Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

37. f(x,y,z) = xyz, g(x,y,z) = x +2y + 3z = 6 => \lf = {yz,xz,xy) = >..Vg = (>.., 2>.,3>.).
SECTION 14.8 LAGRANGE MULTIPLIERS D 233
Then >. = yz = ~xz = ~xy implies x = 2y, z = %Y· But 2y + 2y + 2y = 6 soy = 1, x = 2, z = ~ and the volume
is V = ~-
39. f(x, y, z) = xyz, g(x, y ,.z) = 4(x + y + z) = c => \7 f = (yz, xz, x y), >.Vg = (4>., 4>., 4>.). Thus
4>. = yz = xz = xy or x = y = z = fie are the dimensions giving the maximum volume.
41 . If the dimensions of the box are given by x, y, and z, then we need to find the maximum value of f(x, y, z) = xyz
[x, y, z > 0) subject to the constraint L = ..jx 2 + y 2 + z 2 or g(x, y, z) = x 2 + y 2 + z 2 = £ 2 . \7 f = ).. \lg =>
yz
xz
(yz , xz, xy ) = >.(2x, 2y, 2z), so yz = 2>.x => >. = - , xz = 2>.y => >. = - , and xy = 2.Xz =>
2 X 2 Y ·
Thus). = yz = xz .=> x 2 = y 2 [since z =f. 0) => x = y and >.= y 2
z = x 2
y. => x = z [since y =f. 0].
~ ~ X Z
Substituting into the constraint equation gives x 2 + x 2 + x 2 = L 2 => x 2 = £ 2 / 3 · => x = L / V3 = y = z and the
maximum volume is (L/V3) 3 = £ 3 / (3 V3).
43. We need to find the extreme values of f(x, y, z) = x 2 + y 2 + z 2 subject to the two constraints g(x, y , z) = x + y + 2z = 2
and h(x, y, z) = x 2 + y~ - z = 0. \lf = ( ~x , 2y, 2z}, >. \lg = (>., >., 2>.) and ~\l h = (2J.Lx_, 2~y , - ~ ). Thus we need
2x = ). + 2J.LX (1), 2y = >. + 2~y (2), 2z = 2>.- f.t (3), x + y + 2z = 2 (4), and x 2 + y 2 - z = 0 (5).
From (l) and (2), 2(x - y) = 2 ~ (x - y), so if x =f. y, ~ = 1. Putting this in (3) give,s 2z = 2>.- 1 or >. = z +~. bu t p ~1ttin g
~ = 1 into (l) says>. = 0. Hence z + ~ = 0 or z = -~. Then (4) and (5) become x + y- 3 = 0 and x 2 + y 2 + ~ ~ 0. The ·
last equation cannot be true, so this case gives no solution. So we must have x = y. Then (4) and (5) become 2x + 2z = 2 and
2x 2 -
z = 0 which imply z = 1 - x and z = 2x 2 . Thus 2x 2 = 1 - x or 2x 2 + x - 1 = (2x - 1)(x + 1) = 0 s o x= ~ or
x = - 1. The two points to check are (t, ~ .
~)and ( - 1, - 1, 2): /(~. ~. ~) = ~and f(-1, -1, 2) = 6. Thus(~,~. ~) is
the point on the ellipse nearest the origin and ( - 1, - 1, 2) is the one farthest from the origin.
45. f(x, y , z) = yex-::, g(x, y, z) = 9x 2 + 4y 2 + 36z 2 = 36, h(x, y , z) = xy + yz = 1. \7 f = >.Vg + J-L\lh =>
(yex-::, ex - z, - yex- z) = >.(18x, 8y, 72z) + i'(V, x + z, y), so yex-z = 18>.x + J.l.V, ex- z = 8>.y + ~(x + z),
- yex- :: = 72>.z + ~y . 9x 2 + 4y 2 + 36z 2 .= 36, x y + yz = 1. Using a CAS to solve these 5 equations simultaneously for x,
y, z , >. , and ~ (in Maple, use the all values command), we get 4 real-yalued solutions:
X~ 0.222444, y ~ · -2.157012, z ~ - 0.686049, >. ~ - 0.200401, ~ ~ 2.108584
X~ - 1.951921, y ~ - 0.545867, z ~ 0.119973, ). ~ 0.003141, ~ ~ - 0.076238
X ~ 0.155142, y ~ 0.904622, z ~ 0.950293, >. ~ - 0.012447, f.' ~ 0.489938
X~ 1.138731, y ~ 1.768057, . z ~ -0.573138, >. ~ 0.317141, f.' ~ 1.862675
Substituting these values into f gives /(0.222444, -2.157012, - 0.686049) ~ -5.3506,
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