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Unformatted text preview: SECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS U 375 4.8 Applications to Business and Economics 1. (a) C(O) represents the ﬁxed costs of production. such as (c) The marginal cost function is C'(:c).
rent. utilities. machinery etc.. which are incurred even We graph it as in Example I in
Section 2.9. when nothing is produced.
(b) The inﬂection point is the point at which C"(m) changes C’ from negative to positive; that is. the marginal cost C(27) changes from decreasing to increasing. Thus. the marginal cost is minimized at the inﬂection point. 2. (a) We graph C" as in Example 1 in Section 2.9. (b) By reading values of C(23) from its
graph. we can plot C(95) = C(sc)/ac. (c) Since the graph in part (b) is decreasing. we estimate that the minimum value of C(ZL‘) occurs at m : 7. The average cost and the marginal cost are equal at that value. See the box preceding Example I. 3. C(10) : 21.4 7 0.00230 and C(l‘) = C(m)/I => C(x) : 21.4w — 0.002272. C'(;c) : 21.4 — 0.004x and
C/(IOOO) : 17.4. This means that the cost of producing the 10015t unit is about $17.40. 4. (a) Proﬁt is maximized when the (b) P(m) : R(ac) 7 C(50) is (c) The marginal proﬁt function is
marginal revenue is equal to the sketched. deﬁned as P’(:c).
marginal cost; that is. when R and
C have equal slopes. See the box preceding Example 2. 376 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 5. (a) The cost function is C(m) : 40.000 + 300:2: + :32. so the cost at a production level of 1000 is C(IOOO) : $1.340.000. The average cost function is C(m) = ﬂ = 40000
as an c(1000) = $1340/unit. The marginal cost function is C'(:c) = 300 + 2x and C'(1000) : $2300/unit. 40.000
x + 300 + x and (b) See the box preceding Example 1. We must have C” (x) : c (x) <:> 300 + 250 2 , 40.000
7 {L +300+x ¢> ac => :02 = 40.000 :> :L' : «40,000 = 200. This gives a minimum value of the average cost 80.000
x3 (c) The minimum average cost is c(200) : $700/unit. function c (cc) since c" (at) = > 0. 6. (a) C(33) : 25000 + 120cc + 0,112.0(1000) : $245,000. c($) : CS”) : 25000
SE e(1000) : $245/unit. C'(;z) : 120 + 0.23:. C’(1000) : $320/unit. I 120 + 0.1x. (b) We must have C’(a:) : C(x) (:> 120 + 0.23; : 25000 + 120 + 0.130 <2; 01$ : 25000
I z
2 . . . . . ,, 50.000
0.1w 2 25.000 => w = «250,000 2 500. This gives a minimum Since c (ac) : 3 > 0,
as (c) The minimum average cost is C(SOO) : $220.00/unit. 7. (a) C(m) : 16.000 I 20010 —I— 4x3/2. C(1000) : 16.000 + 200.000 + 40.000 \/ 10 % 216.000 + 126.491. so
16.000
:0 C'(:c) = 200 + 6x1”. C’(1000) = 200 + 60 m z $389.74/unit. C(1000) % $342,491. C(32) = C(cc)/z = + 200 + 4501/2. C(1000) % $342.49/unit. 16.000
x (b) We must have (I'm) = C(e) <2» 200 + 6001/2 : + 200 + 4301/2 e 2W2 2 16,000 4:) x = (8.000)”3 : 400 units. To check that this is a minimum. we calculate —16.000 . 2 2 3/2
2:2 I J; T x2 (x and positive for Lu > 400. so c is decreasing on (0,400) and increasing on (400. 00). Thus. 6 has an absolute c'(m) _ 8000). This is negative for :c < (8000)2/3 : 400. zero at a: : 400. minimum at x : 400. [Note c”(:c) is not positive for all x > 0.]
