Schaum's Outline of Theory and Problems of Beginning Calculus

110 MAXIMUM AND MINIMUM PROBLEMS [CHAP. 14Let x be the side of the square that is removed from each corner. The volume V = Iwh, where I, w, andh are the length, width, and height of the box. Now 1 = 8 - 2x, w = 3 - 2x, and h = x, givingV(X) = (8- 2xK3 - 2x)x = (4x2- 2 2 + ~ 24)~ = 42 - 22x2 + 2 4 ~The width w must be positive. Hence,3-2x>O or 3>2x or 3>xFurthermore, x > 0. But we also can admit the values x = 0 and x = 3, which make V = 0 and which,therefore, cannot yield the maximum volume. Thus, we have to maximize V(x) on the interval CO, 31. SincedV 12x2 - 44~+ 24-=dxthe critical numbers are the solutions of12x2 - 44~+ 24 = 03x2 - llx + 6 = 0(3x - 2xx - 3) = 03x-2=0 or x-3=03x = 2 or x=3x=j or x=3The only critical number in (0,j) is 3. Hence, the volume is greatest when x = 3.14.5 A manufacturer sells each of his TV sets for $85. The cost C (in dollars) of manufacturing andselling x TV sets per week isC = 1500 + 10x + 0.005~~If at most loo00 sets can be produced per week, how many sets should be made and sold tomaximize the weekly profit ?For x sets per week, the total income is 85x. The profit is the income minus the cost,P = 8 5 - ~ (1500 + 1 0 + ~ 0.005~~) = 7 5 - ~ 1500 - 0.005~~We wish to maximize P on the interval CO, lOOOO], since the output is at most 1OOOO.and the critical number is the solution ofWe now construct Table 14-5:-- dP- 75 - 0.01xdx75 - 0.01x = 00.01x = 75x=-- 75 - 75000.01P(7500) = 75(7500) - 1500 - 0.0005(7500)2= 562 500 - 1500 - 0.0005(56 250000)= 561 OOO - 281 250 = 279 750P(0) = 75(0) - 1500 - 0.0005(0)2 = - 1500P(1OOOO) = 75(10000) - 1500 - 0.0005(10000)2750 OOO - 1500 - O.OOOS( 100 OOO OOO)= 748 500 - 500000 = 248 500Thus, the maximum profit is achieved when 7500 TV sets are produced and sold per week.