2.7 Applications of Derivatives to Business

80 CHAPTER 2 Applications of the Derivative 2.7 Applications of Derivatives to Business and Economics Cost = C() In recent ears, economic decision making has become more and more mathematicall oriented. Faced with huge masses of statistical data, depending on hundreds or even thousands of different variables, business analsts and economists have increasingl turned to mathematical methods to help them describe what is happening, predict the effects of various polic alternatives, and choose reasonable courses of action from the mriad of possibilities. Among the mathematical methods emploed is calculus. In this section we illustrate just a few of the man applications of calculus to business and economics. All our applications will center on what economists call the theor of the firm. In other words, we stud the activit of a business (or possibl a whole industr) and restrict our analsis to a time period during which background conditions (such as supplies of raw materials, wage rates, and taes) are fairl constant. We then show how derivatives can help the management of such a firm make vital production decisions. Management, whether or not it knows calculus, utilizes man functions of the sort we have been considering. Eamples of such functions are Cost 5 0 Production level (a) = C() 5 0 Production level (b) Figure A cost function. EXAMPLE C() = cost of producing units of the product, R() = revenue generated b selling units of the product, P () =R() C() = the profit (or loss) generated b producing and (selling units of the product.) Note that the functions C(), R(), and P () are often defined onl for nonnegative integers, that is, for = 0,, 2, 3,... The reason is that it does not make sense to speak about the cost of producing cars or the revenue generated b selling 3.62 refrigerators. Thus, each function ma give rise to a set of discrete points on a graph, as in Fig. (a). In studing these functions, however, economists usuall draw a smooth curve through the points and assume that C() is actuall defined for all positive. Of course, we must often interpret answers to problems in light of the fact that is, in most cases, a nonnegative integer. Cost Functions If we assume that a cost function, C(), has a smooth graph as in Fig. (b), we can use the tools of calculus to stud it. A tpical cost function is analzed in Eample. Marginal Cost Analsis Suppose that the cost function for a manufacturer is given b C() = (0 6 ) 3.003 2 +5 + 000 dollars. (a) Describe the behavior of the marginal cost. (b) Sketch the graph of C(). The first two derivatives of C() are given b C () =(3 0 6 ) 2.006 +5 C () =(6 0 6 ).006. Let us sketch the marginal cost C () first. From the behavior of C (), we will be able to graph C(). The marginal cost function =(3 0 6 ) 2.006 + 5 has as its graph a parabola that opens upward. Since = C () =.000006( 000), we see that the parabola has a horizontal tangent at = 000. So the minimum value of C () occurs at = 000. The corresponding -coordinate is (3 0 6 )(000) 2.006 (000) + 5 = 3 6+5=2. The graph of = C () is shown in Fig. 2. Consequentl, at first, the marginal cost decreases. It reaches a minimum of 2 at production level 000 and increases thereafter.

2.7 Applications of Derivatives to Business and Economics 8 This answers part (a). Let us now graph C(). Since the graph shown in Fig. 2 is the graph of the derivative of C(), we see that C () is never zero, so there are no relative etreme points. Since C () is alwas positive, C() is alwas increasing (as an cost curve should). Moreover, since C () decreases for less than 000 and increases for greater than 000, we see that C() is concave down for less than 000, is concave up for greater than 000, and has an inflection point at = 000. The graph of C() is drawn in Fig. 3. Note that the inflection point of C() occurs at the value of for which marginal cost is a minimum. = C() (000, 2) = C () 000 Figure 2 A marginal cost function. 000 Figure 3 A cost function. Now Tr Eercise Actuall, most marginal cost functions have the same general shape as the marginal cost curve of Eample. For when is small, production of additional units is subject to economies of production, which lowers unit costs. Thus, for small, marginal cost decreases. However, increased production eventuall leads to overtime, use of less efficient, older plants, and competition for scarce raw materials. As a result, the cost of additional units will increase for ver large. So we see that C () initiall decreases and then increases. Revenue Functions In general, a business is concerned not onl with its costs, but also with its revenues. Recall that, if R() is the revenue received from the sale of units of some commodit, then the derivative R () is called the marginal revenue. Economists use this to measure the rate of increase in revenue per unit increase in sales. If units of a product are sold at a price p per unit, the total revenue R() is given b R() = p. If a firm is small and is in competition with man other companies, its sales have little effect on the market price. Then, since the price is constant as far as the one firm is concerned, the marginal revenue R () equals the price p [that is, R () is the amount that the firm receives from the sale of one additional unit]. In this case, the revenue function will have a graph as in Fig. 4. Revenue R() = p Figure 4 A revenue curve. Quantit An interesting problem arises when a single firm is the onl supplier of a certain product or service, that is, when the firm has a monopol. Consumers will bu large amounts of the commodit if the price per unit is low and less if the price is raised.