(c) The minimum average cost is C(400) : 40 + 200 + 80 : $320/unit. 8. (a) C(03) = 10.000 + 340:0 , 0.3x2 + 0.0001x3. C(1000) : $150,000. C(ac) : C(x)/ac : 10:00 C'(m) : 340 , 0.650 + 0000322011000) : $40/unit. + 340 — 0.330 + 0.0001x2. e(1000) : $150/unit. 10000
(b) We must have C'(:c) = C(x) 4:» 340 — 0.63: + 0.0003aa2 : a: + 340 — 0.31: + 0.0001ae2 4:»
2 10.000 3 , 2 7 3 2 7
0.000210 : + 0.320 <:> 0.00021E — 0.3.7: 7 10.000 — 0 <=> z — 1500$ ~ 50.000000 1 0
a:
. . . . . . ,, 20.000
:> a: w 1521.60 % 1522 units. This gives a minimum smce c (x) = x3 + 0.0002 > 0. (c) The minimum average cost is about 00521.60) % $121.62/unit. 9. (a) C(ac) : 3700 + 51' — 0.04252 + 0.0003;c3 :> C/(ac) = 5 1 0.08.1: + 0.0009502 (marginal cost).
C(I) : C(m) : 3700
it IL‘ + 5 7 0.04m + 0.00033:2 (average cost). SECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS 377 (b) 100 The graphs intersect at (208.51, 27.45), so the production level that minimizes average cost is about 209 units. 0 500 (c) c’(w) : —3700 — 0.04 + 0.000613 2 0 => 3700 + 0.04m2 — 0.0006953 : 0 2» m1 2 208.51. $2 C((L‘l) m $27.45/unit. (d) The marginal cost is given by C’(m). so to ﬁnd its minimum value we’ll ﬁnd the derivative of C"; that is. C".
C"(x) : —0.08 + 0.00183: 2 0 :> 361 = 81—0; : 44.44. C’(m1) : $3.22/unn. C'”'(J:) 2 0.0018 > 0 for all at, so this is the minimum marginal cost. 0’" is the second derivative of C”. 10. (a) C(50) : 339 + 25x — 0.09:1:2 + 0.0004x3 2 C(36) : 25 — 0.1836 + 0.0012932 (marginal cost).
C(12) ¥ 339 C(x) 2 — — + 25 2 0.0950 + 0.0004502 (average cost).
x .T
(b) [00 The graphs intersect at (135.567 2265). so the production
I level that minimizes average cost is about 136 units.
0 500
339 (c) c/(x) : ‘F 2 0.09 + 0.000824 2 0 => 271 s 135.56. C(m1)% $22.65/unit. (d) 0’05) : —0.18 + 0.002433 : 0 :> 1: = % = 75. C’(75) : $18.25/unit.
C”'(ac) 2 0.0024 > 0 for all 11:. so this is the minimum marginal cost. 11. C(z) 2 680 + 4x —I— 0.01302. p(:c) 2 12 2> Rm) 2 mp(m) 2 122B. If the proﬁt is maximum. then
R’(x) 2 C’(z) 2> 12 2 4 + 0.022: 2 0.023: 2 8 2 ac 2 400. The proﬁt is maximized if P”(:c) < 0.
but since P"(:I:) 2 R"(;v) — Cl/(z). we can just check the condition R"(z) < C’"(m). Now
R”(;1:) 2 O < 0.02 2 C”(z). so :6 2 400 gives a maximum. 12. C(23) 2 680 + 43: + 0.01.12. p(:t:) 2 12 — m/SOO. Then R(.z') 2 xp(a:) 2 12¢ — 2:2/500. If the proﬁt is
maximum. then R'(;c) 2 C’(0c) [See the box preceding Example 2.] 42 12 ~ 02/250 2 4 + 0.02m 42
8 2 0.02400 <2 cc 2 8/0024 2 %. The proﬁt is maximized if P”(;c) < 0. but since P”(m) 2 R”($) — C'”($). we can just check the condition R”(:c) < C”(w). Now R”(:c) 2 —% < 0.02 2 0"(512). so a: 2 % gives a maximum. 13. C(23) : 1450 + 36x — x2 + 0.001953. 11(3)) : 60 — 001$. Then Rm) : mm) = 60:15 — 0.01m2. If the proﬁt is
maximum, then mm) : C/(x) 4: 60 — 0.02:1: : 36 — 2m + 0.003952 2 0.003352 — 1.98:5 — 24 : 0. By 378 3 CHAPTER4 APPLICATIONS OF DIFFERENTIATION 1.98 :t (—1.98)? + 4(0.003)(24) _ 1.98 i 4.2084
2(0003) _ 0.006 an a“ (1.98 + 2.05)/0.006 z 672. Now R”(x) = —0.02 and 0%) = 72 + 0.00650 2 C”(672) : 2032 => R"(672) < C'"(672) => there is a maximum at :1: = 672. the quadratic formula. cc 7 . Since cc > 0. 14. C(36) : 16.000 + 50010 — 1.6302 + 0.004363. p(:r) : 1700 — 793. Then Rh?) 2 xp(:c) : 1700m  718. If the
proﬁt is maximum. then R'(ac) : C’(w) 4:) 1700 7 14m : 500 — 3.256 + 0.012302 4:)
0.012352 +10.8m , 1200 = 0 a» :02 + 90036 — 100.000 2 0 e (a: +1000)(x , 100) = 0 4:) :1: : 100
(since as > 0). The proﬁt is maximized if P”(m) < 0. but since P"(x) : R"(;c) — C"(:c). we can just check the
condition R"(:c) < C"(ac). Now R"(a:) = 714 < —3.2 + 0.024m = C"(:c) for ac > 0. so there is a maximum at
ac : 100. 15. C(x) : 0.001953 1 03:52 + 69: + 900. The marginal cost is C"(1:) = 0.003m2 7 0.69: + 6.
