2) f( 2, 2) 4 and f(2, 2) f( 2, 2) 4 it follows that when 2 2 8, the maimum value of f(, ) is 4, which occurs at the points (2, 2) and ( 2, 2), and the minimum value is 4, which occurs at (2, 2) and ( 2, 2). For practice, check these answers b solving the optimization problem using the methods of Chapter 3. Note In each of the preceding eamples, the first two Lagrange equations were used to eliminate the new variable, and then the resulting epression relating and was substituted into the constraint equation. For most constrained optimization problems ou encounter, this particular sequence of steps will often lead quickl to the desired solution. LAGRANGE MULTIPLIERS FOR FUNCTIONS OF THREE VARIABLES The method of Lagrange multipliers can be etended to constrained optimization problems involving functions of more than two variables and more than one constraint. For instance, to optimize f(,, z) subject to the constraint g(,, z) k, ou solve f g f g f z g z and g k Here is an eample of a problem involving this kind of constrained optimization. EXAMPLE 4.3 A jewel bo is to be constructed of material that costs $1 per square inch for the bottom, $2 per square inch for the sides, and $5 per square inch for the top. If the total volume is to be 96 in. 3 what dimensions will minimize the total cost of construction? z Solution Let the bo be inches deep, inches long, and z inches wide, as indicated in the accompaning figure. Then the volume of the bo is V z and the total cost of construction is given b C 1z 2(2 2z) 5z 6z 4 4z abc Bottom adddegddbgdddddec Sides abc Top

6 556 Chapter 7 Calculus of Several Variables You wish to minimize C 6z 4 4z subject to V z 96. The Lagrange equations are C V or 4 4z (z) C V or 6z 4 (z) C z V z or 6 4 () and z 96. Solving each of the first three equations for, ou get 4 4z z B multipling across each equation, ou obtain 4z 4z 2 6z 2 4z 4 2 4z 6 2 z 4z 6z 4 2 6z 4 2 z which can be further simplified b first subtracting the common z terms from both sides of each equation 4z 2 6z z z and then dividing z 2 from both sides of the first equation, 2 from the second, and 2 from the third to get z so that z 3 4 4z Finall, b substituting these epressions into the constraint equation z 96, ou find that z 4 z then z (6) 4 3 Thus, the minimal cost occurs when the jewel bo is 6 inches deep with a square base, 4 inches on a side. 6 4 so 6

7 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 557 APPLICATIONS TO ECONOMICS MAXIMIZATION OF UTILITY In the net two eamples, the method of Lagrange multipliers is used to solve constrained optimization problems from economics. A utilit function U(, ) measures the total satisfaction or utilit a consumer receives from having units of one particular commodit and units of another. The problem is to determine how man units of each commodit the consumer should bu to maimize utilit while staing within a fied budget. The application of the method of Lagrange multipliers to the utilit problem is illustrated in the net eample. EXAMPLE 4.4 A consumer has $600 to spend on two commodities, the first of which costs $20 per unit and the second $30 per unit. Suppose that the utilit derived b the consumer from units of the first commodit and units of the second commodit is given b the Cobb-Douglas utilit function U(, ) How man units of each commodit should the consumer bu to maimize utilit? Solution The total cost of buing units of the first commodit at $20 per unit and units of the second at $30 per unit is Since the consumer has onl $600 to spend, the goal is to maimize utilit U(, ) subject to the budgetar constraint that The three Lagrange equations are and From the first two equations ou get or 9 Substituting this into the third equation, ou get from which it follows that 18 and 4 (18) 8 9 That is, to maimize utilit, the consumer should bu 18 units of the first commodit and 8 units of the second.

8 558 Chapter 7 Calculus of Several Variables Optimal indifference curve: U(, ) = C 20 (18, 8) Budget line: = FIGURE 7.22 Budgetar constraint and optimal indifference curve. Recall from Section 1 that the level curves of a utilit function are known as indifference curves. A graph showing the relationship between the optimal indifference curve U(, ) C, where C U(18, 8) and the budgetar constraint , is sketched in Figure ALLOCATION OF RESOURCES An important class of problems in business and economics involves determining an optimal allocation of resources subject to a constraint on those resources. Here is an eample in which sales are maimized subject to a budgetar constraint. EXAMPLE 4.5 An editor has been allotted $60,000 to spend on the development and promotion of a new book. It is estimated that if thousand dollars is spent on development and thousand on promotion, approimatel f(, ) 20 3/2 copies of the book will be sold. How much mone should the editor allocate to development and how much to promotion in order to maimize sales? Solution The goal is to maimize the function f(, ) 20 3/2 subject to the constraint g(, ) 60, where g(, ). The corresponding Lagrange equations are 30 1/2 20 3/2 and 60