82 CHAPTER 2 Applications of the Derivative Price p p = f() Quantit Figure 5 Ademandcurve. EXAMPLE 2 For each quantit, letf() be the highest price per unit that can be set to sell all units to customers. Since selling greater quantities requires a lowering of the price, f() will be a decreasing function. Figure 5 shows a tpical demand curve that relates the quantit demanded,, to the price, p = f(). The demand equation p = f() determines the total revenue function. If the firm wants to sell units, the highest price it can set is f() dollars per unit, and so the total revenue from the sale of units is R() = p = f(). () The concept of a demand curve applies to an entire industr (with man producers) as well as to a single monopolistic firm. In this case, man producers offer the same product for sale. If denotes the total output of the industr, f() is the market price per unit of output and f() is the total revenue earned from the sale of the units. Maimizing Revenue The demand equation for a certain product is p =6 2 dollars. Find the level of production that results in maimum revenue. Revenue R 6 R() = 6 2 2 (6, 8) Figure 6 Maimizing revenue. In this case, the revenue function R() is R() = p = (6 2 ) =6 2 2 dollars. The marginal revenue is given b R () =6. The graph of R() is a parabola that opens downward. (See Fig. 6.) It has a horizontal tangent precisel at those for which R () = 0 that is, for those at which marginal revenue is 0. The onl such is = 6. The corresponding value of revenue is R(6) = 6 6 2 (6)2 = 8 dollars. Thus, the rate of production resulting in maimum revenue is = 6, which results in total revenue of 8 dollars. Now Tr Eercise 3 Ticket price p EXAMPLE 3 500 000 Customers (000, 7) (200, 6) 500 Figure 7 Ademandcurve. Setting Up a Demand Equation The WMA Bus Lines offers sightseeing tours of Washington, D.C. One tour, priced at $7 per person, had an average demand of about 000 customers per week. When the price was lowered to $6, the weekl demand jumped to about 200 customers. Assuming that the demand equation is linear, find the tour price that should be charged per person to maimize the total revenue each week. First, we must find the demand equation. Let be the number of customers per week and let p be the price of a tour ticket. Then (, p) = (000, 7) and (, p) = (200, 6) are on the demand curve. (See Fig. 7.) Using the point slope formula for the line through these two points, we have 7 6 p 7= ( 000) = ( 000) = 000 200 200 200 +5, so p =2. (2) 200 From equation (), we obtain the revenue function: ( R() = 2 ) 200 =2 200 2. The marginal revenue function is R () =2 00 = ( 200). 00

2.7 Applications of Derivatives to Business and Economics 83 Using R() andr (), we can sketch the graph of R(). (See Fig. 8.) The maimum revenue occurs when the marginal revenue is zero, that is, when = 200. The price corresponding to this number of customers is found from demand equation (2): p =2 (200) = 6 dollars. 200 Thus, the price of $6 is most likel to bring the greatest revenue per week. R (200, 7200) Revenue R() = 2 200 2 Figure 8 Maimizing revenue. 200 Now Tr Eercise Profit Functions Once we know the cost function C() and the revenue function R(), we can compute the profit function P () from P () =R() C(). EXAMPLE 4 Maimizing Profits Suppose that the demand equation for a monopolist is p = 00.0 and the cost function is C() =50 +0,000. Find the value of that maimizes the profit and determine the corresponding price and total profit for this level of production. (See Fig. 9.) p p = 00.0 Price Figure 9 Ademandcurve. The total revenue function is Quantit R() = p = (00.0) = 00.0 2. Hence, the profit function is P () =R() C() = 00.0 2 (50 +0,000) =.0 2 +50 0,000. The graph of this function is a parabola that opens downward. (See Fig. 0.) Its highest point will be where the curve has zero slope, that is, where the marginal profit P () is zero. Now, P () =.02 +50=.02( 2500).

