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Chapter 4 116

# Chapter 4 116 - 378 CHAPTER 4 APPLlCATIONS 0F...

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Unformatted text preview: 378 CHAPTER 4 APPLlCATIONS 0F DlFFERENTlATIUN 1.98 d: (—1.98)2 + 4(0.003)(24) _ 1.98 :1: 4.2084 2(0003) 0.006 m z (1.98 + 2.05)/0.006 % 672. Now R”(m) : 40.02 and 0"(516): —2 + 0.0061: => C"(672) = 2.032 2 R"(672) < C”(672) ﬁ there is a maximum at m : 672. the quadratic formula. 13 — . Since 90 > 0 14. C(93) = 16.000 + 50013 4 1.6952 + 0.00453. 12(5) : 1700 1 7w. Then Rm 2 515(5) 2 1700e — 752. If the proﬁt is maximum. then R’(:l:) = C’(m) (I) 1700 — 14:1: : 500 — 3.21: + 0.0123:2 c) 0.012ac2 +1082 — 1200 : 0 4:» x2 + 90017 — 100.000 : 0 4: (ac + 1000)(m — 100) = 0 4:» a; = 100 (since :5 > 0). The proﬁt is maximized if P”(m) < 0. but since P"(m) : R"(m) — C”(a:). we can just check the condition R”(m) < 0"(22). Now R"(9:) = 714 < —3.2 + 0.02413 2 0"(33) for as > 0. so there is a maximum at 00 : 100. 15. C(50) : 0.001953 1 0.3952 + 65; + 900. The marginal cost is C'(w) = 0.00392 4 0.62: + 6. C'(a:) is increasing when C"(:c) > 0 <=> 0.00690 — 0.6 > 0 <=> m > 0.6/0.006 = 100. S0 C’(ac) starts to increase when :0 : 100. 16. C(ze) : 0.000253 — 0.25:1;2 + 42: + 1500. The marginal cost is C’(:c) : 0.0006322 4 0.5095 + 4. C(90) is increasing when C"(m) > 0 <:> 0.00123: — 0.5 > 0 42> m > 0.5/0.0012 % 417. So C’(m) starts to increase when at = 417. 17. (a) C(m) : 1200 + 122: - 0.1ac2 + 0.000529. 10.000 R(:l:) : 3312(30) : 29:5 - 0.00021m2. ’ Since the proﬁt is maximized when R737) : C/(cc). we examine the curves R and C in the ﬁgure. looking for m—values at which the slopes of the tangent lines are equal. It appears that :0 = 200 is 0’ 400 a good estimate. (b) R'(:l:) : 0’03) :> 29 — 0.000422: : 12 — 0.290 + 000151:2 :> 0.001552 1 0.1995890 4 17 : 0 => 1: w 192.06 (for as > 0). As in Exercise 11. R”(\$) < C"(a:) => —0.00042 < —0.2 + 0.00300 (2 0.00330 > 0.19958 4:) :c > 66.5. Our value of 192 is in this range. so we have a maximum proﬁt when we produce 192 yards of fabric. 18. (a) Cost : setup cost + manufacturing cost => C(93) : 500 + m(a:) : 500 + 20m # 5333/4 + 0.01332. We can solve m(p) : 320 — 7.71) for p in terms of m to ﬁnd the demand (or price) function. 320 4 320 — 2 a: : 320 7 7.7p 7.7p # 320 m p(a:) # m. R(cc) : \$p(m) = ——\$-—-x—- 7.7 7.7 320 # 2:3 (b) C'(9c) : R'(m) => 20 — lfe—l/4 + 0.0235 : 7 7 .e x 81.53 planes. and p(:c) : \$30.97 million. The maximum proﬁt associated with these values is about \$463.59 million. 19. (a) We are given that the demand function 1) is linear and p(27.000) 2 10, p(33.000) : 8. so the slope is m : -ﬁ and an equation of the line is y — 10 = (#50170) (:0 — 27.000) => y : p(w) : -§01—05:c + 19 : 19 4 (13/3000). (b) The revenue is R(m) = wp(x) : 19\$ — (222/3000) :> R’(a:) : 19 — (93/1500) 2 0 when m : 28,500. Since R"(a:) : —1 / 1500 < 0, the maximum revenue occurs when :1: : 28.500 :> the price is p(28,500) = \$9.50. ...
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