9 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 559 From the first two equations ou get 30 1/2 20 3/2 Since the maimum value of f clearl does not occur when 0, ou ma assume that 0 and divide both sides of this equation b 30 1/2 to get 2 3 Substituting this epression into the third equation, ou get 2 5 or from which it follows that 2 36 and (36) 24 3 That is, to maimize sales, the editor should spend $36,000 on development and $24,000 on promotion. If this is done, approimatel f(36, 24) 103,680 copies of the book will be sold. A graph showing the relationship between the budgetar constraint and the level curve for optimal sales is shown in Figure Optimal sales level: f(, ) = 103, (36, 24) Budget line: + = FIGURE 7.23 Budgetar constraint and optimal sales level.

10 560 Chapter 7 Calculus of Several Variables THE SIGNIFICANCE OF THE LAGRANGE MULTIPLIER You can solve most constrained optimization problems b the method of Lagrange multipliers without actuall obtaining a numerical value for the Lagrange multiplier. In some problems, however, ou ma want to compute. This is because has the following useful interpretation. The Lagrange Multiplier Suppose M is the maimum (or minimum) value of f(, ), subject to the constraint g(, ) k. The Lagrange multiplier is the rate of change of M with respect to k. That is, Hence, dm dk change in M resulting from a 1-unit increase in k EXAMPLE 4.6 Suppose the editor in Eample 4.5 is allotted $61,000 instead of $60,000 to spend on the development and promotion of the new book. Estimate how the additional $1,000 will affect the maimum sales level. Solution In Eample 4.5, ou solved the three Lagrange equations 30 1/2 20 3/2 and 60 to find that the maimum value M of f(, ) subject to the constraint 60 occurred when 36 and 24. To find, substitute these values of and into the first or second Lagrange equation. Using the second equation, ou get 20(36) 3/2 4,320 which means that maimal sales will increase b approimatel 4,320 copies (from 103,680 to 108,000) if the budget is increased from $60,000 to $61,000. WHY THE METHOD OF LAGRANGE MULTIPLIERS WORKS Although a rigorous eplanation of wh the method of Lagrange multipliers works involves advanced ideas beond the scope of this tet, there is a rather simple geometric argument that ou should find convincing. This argument depends on the fact that for the level curve F(, ) C, the slope is given b d d F F

11 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 561 This result is true for an level curve of a function F whose partial derivatives eist (provided F 0). An eample justifing the formula is outlined in Problem 48. Now, consider the constrained optimization problem: Maimize f(, ) subject to g(, ) k Geometricall, this means ou must find the highest level curve of f that intersects the constraint curve g(, ) k. As Figure 7.24 suggests, the critical intersection will occur at a point where the constraint curve is tangent to a level curve; that is, where the slope of the constraint curve g(, ) k is equal to the slope of a level curve f(, ) C. Direction in which C increases Point of tangenc Constraint curve: g(, ) = k Highest level curve, f(, ) = C, intersecting the constraint FIGURE 7.24 Increasing level curves and the constraint curve. According to the formula stated at the beginning of this discussion, ou have Slope of constraint curve Slope of level curve or, equivalentl, f g f g If ou let denote this common ratio, then g g f f f g and f g

