Solução_Calculo_Stewart_6e

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F. TX.10 CHAPTER 15 REVIEW ET CHAPTER 14 ¤ 219 xy 2 z 3 =2so x and z must have the same sign, that is, x = 1 √ 3 z.Thusg(x, y, z) =2implies 1 √ 3 z 2 3 z2 z 3 =2or z = ±3 1/4 and the possible points are (±3 −1/4 , 3 −1/4√ 2, ±3 1/4 ), (±3 −1/4 , −3 −1/4√ 2, ±3 1/4 ). However at each of these points f takes on the same value, 2 √ 3.But(2, 1, 1) also satisfies g(x, y, z) =2and f(2, 1, 1) = 6 > 2 √ 3. Thus f has an absolute minimum value of 2 √ 3 and no absolute maximum subject to the constraint xy 2 z 3 =2. Alternate solution: g(x, y, z) =xy 2 z 3 =2implies y 2 = 2 xz 3 , so minimize f(x, z) =x2 + 2 xz 3 + z2 .Then f x =2x − 2 x 2 z 3 , f z = − 6 xz 4 +2z, f xx =2+ 4 x 3 z 3 , f zz = 24 xz 5 +2and f xz = 6 x 2 z 4 .Nowf x =0implies 2x 3 z 3 − 2=0or z =1/x. Substituting into f y =0implies −6x 3 +2x −1 =0or x = ± √ 1 , ± 4√ 3 .ThenD ± 1 4 4 3 √ , ± 4√ 3 3 =(2+4) 2+ 24 3 1 4 √ 3 ,sothetwocriticalpointsare 2 − √ 6 3 > 0 and fxx ± √ 1 , ± 4√ 3 =6> 0, so each point 4 3 is a minimum. Finally, y 2 = 2 xz , so the four points closest to the origin are ± 1 3 √ , √ 2 4 3 4√ , ± 4√ 3 , ± √ 1 , − √ 2 3 4 3 4√ , ± 4√ 3 . 3 65. The area of the triangle is 1 2 ca sin θ and the area of the rectangle is bc. Thus, the area of the whole object is f(a, b, c) = 1 ca sin θ + bc. The perimeter of the object 2 is g(a, b, c) =2a +2b + c = P . To simplify sin θ in terms of a, b,andc notice that a 2 sin 2 θ + 1 2 c2 = a 2 ⇒ sin θ = 1 √ 4a2 − c 2a 2 .Thus f(a, b, c) = c 4 √ 4a2 − c 2 + bc. (Instead of using θ, we could just have used the Pythagorean Theorem.) As a result, by Lagrange’s method, we must find a, b, c,andλ by solving ∇f = λ∇g which gives the following equations: ca(4a 2 − c 2 ) −1/2 =2λ (1), c =2λ (2), 1 4 (4a2 − c 2 ) 1/2 − 1 4 c2 (4a 2 − c 2 ) −1/2 + b = λ (3),and 2a +2b + c = P (4). From(2), λ = 1 2 c and so (1) produces ca(4a2 − c 2 ) −1/2 = c ⇒ (4a 2 − c 2 ) 1/2 = a ⇒ 4a 2 − c 2 = a 2 ⇒ c = √ 3 a (5). Similarly, since 4a 2 − c 21/2 = a and λ = 1 c, (3) gives a 2 4 − c2 4a + b = c , so from 2 (5), a 4 − 3a 4 + b = √ 3 a 2 ⇒ − a 2 − √ 3 a 2 = −b ⇒ b = a 2 1+ √ 3 (6). Substituting (5) and (6) into (4) we get: 2a + a 1+ √ 3 + √ 3 a = P ⇒ 3a +2 √ P 3 a = P ⇒ a = 3+2 √ 3 = 2 √ 3 − 3 P and thus 3 √ √ 2 3 − 3 1+ 3 b = P = 3 − √ 3 P and c = 2 − √ 3 P . 6 6
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