Chapter 4 114

# Chapter 4 114 - 376 D CHAPTER4 APPLICATIONS OF...

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Unformatted text preview: 376 D CHAPTER4 APPLICATIONS OF DIFFERENTIATION 5. (a) The cost function is C(33) = 40.000 —I— 3000: + 232. so the cost at a production level of 1000 is C(m) _ 40.000 m a: C(1000) = \$1340/unit. The marginal cost function is C’(m) : 300 + 2:0 and C’(1000) = \$2300/unit. 40.000 x C(1000) = \$1.340.000. The average cost function is C(30) 2 + 300 —I— w and (b) See the box preceding Example I. We must have C" (w) = c (x) {I} 300 + 21: = _ 40.000 — iL' +300—I—m <=> (E § 9:2 : 40.000 => as = «40,000 = 200. This gives a minimum value of the average cost 80.000 3:3 (c) The minimum average cost is c(200) : \$700/ unit. function 0 (:13) since c" (m) : > 0. 6. (a) C(m) : 25.000 +120m + 0.112.0(1000) : \$245000. C(53) : CI’J) : 25900 1' 32' e(1000) : \$245/unit. C'(a:) : 120 + 0.21:. C’(1000) : \$320/unit. 25000 + 120 + 0.13: 4:) 01:3 = 25000 cc x 0.1m2 2 25.000 => :1: = «250,000 : 500. This gives a minimum since 0"(20) : 500300 > 0. w —I— 120 + 0.1117, (b) We must have C'(a:) : C(ZL‘) (i) 120 + 0.200 : (c) The minimum average cost is c(500) : \$220.00/unit. 7. (a) C(37) : 16.000 + 200w -I— 4123/2. C(1000) : 16.000 + 200.000 —I— 40.000 \/ 10 s 216.000 —I— 126,491. so 16.000 ac C(30) = 200 + 6321/2. C’(1000) = 200 + 60 m % \$389.74/unit. C(1000) % \$342491. C(ac) = C(cc)/z : + 200 + 41:1/2. C(1000) % \$342.49/unit. 16.000 x (b) We must have C’(m) 2 C(33) {I} 200 + 6581/2 : + 200 + 401:”2 (I) 2203/2 = 16.000 (1} x = (8.000)”3 : 400 units. To check that this is a minimum. we calculate —16.000 2 2 3/2 9:2 + ﬂ 7 362 (m and positive for a: > 400. so 0 is decreasing on (0,400) and increasing on (400. 00). Thus. 0 has an absolute c'(:c) _ 8000). This is negative for :c < (8000)2/3 : 400. zero at x : 400. minimum at m : 400. [Notes 0"(30) is not positive for all x > 0.] (c) The minimum average cost is C(400) : 40 + 200 —I— 80 : \$320/unit. 3. (3) C(33) = 10.000 + 3409; , 0.3g;2 + 0.0001cc3. C(1000) : \$150000. C(w) : C(ac)/;I: : 10:00 C’(:I:) : 340 , 0.620 + 0.0003302. C’(1000) : \$40/unit. + 340 — 0.32: + 0.0001m2. e(1000) : \$150/unit. 10.000 (b) We must have C'(:c) = C(x) 4:» 340 — 0.6a: + 0.0003902 : I + 340 — 0.33: + 0.0001902 41» 0.00029:2 : 10000 + 0.31: 4:) 0.0002583 — 0.33:2 , 10.000 : 0 4:» :53 — 1500:52 7 50000000 : 0 33 . . . . . . ,, 20000 :> cc % 1521.60 % 1522 units. This gives a minimum Since 6 (LE) : \$3 + 0.0002 > 0. (c) The minimum average cost is about 00521.60) % \$121.62/unit. 9. (a) C(30) : 3700 + 5m — 0.043;2 + 0.0003953 :s C’(:c) = 5 A 00800 + 0.00093;2 (marginal cost). em : C(27) : 3700 (C II? —I— 5 i 0.04:6 + 0.00033:2 (average cost). ...
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