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Chapter 4 103

# Chapter 4 103 - SECTION 4.7 OPTIMIZATION PROBLEMS 365 d5...

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Unformatted text preview: SECTION 4.7 OPTIMIZATION PROBLEMS 365 d5 1 (3056 1 . . . —— - , 0 — —> - 3 . — 0 —> c050 = —. The First Derivative (b) d6 —0 when cscO— x/gcot 0 sinB \/— smi9 J5 Test shows that the minimum surface area occurs when 0 : c0s’1(%) z 55°. (c) If cos 0 = i. then cot t9 : «ii and csc 0 = % so the surface area is ,, . 3.21 2&3; ___3_2 i2 ‘/3 \/2 Szﬁsh—55ﬁ+3\$ 2%—63h 2\/§Ls+2\/§s :6sh+ nfs 21:63(h+ 2—fs) l 40. 15 km/h N Let t be the time, in hours. after 2:00 RM. The position of the boat heading W E south at time t is (07 720t). The position of the boat heading east at time t is (A15 + 1515. 0). If D(t) is the distance between the boats at time t, we minimize f(t) : [pm]2 : 2022:2 + 15% —1)2. 20 Mb I E f’(t) = 800t + 450(t , 1) _ 1250i: — 450: 0 when t: {12—5050 , _.0 36 h. S 0.36 h x goh'ﬂ = 21.6 min = 21 min 36 5. Since f"(t) > 0. this gives a minimum, so the boats are closest together at 2:21 :36 RM. / 2 2 7 1 41. HereT(m):\$—+5+5 (m)— ‘T _0 9 8m_6i/x2+25 6 8 61/302 + 25 8 x 161’2 — 9(272 I 25) 1-5 a:- 3/51. But 3/5— > 5 s0 T has no critical number. Since T(O) % 1.46 and T(5) % 118. he should row directly to B. 42. B In isosceles triangle AOB. LO 2 1800 i 0 i 0. so ZBOC : 20. The ‘ distance rowed is 4 cos 0 while the distance walked is the length of arc BC : 2(26) : 40. The time taken is given by A C 4 T(0) = C356 +416 :2c056+0. 0363 g. T’(0):i2sin0+1=0 <=> sinO: => 92’ l 1 2 6 ' Check the value ofT at 0 : g and at the endpoints of the domain of T; that is, 9 : 0 and 0 = %. T(O) : 2, T(%) : \/§ +% m 2.26, and T(%) : g m 1.57. Therefore, the minimum value ofT is g when 0 : g; that is, the woman should walk all the way. Note that T"(6) = —2 cos 9 < 0 for 0 S 9 < 325. so 19 : % gives a maximum time. 43. The total illumination is 1(ac) = % + —k— 0 < a: < 10. Then 3k m2 (10 — ac)2 k —6l~c 2k: {g—Xﬂmoiiﬁg I’(m):?+m:0 :> 6k(10—x)3:2km3 :> ‘0 3(10—m)3:x3 => \3/§(107\$):x 2 ioV—S/Emmc =>10\3/?_)=\$+\3/§\$ => 10%:(1+\P/§)m => 109/3 1+??? O<ac<10. x: R: 5.9 ft. This gives a minimum since 1"(x) > 0 for ...
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