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Chapter 4 105 - SECTION“ OPTIMIZATION PROBLEMS U 367 47...

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Unformatted text preview: SECTION“ OPTIMIZATION PROBLEMS U 367 47. Here 52 : h2 + b2/4. $0112 2 52 ~ b2/4. The area is A = %b Ms2 — b2/4. S 5 Let the perimeter be p, so 25 + b = p or s = (p — b)/‘2 : A(b) = gin/(p — (2)2/4 - b2/4 = b Mp? — 2pb/4. Now %r_—/ / 2 _ g ‘2 b A’(b) = __p_2pb — JL — fl. Therefore. A’(b) = 0 => 4 w/pz—Qpb—4 p2—2pb —3pb + p2 = 0 => b : p/3. Since A’(b) > 0 for b < 17/3 and A’(b) < O for b > 19/3. there is an absolute maximum when b : p/3. But then 25 +p/3 — p. so 3 — p/3 —,> s — b the triangle is equilateral. 48. See the figure. The area is given by A(m) : ;(2m)x+ azmxm) : W(w+ 3:2 +b2 _ a2) for ogxgaNowA’m: a2 w2<ll m )l(ml :02lb2 (mi—:0 41> /w2+b2—a2 (12-352 Except for the trivial case where as = O. a = b and A(m) = O. we have ac + \/ 9:2 + b2 — a2 > 0. Hence. cancelling this factor gives 2 2 ,,,,AT,,.,,, :c x/a —x . (0. x2), _ mx/zr2+b2~a2:a2ix2 :> Va?‘ —:c2 $162-I-b2 —a2 x/x2+b2—7 x2 (x2 + b2 7 a2) : a4 i 2a2122 + $4 => x2 (b2 — a2) : a4 i 2(12932 (12 Now we must check the value of A at this point as well as at the endpoints of the domain to see which gives the :> w2(b2+a2):a4 :> 3:: maximum value. A(O) : (Ix/b2 — a2. A(a) : 0 and ( “2 I 2 (#2 >2 “2 i “2 )2 A —— , a : +b2wa2 ¢a2+b2 a2+b2 m m ab [ a2 I b2 ab(a2 + b2) b 7 I 7 a /a2 + b2 /a2 + b2 \/a2 + b2 a2 + b2 2 Since b 2 \/ b2 — a2. A(a2 V a2 + b2 ) 2 A (0). So there is an absolute maximum when a: : fl. In this a case the horizontal iece should be ib— d th t’ l ' ‘h 1d b a2 + b2 ~ 2 b2 p Wan everica pieces ou eWiVa + ‘ ...
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