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2 The only critical number in 0, 3 is 3 . Hence, the volume is greatest when x = 2 . 2 3
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14.5 A manufacturer sells each of his TV sets for \$85. The cost C (in dollars) of manufacturing and selling x TV sets per week is C = 1500 + 10x + 0.005x 2 If at most 10 000 sets can be produced per week, how many sets should be made and sold to maximize the weekly pro t
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For x sets per week, the total income is 85x. The pro t is the income minus the cost, P = 85x (1500 + 10x + 0.005x 2 ) = 75x 1500 0.005x 2 We wish to maximize P on the interval [0, 10 000], since the output is at most 10 000. dP = 75 0.01x dx and the critical number is the solution of 75 0.01x = 0 0.01x = 75 75 x= = 7500 0.01 We now construct Table 14-5 (see below): P(7500) = 75(7500) 1500 0.005(7500)2 = 562 500 1500 0.005(56 250 000) = 561 000 281 250 = 279 750 P(0) = 75(0) 1500 0.005(0)2 = 1500 P(10 000) = 75(10 000) 1500 0.005(10 000)2 = 750 000 1500 0.005(100 000 000) = 748 500 500 000 = 248 500 Thus, the maximum pro t is achieved when 7500 TV sets are produced and sold per week. Table 14-5 x 7500 0 10 000 P(x) 279 750 1500 248 500
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14.6 An orchard has an average yield of 25 bushels per tree when there are at most 40 trees per acre. When there are more than 40 trees per acre, the average yield decreases by 1 bushel per tree for every tree over 40. Find the 2 number of trees per acre that will give the greatest yield per acre.
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Let x be the number of trees per acre, and let f (x) be the total yield in bushels per acre. When 0 x 40, f (x) = 25x. If x > 40, the number of bushels produced by each tree becomes 25 1 (x 40). [Here x 40 is the number of trees over 2 40, and 1 (x 40) is the corresponding decrease in bushels per tree.] Hence, for x > 40, f (x) is given by 2 25 1 x 40 2 1 1 x x = 25 x + 20 x = 45 x x = (90 x) 2 2 2
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Thus, f (x) =
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25x if 0 x 40 x (90 x) if x > 40 2
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x f (x) is continuous everywhere, since 25x = 2 (90 x) when x = 40. Clearly, f (x) < 0 when x > 90. Hence, we may restrict attention to the interval [0, 90]. For 0 < x < 40, f (x) = 25x, and f (x) = 25. Thus, there are no critical numbers in the open interval (0, 40). For 40 < x < 90,
f (x) =
x x2 (90 x) = 45x 2 2
and f (x) = 45 x
Thus, x = 45 is a critical number. In addition, 40 is also a critical number since f (40) happens not to exist. We do not have to verify this fact, since there is no harm in adding 40 [or any other number in (0, 90)] to the list for which we compute f (x). We now construct Table 14-6: f (45) = 45 45 2025 (90 45) = (45) = = 1012.5 2 2 2 f (40) = 25(40) = 1000 f (0) = 25(0) = 0 90 90 f (90) = (90 90) = (0) = 0 2 2 The maximum yield per acre is realized when there are 45 trees per acre. Table 14-6 x 45 40 0 90 f (x) 1012.5 1000 0 0
Supplementary Problems
14.7 Find the absolute maxima and minima of the following functions on the indicated intervals: (a) f (x) = 4x + 5 on [ 2, 3] on [ 1, 1] on [0, 4] (b) f (x) = 2x 2 7x 10 (d) f (x) = 4x 3 8x 2 + 1 on [ 1, 3]