Chapter 4 108

# Chapter 4 108 - 370 D CHAPTEB4 APPLICATIONS OF...

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Unformatted text preview: 370 D CHAPTEB4 APPLICATIONS OF DIFFERENTIATION 55. 56. 57. It sufﬁces to maximize tan 6. Now 315 - —=tan(w+6)= tanw—I—tan6 : t+tain6. 1 litany} ta116 1~ttan6 So3t(1—ttan6)=t+tan6 ;» 2t=(1+3t2)tan6 2 2t 2t t‘ 6: Lt tzt 6: an ef() an 1+3t2 3r__,. 1+3t2' , _ 2(1+3t2) —2t(6t) _ 2(1—3t2) _ f(t)— (1+3t2)2 _ (H31?)2 _0 a} 1—3t2:0 c) tzﬁsincetZO. Now f’(t) > 0for0 S t < T}? and f’(t) < 0 fort > i so f has an absolute maximum when t : i 20%) 1+3(1/\/§)2 x/g:tan('¢2+%) => 1/):g. :; and tan 6 — Jg 6 i 2'. Substituting for t and 6 in 315 : tan(1/1 + 6) gives us We maximize the cross—sectional area A (0) : 10h + 2(gdh) : 10h + dh : 10(10 sin 0) + (10cos 6)(10 sin6) =100(sin6 + sin6 cos 6). 0 S 6 S g A] (6) : 100(cos6 + cos2 6 ~ sin2 6) : 100(cos6 + 2 cos2 6 i 1) : 100(2cos 6 — 1)(cos 6 + 1) :Owhencos621 c> 6:? (c0367é—1since0g6gg.) [0| Now A(0) : 0.14%) : 100 and A(§) : 75 \/§ m 129.9. so the maximum occurs when 6 : g 5 2 . From the ﬁgure. tana : -— and tanﬁ : . Since 1: 3 — z 0 ,1 5 _1 2 a+5+6v180 77n6—7r tan tan :> 5 a: 3 w 2: d6 1 < 5 > 1 [ 2 J 2 d2: ’ 2 2 2 2 5 13 2 (3 — x) 1 — 1 A P B 1 x2 5 (3 1 m2 2 _z2—I—25 2:2 (373:)2+4 (3—213)2 d6 5 2 2 2 — — : 2 I 50 — 5 i 30 65 :> NOW dzr 0 \$24—25 m2—6\$+13 a: a: \$+ 3x2 1 3030 + 15 : 0 :5 m2 —10:c + 5 : 0 => :17 : 5 :: 2 i/S. We reject the root with the + sign, since it is larger than 3. d6/dw > 0 for m < 5 A 2 \/5 and d6/da: < 0 for z > 5 — 2 \/§ so 6 is maximized when |API = is : 5 , 2\/5 m 0.53. ...
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