84 CHAPTER 2 Applications of the Derivative P Profit (2500, 52,500) P() =.0 2 + 50 0,000 Figure 0 Maimizing profit. 2500 So P () = 0 when = 2500. The profit for this level of production is P (2500) =.0(2500) 2 + 50(2500) 0,000 = 52,500 dollars. Finall, we return to the demand equation to find the highest price that can be charged per unit to sell all 2500 units: p = 00.0(2500) = 00 25 = 75 dollars. Thus, to maimize the profit, produce 2500 units and sell them at $75 per unit. The profit will be $52,500. Now Tr Eercise 7 EXAMPLE 5 Rework Eample 4 under the condition that the government has imposed an ecise ta of $0 per unit. For each unit sold, the manufacturer will have to pa $0 to the government. In other words, 0 dollars are added to the cost of producing and selling units. The cost function is now C() = (50 +0,000) + 0 =60 +0,000. The demand equation is unchanged b this ta, so the revenue is still Proceeding as before, we have R() = 00.0 2. P () =R() C() = 00.0 2 (60 +0,000) =.0 2 +40 0,000. P () =.02 +40=.02( 2000). The graph of P () is still a parabola that opens downward, and the highest point is where P () = 0, that is, where = 2000. (See Fig..) The corresponding profit is P (2000) =.0(2000) 2 + 40(2000) 0,000 = 30,000 dollars. P P() =.0 2 + 40 0,000 Profit (2000, 30,000) Figure Profit after an ecise ta. 2000

2.7 Applications of Derivatives to Business and Economics 85 From the demand equation, p = 00.0, we find the price that corresponds to = 2000: p = 00.0(2000) = 80 dollars. To maimize profit, produce 2000 units and sell them at $80 per unit. The profit will be $30,000. Notice in Eample 5 that the optimal price is raised from $75 to $80. If the monopolist wishes to maimize profits, he or she should pass onl half the $0 ta on to the customer. The monopolist cannot avoid the fact that profits will be substantiall lowered b the imposition of the ta. This is one reason wh industries lobb against taation. Setting Production Levels Suppose that a firm has cost function C() and revenue function R(). In a free-enterprise econom the firm will set production in such a wa as to maimize the profit function P () =R() C(). We have seen that if P () has a maimum at = a, then P (a) = 0. In other words, since P () =R () C (), R (a) C (a) =0 R (a) =C (a). Thus, profit is maimized at a production level for which marginal revenue equals marginal cost. (See Fig. 2.) = R() = C() Figure 2 a = optimal production level Check Your Understanding 2.7. Rework Eample 4 b finding the production level at which marginal revenue equals marginal cost. 2. Rework Eample 4 under the condition that the fied cost is increased from $0,000 to $5,000. 3. On a certain route, a regional airline carries 8000 passengers per month, each paing $50. The airline wants to increase the fare. However, the market research department estimates that for each $ increase in fare the airline will lose 00 passengers. Determine the price that maimizes the airline s revenue.