C'(m) is increasing when C”(l‘) > 0 <: 0.00630 — 0.6 > 0 4:, a: > 06/0006 2 100. So C'(a:) starts to
increase when :0 : 100. 16. C(ac) : 0.0002353 — 0.25.7102 + 42: + 1500. The marginal cost is C'(m) : 0.0006302 7 0.5010 + 4.
C(18) is increasing when C"(x) > 0 (it 0.00123: — 0.5 > 0 4:) m > 0.5/0.0012 % 417. So C'(:c) starts to increase when m = 417. 17. (a) 0(a) : 1200 + 12x — 0.1352 + 0.0005253. 10.000
R(:c) : “2(a) : 29:6 — 000021232. 7
Since the proﬁt is maximized when R’($) : C/(x).
we examine the curves R and C in the ﬁgure. looking for :c—values at
which the slopes of the tangent lines are equal. It appears that so = 200 is V
0 400 a good estimate. (b) R'm : C’(a:) :> 29 — 0.00042x : 12 — 0.230 + 0.0015352 :~ 0.0015:c2 ~ 0.1995850 4 17 : 0 :> a: w 192.06 (for as > 0). As in Exercise 11.R"(a:) < C"(x) : —0.00042 < —0.2 + 0.003ac 4:)
0.003% > 0.19958 c» ac > 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(w) : 500 + m(z‘) : 500 + 20x # 51:3/4 + 0.015c2.
We can solve $(p) : 320 — 7.71) for p in terms of :c to ﬁnd the demand (or price) function. 320 1 3207: — x2
x:32077.7p 2 7.7p:320—9: => 19(13): 77 “”0. R(z) : wp(m) = —7—?—.
320 9 2m (b) C'(ac) : R'(1:) => 20 — lfaz—l/4 + 0.02% : 7 7 x m 81.53 planes. and p(:c) : $30.97 million. The maximum proﬁt associated with these values is about $463.59 million. 19. (a) We are given that the demand function p is linear and p(27.000) : 10. p(33.000) : 8. so the slope is m : ﬁ and an equation of the line is y — 10 = (! 30100)(:c — 27.000) =>
3/ : 10(53): 73—059: + 19 = 19 1 (58/3000). (b) The revenue is R(x) = wp(sc) : 19:1:  (53/3000) :> R'(:c) : 19 — (cc/1500) = 0 when so : 28.500.
Since R"($) : —1/1500 < 0. the maximum revenue occurs when ac : 28.500 :> the price is p(28.500) = $9.50. SECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS 379 20. (a) Let p(a:) be the demand function. Then p(:c) is linear and y 2 p(z) passes through (20, 10) and (18. 11). so the slope is 7% and an equation of the line is y — 10 2 —%($ — 20) (2 y 2 —%m + 20. Thus. the demand is
p(;c) 2 ~§sc + 20 and the revenue is R(:c) 2 xp($) 2 —%$2 + 20:0. (b) The cost is C(x) 2 635. so the proﬁt is P(m) 2 R(:c) — C(33) 2 —%m2 + 1450. Then 0 2 P’(m) 2 2$ +14
(14) + 20 = $13. [ssh 2> :L’ 2 14. Since P”(x) 2 ~1 < 0. the selling price for maximum proﬁt is p(14) 2 — 21. (a) As in Example 3. we see that the demand function p is linear. We are given that p(1000) 2 450 and deduce that
p(1100) 2 440. since a $10 reduction in price increases sales by 100 per week. The slope forp is m 2 —1—10. so an equation isp — 450 2 —1—10(x — 1000) 0rp(a:) 2 —ﬁx + 550. (b) R(ac) =. mp(m) : —1—10$2 + 550:0. R’(x) : ~§$ + 550 = 0 when x = 5(550) : 2750.
p(2750) 2 275. so the rebate should be 450 — 275 2 $175.