12 562 Chapter 7 Calculus of Several Variables from which ou get the first two Lagrange equations f g and f g The third Lagrange equation g(, ) k is simpl a statement of the fact that the point of tangenc actuall lies on the constraint curve. P. R. O. B. L. E. M. S 7.4 P. R. O. B. L. E. M. S 7.4 In Problems 1 through 16, use the method of Lagrange multipliers to find the indicated etremum. You ma assume the etremum eists. 1. Find the maimum value of f(, ) subject to the constraint Find the maimum and minimum values of the function f(, ) subject to the constraint Find the minimum value of the function f(, ) 2 2 subject to the constraint Find the minimum value of the function f(, ) subject to the constraint Find the minimum value of f(, ) 2 2 subject to the constraint Let f(, ) Find the maimum and minimum values of the function f(, ) subject to the constraint Let f(, ) Find the maimum and minimum values of the function f(, ) subject to the constraint Find the maimum value of f(, ) 2 subject to the constraint Let f(, ) Find the minimum value of the function f(, ) subject to the constraint Let f(, ) Find the minimum value of the function f(, ) subject to the constraint Find the maimum and minimum values of f(, ) e subject to Find the maimum value of f(, ) ln ( 2 ) subject to for 0 and Find the maimum value of f(,, z) z subject to 2 3z Find the maimum and minimum values of f(,, z) 3 z subject to z Find the maimum and minimum values of f(,, z) 2 3z subject to 2 2 z 2 16.

13 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 563 CONSTRUCTION CONSTRUCTION POSTAL PACKAGING 16. Find the minimum value of f(,, z) 2 2 z 2 subject to z A farmer wishes to fence off a rectangular pasture along the bank of a river. The area of the pasture is to be 3,200 square meters, and no fencing is needed along the river bank. Find the dimensions of the pasture that will require the least amount of fencing. 18. There are 320 meters of fencing available to enclose a rectangular field. How should the fencing be used so that the enclosed area is as large as possible? 19. According to postal regulations, the girth plus length of parcels sent b fourth-class mail ma not eceed 108 inches. What is the largest possible volume of a rectangular parcel with two square sides that can be sent b fourth-class mail? Girth = 4 Girth = 2π PROBLEM 19 PROBLEM 20 POSTAL PACKAGING PACKAGING PACKAGING ALLOCATION OF FUNDS 20. According to the postal regulation given in Problem 19, what is the largest volume of a clindrical can that can be sent b fourth-class mail? (A clinder of radius R and length H has volume R 2 H.) 21. Use the fact that 12 fluid ounces is (approimatel) 6.89 cubic inches to find the dimensions of the 12-ounce soda can that can be constructed using the least amount of metal. (Recall that the volume of a clinder of radius r and height h is r 2 h, that the circumference of a circle of radius r is 2 r, and that the area of a circle of radius r is r 2.) 22. A clindrical can is to hold 4 cubic inches of frozen orange juice. The cost per square inch of constructing the metal top and bottom is twice the cost per square inch of constructing the cardboard side. What are the dimensions of the least epensive can? 23. A manufacturer has $8,000 to spend on the development and promotion of a new product. It is estimated that if thousand dollars is spent on development and thousand is spent on promotion, sales will be approimatel f(, ) 50 1/2 3/2 units. How much mone should the manufacturer allocate to development and how much to promotion to maimize sales?

14 564 Chapter 7 Calculus of Several Variables ALLOCATION OF FUNDS MARGINAL ANALYSIS SURFACE AREA OF THE HUMAN BODY MICROBIOLOGY 24. If thousand dollars is spent on labor and thousand dollars is spent on equipment, the output at a certain factor will be Q(, ) 60 1/3 2/3 units. If $120,000 is available, how should this be allocated between labor and equipment to generate the largest possible output? 25. Use the Lagrange multiplier to estimate the change in the maimum output of the factor in Problem 24 that will result if the mone available for labor and equipment is increased b $1, Recall from Problem 35 of Section 1 that an empirical formula for the surface area of a person s bod is S(W, H) W H where W (kg) is the person s weight and H (cm) is his or her height. Suppose for a short period of time, Maria s weight adjusts as she grows taller so that W H 160. With this constraint, what height and weight will maimize the surface area of Maria s bod? In Problems 27 and 28, ou will need to know that a closed clinder of radius R and length L has volume V R 2 L and surface area S 2 RL 2 R 2. The 2 volume of a hemisphere of radius R is V R 3 and its surface area is S 2 R A bacterium is shaped like a clindrical rod. If the volume of the bacterium is fied, what relationship between the radius R and length H of the bacterium will result in minimum surface area? H R Radius R L PROBLEM 27 PROBLEM 28 MICROBIOLOGY OPTICS 28. A bacterium is shaped like a clindrical rod with two hemispherical caps on the ends. If the volume of the bacterium is fied, what must be true about its radius R and length L to achieve minimum surface area? 29. The thin lens formula in optics sas that the focal length L of a thin lens is related to the object distance d o and image distance d i b the equation 1 d o 1 d i 1 L If L is fied, what is the maimum distance s d o d i between the object and the image?