86 CHAPTER 2 Applications of the Derivative EXERCISES 2.7. Minimizing Marginal Cost Given the cost function C() = 3 6 2 +3 + 5, find the minimum marginal cost. $ 2. Minimizing Marginal Cost If a total cost function is C() =.000 3.06 2 +2 + 00, is the marginal cost increasing, decreasing, or not changing at = 00? Find the minimum marginal cost. 3. Maimizing Revenue Cost The revenue function for a oneproduct firm is R() = 200 600 +8. Find the value of that results in maimum revenue. 32 4. Maimizing Revenue The revenue function for a particular product is R() =(4.000). Find the largest possible revenue. R(20,000) = 40,000 is maimum possible. 5. Cost and Profit A one-product firm estimates that its dail total cost function (in suitable units) is C() = 3 6 2 +3 + 5 and its total revenue function is R() =28. Find the value of that maimizes the dail profit. 5 6. Maimizing Profit A small tie shop sells ties for $3.50 each. The dail cost function is estimated to be C() dollars, where is the number of ties sold on a tpical da and C() =.0006 3.03 2 +2+20. Find the value of that will maimize the store s dail profit. Maimum occurs at = 50. 7. Demand and Revenue The demand equation for a certain commodit is p = 2 2 0 + 300, 0 60. Find the value of and the corresponding price p that maimize the revenue. = 20 units, p = $33.33 8. Maimizing Revenue The demand equation for a product is p =2.00. Find the value of and the corresponding price p that maimize the revenue. p = $, = 000 9. Profit Some ears ago it was estimated that the demand for steel approimatel satisfied the equation p = 256 50, and the total cost of producing units of steel was C() = 82 + 56. (Thequantit was measured in millions of tons and the price and total cost were measured in millions of dollars.) Determine the level of production and the corresponding price that maimize the profits. 2 million tons, $56 per ton 0. Maimizing Area Consider a rectangle in the -plane, with corners at (0, 0), (a, 0), (0,b), and (a, b). If (a, b) lies on the graph of the equation =30, find a and b such that the area of the rectangle is maimized. What economic interpretations can be given to our answer if the equation =30 represents a demand curve and is the price corresponding to the demand? = 5, = 5.. Demand, Revenue, and Profit Until recentl hamburgers at the cit sports arena cost $4 each. The food concessionaire sold an average of 0,000 hamburgers on a game night. When the price was raised to $4.40, hamburger sales dropped off to an average of 8000 per night. (a) Assuming a linear demand curve, find the price of a hamburger that will maimize the nightl hamburger revenue. $3.00 (b) If the concessionaire has fied costs of $000 per night and the variable cost is $.60 per hamburger, find the price of a hamburger that will maimize the nightl hamburger profit. $3.30 2. Demand and Revenue The average ticket price for a concert at the opera house was $50. The average attendance was 4000. When the ticket price was raised to $52, attendance declined to an average of 3800 persons per performance. What should the ticket price be to maimize the revenue for the opera house? (Assume a linear demand curve.) $45 per ticket 3. Demand and Revenue An artist is planning to sell signed prints of her latest work. If 50 prints are offered for sale, she can charge $400 each. However, if she makes more than 50 prints, she must lower the price of all the prints b $5 for each print in ecess of the 50. How man prints should the artist make to maimize her revenue? 4. Demand and Revenue A swimming club offers memberships at the rate of $200, provided that a minimum of 00 people join. For each member in ecess of 00, the membership fee will be reduced $ per person (for each member). At most, 60 memberships will be sold. How man memberships should the club tr to sell to maimize its revenue? = 50 memberships 5. Profit In the planning of a sidewalk café, it is estimated that for 2 tables, the dail profit will be $0 per table. Because of overcrowding, for each additional table the profit per table (for ever table in the café) will be reduced b $.50. How man tables should be provided to maimize the profit from the café? 6. Demand and Revenue A certain toll road averages 36,000 cars per da when charging $ per car. A surve concludes that increasing the toll will result in 300 fewer cars for each cent of increase. What toll should be charged to maimize the revenue? Toll should be $.0. 7. Price Setting The monthl demand equation for an electric utilit compan is estimated to be p =60 (0 5 ), where p is measured in dollars and is measured in thousands of kilowatt-hours. The utilit has fied costs of 7 million dollars per month and variable costs of $30 per 000 kilowatt-hours of electricit generated, so the cost function is C() =7 0 6 +30. (a) Find the value of and the corresponding price for 000 kilowatt-hours that maimize the utilit s profit. =5 0 5, p = $45. indicates answers that are in the back of the book. 2. Marginal cost is decreasing at = 00. M (200) = 0 is the minimal marginal cost.