(c) C(95) : 68,000 + 15010 :>
P(a.~) : Rog) — C(x) : +0352 + 550x — 68.000 — 1503: : —%cc2 + 400:: — 68.000.
P’(m) 2 2%90 + 400 2 0 when x 2 2000. p(2000) 2 350. Therefore. the rebate to maximize proﬁts should
be 450 — 350 2 $100.
22. Let 3: denote the number of $10 increases in rent. Then the price is p(;c) 2 800 + 1090. and the number of units occupied is 100 — .7}. Now the revenue is R(m) 2 (rental price per unit) X (number of units rented)
= (800 +10$)(100 — 5c) : —10$2 + 20030 + 80000 for 0 g :c g 100 :> R’(z) 2 —20x + 200 2 0 42 a: 2 10. This is a maximum since R”(x) 2 ~20 < 0 for all :0. Now we must
check the value of R(x) 2 (800 + 10:3) (100 ~ :12) at a: 2 10 and at the endpoints of the domain to see which value
ofx gives the maximum value of R. R(0) 2 80.000. R(10) 2 (900)(90) 2 81.000. and R(100) 2 (1800)(0) 2 0‘ Thus. the maximum revenue of $81.000/week occurs when 90 units are occupied at a
rent of $900 / week. 23. If the reorder quantity is m. then the manager places @ orders per year. Storage costs for the year are
x $2: ~ 4 2 2:0 dollars. Handling costs are $100 per delivery. for a total of y  100 2 80000
at :0 dollars. The total 80000 costs for the year are C(m) 2 22: +
.7; . To minimize C(90). we calculate 80.000 2 C’(m) 2 2 — 2 F0732 ~ 40,000) This is negative when :0 < 200. zero when 2: 2 200. and positive when $2
ac > 200. so C is decreasing on (0, 200) and increasing on (200. 00). reaching its absolute minimum when w 2 200. Thus. the optimal reorder quantity is 200 cases. The manager will place 4 orders per year for a total cost
of C(200) 2 $800. 380 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 24. She will have A / n dollars after each withdrawal and 0 dollars just before the next withdrawal. so her average cash
balance at any given time is %(A/n + 0) = A/(2n). The transaction costs for n withdrawals are nT. The lost interest cost on the average cash balance is [A/(2n)]  R. Thus. the total cost for 71 transactions is AR AR AR AR
C : T —. N C’ : a __ . ' __ 2
(n) n + 2n ow (n) T 2712 and C (n) 0 2712 T > n 2T >
AR . . . . H AR .
n : 27‘ the value of n that minimizes total costs Since C (n) : ——3 < 0. Thus. the optimal average cash
71 A Ax/2T _ x/AT _ AT balance is — 2n 2 2 x/AR _ x/2R 2R" 4.9 Newton's Method 1. (a) The tangent line at m : 1 intersects the arr—axis at
as m 2.3. so 382 z 2.3. The tangent line at
ac = 2.3 intersects the x—axis at cc 2 3. so x3 x 3.0. (b) :31 : 5 would not be a better ﬁrst approximation than :61 : 1 since the tangent line is nearly horizontal. In fact. the second approximation for $1 : 5 appears to be to the left of a: = 1. The tangent line at :L' : 9 intersects the m—axis at ac % 6.0. so
mg m 6.0. The tangent line at as : 6.0 intersects the ac—axis at cc % 8.0. so x3 x 8.0. 3. Since 2:1 : 3 and y : 5:5 ~ 4 is tangent to y : ﬂan) at cc 2 3. we simply need to ﬁnd where the tangent line intersects the m—axis. y — O 5332 4 R 0 5 x2 ‘ g 4. (a) (b) If :51 : 0. then :02 is negative. and $3 is even If :31 z 1. the tangent line is horizontal and more negative. The sequence of approximations Newton’s method fails. does not converge. that is. Newton‘s method fails. ...
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 Fall '08
 Dr.Tahir
 Math

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