15 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 565 CONSTRUCTION 30. A jewelr bo is constructed b partitioning a bo with a square base as shown in the accompaning figure. If the bo is designed to have volume 800 cm 3, what dimensions should it have to minimize its total surface area (top, bottom, sides, and interior partitions). Notice that we have said nothing about where the partitions are located. Does it matter? Top view CONSTRUCTION SPY STORY 31. Suppose the jewelr bo in Problem 30 is designed so that the material in the top costs twice as much as the material in the bottom and sides and three times as much as the material in the interior partitions. Find the dimensions that minimize the total cost of constructing the bo. 32. Having disposed of Scélérat s gunmen (Problem 51 in Section 4, Chapter 6), the sp goes looking for his enem. He enters a room and the door slams behind him. Immediatel, he begins to feel warm, and too late, he realizes he is trapped inside Scélérat s dreaded broiler room. Searching desperatel for a wa to survive, he notices that the room is shaped like the circle and that he is standing at the center (0, 0). He presses the stem on his special heat-detecting wristwatch and sees that the temperature at each point (, ) in the room is given b T(, ) PARTICLE PHYSICS From an informant s report, he knows that somewhere in this room there is a trap door leading outside the castle, and he reasons that it must be located at the coolest point in the room. Where is it? Just how cool will the sp be when he gets there? 33. A particle of mass m in a rectangular bo with dimensions,, and z has ground state energ E(,, z) 8m k z 2 where k is a phsical constant. In Problem 29, Section 3, ou were asked to minimize the ground state energ subject to the fied volume constraint V 0 z using substitution. Solve the same constrained optimization problem using the method of Lagrange multipliers.

16 566 Chapter 7 Calculus of Several Variables CONSTRUCTION CONSTRUCTION ALLOCATION OF FUNDS MARGINAL ANALYSIS ALLOCATION OF UNRESTRICTED FUNDS UTILITY UTILITY 34. A rectangular building is to be constructed of material that costs $31 per square foot for the roof, $27 per square foot for the two sides and the back, and $55 per square foot for the fanc facing and glass used in constructing the front. If the building is to have a volume of 16,000 ft 3, what dimensions will minimize the total cost of construction? 35. A storage shed is to be constructed of material that costs $15 per square foot for the roof, $12 per square foot for the two sides and back, and $20 per square foot for the front. What are the dimensions of the largest shed (in volume) that can be constructed for $8,000? 36. A manufacturer is planning to sell a new product at the price of $150 per unit and estimates that if thousand dollars is spent on development and thousand 320 dollars is spent on promotion, approimatel units of the product will be sold. The cost of manufacturing the product is $50 per unit. If the manufacturer has a total of $8,000 to spend on development and promotion, how should this mone be allocated to generate the largest possible profit? [Hint: Profit (number of units)(price per unit cost per unit) total amount spent on development and promotion.] 37. Suppose the manufacturer in Problem 36 decides to spend $8,100 instead of $8,000 on the development and promotion of the new product. Use the Lagrange multiplier to estimate how this change will affect the maimum possible profit. 38. (a) If unlimited funds are available, how much should the manufacturer in Problem 36 spend on development and how much on promotion in order to generate the largest possible profit? [Hint: Use the methods of Section 3.] (b) What is the value of the Lagrange multiplier that corresponds to the optimal budget in part (a)? Eplain our answer in light of the interpretation of dm as. dk (c) Your answer to part (b) should suggest another method for solving the problem in part (a). Solve the problem using this new method. 39. A consumer has $280 to spend on two commodities, the first of which costs $2 per unit and the second $5 per unit. Suppose that the utilit derived b the consumer from units of the first commodit and units of the second is U(, ) (a) How man units of each commodit should the consumer bu to maimize utilit? (b) Compute the marginal utilit of mone and interpret the result in economic terms. 40. A consumer has k dollars to spend on two commodities, the first of which costs a dollars per unit and the second b dollars per unit. Suppose that the utilit derived b