Eercises 2.7 87 (b) Suppose that rising fuel costs increase the utilit s variable costs from $30 to $40, so its new cost function is C () =7 0 6 +40. Should the utilit pass all this increase of $0 per thousand kilowatt-hours on to consumers? Eplain our answer. No. Profit is maimized when price is increased to $50. 8. Taes, Profit, and Revenue The demand equation for a compan is p = 200 3, and the cost function is C() =75+80 2, 0 40. (a) Determine the value of and the corresponding price that maimize the profit. = 30, p = $0 (b) If the government imposes a ta on the compan of $4 per unit quantit produced, determine the new price that maimizes the profit. $3 (c) The government imposes a ta of T dollars per unit quantit produced (where 0 T 20), so the new cost function is C() = 75 + (80 + T ) 2, 0 40. Determine the new value of that maimizes the compan s profit as a function of T. Assuming that the compan cuts back production to this level, epress the ta revenues received b the government as a function of T. Finall, determine the value of T that will maimize the ta revenue received b the government. =30 T/4, T = $60/unit 9. Interest Rate A savings and loan association estimates that the amount of mone on deposit will be million times the percentage rate of interest. For instance, a 4% interest rate will generate $4 million in deposits. If the savings and loan association can loan all the mone it takes in at 0% interest, what interest rate on deposits generates the greatest profit? 5% 20. Analzing Profit Let P () be the annual profit for a certain product, where is the amount of mone spent on advertising. (See Fig. 3.) (a) Interpret P (0). (b) Describe how the marginal profit changes as the amount of mone spent on advertising increases. (c) Eplain the economic significance of the inflection point. P 2. Revenue The revenue for a manufacturer is R() thousand dollars, where is the number of units of goods produced (and sold) and R and R are the functions given in Figs. 4(a) and 4(b). 80 70 60 50 40 30 20 0.8.6 2.4 3.2 = R() 0 20 (a) 0 20 (b) 30 3.2 = R () 2.4.6.8 30 40 40 Figure 4 Revenue function and its first derivative. (a) What is the revenue from producing 40 units of goods? $75,000 (b) What is the marginal revenue when 7.5 units of goods are produced? $3200 per unit (c) At what level of production is the revenue $45,000? 5 units (d) At what level(s) of production is the marginal revenue $800? 32.5 units (e) At what level of production is the revenue greatest? 35 units 20. (a) P (0) is the profit with no advertising budget (b) As mone is spent on advertising, the marginal profit initiall increases. However, at some point the marginal profit begins to decrease. (c) Additional mone spent on advertising is most advantageous at the inflection point. Annual profit = P() Mone spent on advertising Figure 3 Profit as a function of advertising.

88 CHAPTER 2 Applications of the Derivative 22. Cost and Original Cost The cost function for a manufacturer is C() dollars, where is the number of units of goods produced and C, C,andC are the functions given in Fig. 5. (a) What is the cost of manufacturing 60 units of goods? $,00 (b) What is the marginal cost when 40 units of goods are manufactured? $2.5 per unit (c) At what level of production is the cost $200? 00 units (d) At what level(s) of production is the marginal cost $22.50? 20 units and 40 units (e) At what level of production does the marginal cost have the least value? What is the marginal cost at this level of production? 80 units, $5 per unit. 500 000 = C() 30 20 = C () 500 0 = C () 50 00 (a) 50 50 00 (b) 50 Figure 5 Cost function and its derivatives. Solutions to Check Your Understanding 2.7. The revenue function is R() = 00.0 2, so the marginal revenue function is R () = 00.02. The cost function is C() = 50 +0,000, so the marginal cost function is C () = 50. Let us now equate the two marginal functions and solve for : R () =C () 00.02 =50.02 = 50 = 50.02 = 5000 = 2500. 2 Of course, we obtain the same level of production as before. 2. If the fied cost is increased from $0,000 to $5,000, the new cost function will be C() =50 +5,000, but the marginal cost function will still be C () = 50. Therefore, the solution will be the same: 2500 units should be produced and sold at $75 per unit. (Increases in fied costs should not necessaril be passed on to the consumer if the objective is to maimize the profit.) 3. Let denote the number of passengers per month and p the price per ticket. We obtain the number of passengers lost due to a fare increase b multipling the number of dollars of fare increase, p 50, b the number of passengers lost for each dollar of fare increase. So = 8000 (p 50)00 = 00p +3,000. Solving for p, we get the demand equation p = 00 + 30. From equation (), the revenue function is ( R() = p = ) 00 + 30. The graph is a parabola that opens downward, with -intercepts at =0and =3,000. (See Fig. 6.) Its maimum is located at the midpoint of the -intercepts, or = 6500. The price corresponding to this number of passengers is p = (6500)+ 30 = $65. Thus the price 00 of $65 per ticket will bring the highest revenue to the airline compan per month. Revenue in dollars 422,500 0 Figure 6 R() = ( + 30) 00 6500 3,000 Number of passengers