1Slide 2.5 - 1Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyCopyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyMaximum-Minimum Problems; Business and Economics ApplicationsOBJECTIVE Solve maximum and minimum problems using calculus.2.5Slide 2.5 - 3Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyA Strategy for Solving Maximum-Minimum Problems:1. Read the problem carefully. If relevant, make a drawing.2. Make a list of appropriate variables and constants, noting what varies, what stays fixed, and what units are used. Label the measurements on your drawing, if one exists.2.5 Maximum-Minimum Problems; Business and Economics Applications2Slide 2.5 - 4Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyA Strategy for Solving Maximum-Minimum Problems (concluded):3. Translate the problem to an equation involving a quantity Q to be maximized or minimized. Try to represent Q in terms of the variables of step (2).4.Try to express Q as a function of one variable. Use the procedures developed in sections 2.1 – 2.3 to determine the maximum or minimum values and the points at which they occur.2.5 Maximum-Minimum Problems; Business and Economics ApplicationsSlide 2.5 - 5Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2: From a thin piece of cardboard 8 in. by 8 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume?1stmake a drawing in which x is the length of each square to be cut.2.5 Maximum-Minimum Problems; Business and Economics ApplicationsSlide 2.5 - 6Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2 (continued):2ndwrite an equation for the volume of the box.Note that x must be between 0 and 4. So, we need to maximize the volume equation on the interval (0, 4).2.5 Maximum-Minimum Problems; Business and Economics ApplicationsV=l⋅w⋅hVx( )= (8 − 2x) ⋅(8 − 2x) ⋅ xVx( )= (64 − 32x + 4x2)⋅ xVx( )= 4x3− 32x2+ 64x3Slide 2.5 - 7Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2 (continued):is the only critical value in (0, 4). So, we can use the second derivative.2.5 Maximum-Minimum Problems; Business and Economics Applications34′V=12x2−64x+64=03x2−16x + 16 = 0(3x − 4)(x − 4) = 03x − 4 = 0 or x − 4 = 0x =43or x = 4Slide 2.5 - 8Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2 (concluded):Thus, the volume is maximized when the square corners are inches. The maximum volume is2.5 Maximum-Minimum Problems; Business and Economics Applications′′V x()=24x−64′′V43= 2443− 64′′V43= −32 < 034V43= 4433− 32432+ 6443V43= 372527in3Slide 2.5 - 9Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 4: A stereo manufacturer determines that in order to sell x units of a new stereo, the price per unit,in dollars, must be The manufactureralso determines that the total cost of producing x units is given bya) Find the total revenue R(x).b) Find the total profit P(x).c) How many units must the company produce andsell in order to maximize profit?d) What is the maximum profit?e) What price per unit must be charged in order tomake this maximum profit? 2.5 Maximum-Minimum Problems; Business and Economics Applications.1000)( xxp−=.23000)( xxC+=4Slide 2.5 - 10Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 4 (continued):a)b)2.5 Maximum-Minimum Problems; Business and Economics ApplicationsRevenue=quantity⋅priceR(x) = x ⋅ pR(x) = x(1000 − x)R(x) = 1000x − x2Profit=Total Revenue−Total CostP(x) = R x( )− C x( )P(x) = 1000x − x2− 3000 + 20x( )P(x) = −x2+ 980x − 3000Slide 2.5 - 11Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 4 (continued):c)Since there is only one critical value, we can use the second derivative to determine whether or not it yields a maximum or minimum.Since P′′(x) is negative, x = 490 yields a maximum. Thus, profit is maximized when 490 units are bought and sold.2.5 Maximum-Minimum Problems; Business and Economics Applications′P(x)=−2x+980=0−2x = −980x = 490′′P(x)=−2Slide 2.5 - 12Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 4 (concluded):d) The maximum profit is given byThus, the stereo manufacturer makes a maximum profit of $237,100 when 490 units are bought and sold.e) The price per unit to achieve this maximum profit is2.5 Maximum-Minimum Problems; Business and Economics ApplicationsP(490)=−(490)2+980(490)−3000P(490) = $237,100.p(490)=1000−490p(490) = $510.5Slide 2.5 - 13Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyTHEOREM 10Maximum profit occurs at those x-values for whichR′(x) = C′(x) and R′′(x) < C′′(x).2.5 Maximum-Minimum Problems; Business and Economics ApplicationsSlide 2.5 - 14Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 5: Promoters of international fund-raising concerts must walk a fine line between profit and loss, especially when determining the price to charge for admission to closed-circuit TV showings in local theaters. By keeping records, a theater determines that, at an admission price of $26, it averages 1000 people in attendance. For every drop in price of $1, it gains 50 customers. Each customer spends an average of $4 on concessions. What admission price should the theater charge in order to maximize total revenue?2.5 Maximum-Minimum Problems; Business and Economics ApplicationsSlide 2.5 - 15Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 5 (continued):Let x = the number of dollars by which the price of $26 should be decreased (if x is negative, the price should be increased).2.5 Maximum-Minimum Problems; Business and Economics ApplicationsRevenue=Rev. from tickets + Rev. from concessionsR x( )= # of people ⋅ ticket price + # of people ⋅ 4R x( )= (1000 + 50x)(26 − x) + (1000 + 50x)⋅ 4R x( )= 26, 000 −1000x +1300x −