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Chapter 4 98

# Chapter 4 98 - 360 El CHAPTER4 APPLICATIONS OF...

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Unformatted text preview: 360 El CHAPTER4 APPLICATIONS OF DIFFERENTIATION 2 2 The area of the rectangle is (2:5)(2y) = 4mg. Now 50—2 —I— %2- z 1 gives a b . . b y = —\/a2 —- m2, so we max1mize Am) 2 4—3: Va2 — \$2. a 20. a .%(a2 i \$2)_1/2(—2\$) + (a2 — 932)1/2-1] _2_2—1/2_2 2‘2_ 45 2_2 a(a m) [ze—l—a x]———a aZ—m2[a 2.16] So the critical number is :c = ﬁa, and this clearly gives a maximum. Then 3/ : «Lib, so the maximum area is 4(ﬁa) (ﬁb) = 2ab. 21. The height h of the equilateral triangle with sides of length L is g L, - 2 2 _ 2 2 _ 2 1 2 _ 3 2 smceh +(L/2) —L => h —L 7 4L 7 EL => ﬁLV EL _ ﬂ ‘. . . . 2 y _ 2 h — 2 L. Uslng Simllar trlangles, i L /2 x/gm:§L—y => y=§L—\/§\$ => y:l/§-§(L*2:II). =\/§=> The area of the inscribed rectangle is A(1:) : (2x)y : ﬁﬂL 7 2m) : x/ng — 2 V3302, where 0 g m g L/2. Now 0 : A’(:E) : x/gL — 4\/§m => 3: = x/gL/(Zix/g) = L/4. Since A(0) : A(L/2) : 0, the maximum occurs when a: : L/4. and y : éL i -‘:—§L : 39L. so the dimensions are L/2 and éL. The rectangle has area A(m) = 2363; : 23(8 7 3:2) : 16m i 2353. where 0§x§2\/§.NowA’(m) :164522 :0 :> w=2\/§.smce A (0) : A(2\/§) : 0. there is a maximum when as = 2 ﬂ. Then _ 1_6 ' ' 2 E y — 3 . so the rectangle has dimensmns 4 \/: and 3 . 23. T The area of the triangle is I A(\$) = %(2t)(r+m):t(r+w) :x/r2—cc2(r+:c).Then _233 72m O_A’ ” l V73 x2 I w ‘ i (m) Tgm 2 7‘2 — 372 \$2+r\$ 2 _— l 'r a: 73—23 __\$ +T\$ 2m => 132+T\$2r27m2 => 0:2\$2+T\$47‘2=(2\$—T)(\$+T) :> ...
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