17 Chapter 7 Section 4 Constrained Optimization: The Method of Lagrange Multipliers 567 the consumer from units of the first commodit and units of the second commodit is given b the Cobb-Douglas utilit function U(, ), where 0 1 and 1. Show that utilit will be maimized when k and a k. b MINIMUM COST FIXED BUDGET MINIMUM COST HAZARDOUS WASTE MANAGEMENT 41. In Problem 40, how does the maimum output change if k is increased b 1 dollar? In Problems 42 through 44, let Q(, ) be a production function, where and represent units of labor and capital, respectivel. If unit costs of labor and capital are given b p and q, respectivel, then p q represents the total cost of production. 42. Use Lagrange multipliers to show that subject to a fied production level c, the total cost is minimized when Q and Q(, ) c p Q q provided Q and Q are not both 0 and p 0 and q 0. (This is often referred to as the minimum cost problem, and its solution is called the least-cost combination of inputs.) 43. Show that the inputs and that maimize the production level Q(, ) subject to a fied cost k satisf Q with p q k p Q q (Assume that neither p nor q is 0.) This is called a fied budget problem. 44. Show that subject to the fied production level A k, with 1, the cost function C(, ) p q is minimized when A k q p, A k p q 45. Use Lagrange multipliers to find the possible maimum or minimum points on that part of the surface z for which 5 2. Then use our calculator to sketch the curve 5 2 and the level curves to the surface f(, ) and show that the points ou have just found do not represent local maima or minima. What do ou conclude from this observation? 46. A stud conducted at a waste disposal site reveals soil contamination over a region that ma be described roughl as the interior of the ellipse

18 568 Chapter 7 Calculus of Several Variables where and are in miles. The manager of the site plans to build a circular enclosure to contain all polluted territor. (a) If the office at the site is at the point S(1, 1), what is the radius of the smallest circle centered at S that contains the entire contaminated region? [Hint: The function f(, ) ( 1) 2 ( 1) 2 measures the square of the distance from S(1, 1) to the point P(, ). The required radius can be found b maimizing f(, ) subject to a certain constraint.] (b) Read an article on waste management, and write a paragraph on how management decisions are made regarding landfills and other disposal sites.* MARGINAL ANALYSIS 47. Let P(K, L) be a production function, where K and L represent the capital and labor required for a certain manufacturing procedure. Suppose we wish to maimize P(K, L) subject to a cost constraint, C(K, L) A, for constant A. Use the method of Lagrange multipliers to show that optimal production is attained when P P K L C C K L that is, when the ratio of marginal production from capital to the marginal cost of capital equals the ratio of marginal production of labor to the marginal cost of labor. d In Eercises 48 and 49, use the formula for implicitl differentiating d F F the function f() given b the equation F(, ) k. 48. Let F(, ) (a) If F(, ) k for constant k, use the method of implicit differentiation develd oped in Chapter 2 to find. d (b) Find the partial derivatives F and F and verif that d d F F d 49. If F(, ) k, where F(, ) e 2 ln ( ), find. d * An ecellent case stud ma be found in M. D. LaGrega, P. L. Buckingham, and J. C. Evans, Hazardous Waste Management, McGraw-Hill, Inc., New York, 1994, pages

19 Chapter 7 Section 5 Double Integrals over Rectangular Regions Double Integrals over Rectangular Regions In Problems 50 through 53, use the method of Lagrange multipliers to find the indicated maimum or minimum. You will need to use the graphing utilit or the solve application on our calculator. 50. Maimize f(, ) e ln subject to Minimize f(, ) ln ( 2) subject to Minimize f(, ) subject to Maimize f(, ) subject to Similarl, to evaluate e In Chapters 5 and 6, ou integrated a function of one variable f() b reversing the process of differentiation, and a similar procedure can be used to integrate a function of two variables f(, ). However, since two variables are involved, we shall integrate f(, ) b holding one variable fied and integrating with respect to the other. 2 For instance, to evaluate the partial integral 2 d ou would integrate with 1 respect to, using the fundamental theorem of calculus with held constant: 2 2 d (2) (1) d, ou integrate with respect to, holding constant: 1 2 d (1)3 1 3 ( 1)3 2 3 In general, partiall integrating a function f(, ) with respect to results in a function of alone, which can then be integrated as a function of a single variable, thus producing what we call an iterated integral f(, ) d d Similarl, the iterated integral is obtained b first integrating with respect to, holding constant, and then with respect to. For instance, f(, ) d d.