Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 15.8 LAGRANGE MULTIPLIERS ET SECTION 14.8 ¤ 209
25. Let the sides of the rectangle be x and y. Thenf(x, y) =xy, g(x, y) =2x +2y = p ⇒ ∇f(x, y) =hy, xi,
λ∇g = h2λ, 2λi. Thenλ = 1 2 y = 1 2 x implies x = y and the rectangle with maximum area is a square with side length 1 4 p.
27. Let f(x, y, z) =d 2 =(x − 2) 2 +(y − 1) 2 +(z +1) 2 , then we want to minimize f subject to the constraint
g(x, y, z) =x + y − z =1. ∇f = λ∇g ⇒ h2(x − 2), 2(y − 1), 2(z +1)i = λ h1, 1, −1i, sox =(λ +4)/2,
y =(λ +2)/2, z = −(λ +2)/2. Substituting into the constraint equation gives λ +4
2
+ λ +2
2
+ λ +2
2
=1 ⇒
3λ +8=2 ⇒ λ = −2,sox =1, y =0,andz =0. This must correspond to a minimum, so the shortest distance is
d = (1 − 2) 2 +(0− 1) 2 +(0+1) 2 = √ 3.
29. Let f(x, y, z) =d 2 =(x − 4) 2 +(y − 2) 2 + z 2 . Then we want to minimize f subject to the constraint
g (x, y, z) =x 2 + y 2 − z 2 =0. ∇f = λ∇g ⇒ h2(x − 4) , 2(y − 2) , 2zi = h2λx, 2λy, −2λzi,sox − 4=λx,
y − 2=λy,andz = −λz. From the last equation we have z + λz =0 ⇒ z (1 + λ) =0,soeitherz =0or λ = −1.
But from the constraint equation we have z =0 ⇒ x 2 + y 2 =0 ⇒ x = y =0which is not possible from the first two
equations. So λ = −1 and x − 4=λx ⇒ x =2, y − 2=λy ⇒ y =1,andx 2 + y 2 − z 2 =0 ⇒
4+1− z 2 =0 ⇒ z = ± √ 5. This must correspond to a minimum, so the points on the cone closest to (4, 2, 0)
are 2, 1, ± √ 5 .
31. f(x, y, z) =xyz, g(x, y, z) =x + y + z =100 ⇒ ∇f = hyz, xz, xyi = λ∇g = hλ, λ, λi. Thenλ = yz = xz = xy
implies x = y = z = 100
3 .
33. If the dimensions are 2x, 2y, and2z, thenmaximizef(x, y, z) =(2x)(2y)(2z) =8xyz subject to
g(x, y, z) =x 2 + y 2 + z 2 = r 2 (x>0, y>0, z>0). Then ∇f = λ∇g ⇒ h8yz, 8xz, 8xyi = λ h2x, 2y, 2zi ⇒
8yz =2λx, 8xz =2λy,and8xy =2λz,soλ = 4yz
x
= 4xz
y
= 4xy
z .Thisgivesx2 z = y 2 z ⇒ x 2 = y 2 (since z 6= 0)
and xy 2 = xz 2 ⇒ z 2 = y 2 ,sox 2 = y 2 = z 2 ⇒ x = y = z, and substituting into the constraint
equation gives 3x 2 = r 2 ⇒ x = r/ √ 3=y = z. Thus the largest volume of such a box is
r
f √
r
3
, √
r
3
, √
r
3
=8 √ √
r
√
r
3 3 3
= 8
3 √ 3 r3 .
35. f(x, y, z) =xyz, g(x, y, z) =x +2y +3z =6 ⇒ ∇f = hyz, xz, xyi = λ∇g = hλ, 2λ, 3λi.
Then λ = yz = 1 xz = 1 xy implies x =2y, z = 2 y.But2y +2y +2y =6so y =1, x =2, z = 2 and the volume
2 3 3 3
is V = 4 3 .
37. f(x, y, z) =xyz, g(x, y, z) =4(x + y + z) =c ⇒ ∇f = hyz, xz, xyi, λ∇g = h4λ, 4λ, 4λi. Thus
4λ = yz = xz = xy or x = y = z = 1 c are the dimensions giving the maximum volume.
12
39. If the dimensions of the box are given by x, y, andz, thenweneedtofind the maximum value of f(x, y, z) =xyz
[x, y, z > 0] subject to the constraint L = x 2 + y 2 + z 2 or g(x, y, z) =x 2 + y 2 + z 2 = L 2 . ∇f = λ∇g ⇒
hyz, xz, xyi = λh2x, 2y, 2zi,soyz =2λx ⇒ λ = yz
xz
xy
, xz =2λy ⇒ λ = ,andxy =2λz ⇒ λ =
2x 2y 2z . Thus