Solução_Calculo_Stewart_6e

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
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