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Chapter 4 158

# Chapter 4 158 - 420 CHAPTER 4 APPLICATIONS OF...

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Unformatted text preview: 420 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (c) From the ﬁgure in part (a). the width is 2m and the depth is 2y. so the strength is S = k(2:c)(2y)2 2 81¢ny = 8kat(100 — x2) : 800m; — 8km3. 0 g m g 10. dS/da: : 800k — 24km2 = 0 when24km2—800k —> mzﬁigo —> xﬁfg —> y—(/-2%—ﬂj§—x/§m.8ince 5(0) 2 3(10) 2 O. the maximum strength occurs when m = %. The dimensions should be % x 11155 inches by 2%.? m 16.33 inches. 78. (a) .V y = (tan 0)m — ﬂaw—26362. The parabola intersects the 9 2 _113 :> 2122 COS2 0 line when (tan (1)3: : (tan 6)\$ — (tan!) — tan (1)2122 cos2 0 I: 7 A? g . 0 . 2 2 - - 2 Rm): x : 5111 k Slna 21) cos I9 : Sln6 _ sma (c0s0 cosa)2v €056 cos (1 cos 0 cos 05 9 cos a cos 6 cos a 9 cos2 a 2 2 2 2 = (sin6 cosa i sina cos (NM : sin(6 — 00w 9 cos2 a 9 cos2 a (b) R'(6) * 2112 [0050 - c0s(0 i a) —|— sin(9 — a)(— sin 0)] m 2122 cos[0 + (6 i (1)] — 900320 _ gcos2a 2 : 2712 005(20701):Owhencos(2c9—a)=0 => 26—a:% => gcos oz 7r/2 —I— (1 7r (1 . i . . . . 0 2 —— : — + —i The First Derivative Test shows that this gives a max1mum value for R09). 2 4 2 [This could be done without calculus by applying the formula for sin .7: cos y to R(0).] 202 cos 6 sin(6 —I— oz) (c) Replacmg a by —a 1n part (a). we get R09) : g c052 a Proceeding as in part (b). or simply by replacing a by —a in the result of part (b). we see that R09) is maximized when 9 : g i g. k cos 9 k(h/ d) h h h 79. I: : :k—2k—:k_—— :> (a) d2 d2 d3 («402 + h2 )3 (1600 + h2)3/2 d] k (1600 + h2)3/2 , hg (1600 + hZ) “2 ~ 2h k (1600 + h2)1/2 (1600 + h? — 3h2) dh [(1600 + WW] (1600 + h ) k 1 , 2h2 : ﬂWS/l [k is the constant of proportionality] 1600 + Set dI/dh : 0: 1600 2h2 0 h2 800 h M800 20 ﬂ. By the First Derivative Test, I has a local maximum at h : 20 \/§ 2 28 ft. ...
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