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

SECTION 14.8 LAGRANGE MULTIPLIERS 0 231
the possible points are ( ±-/2, ± 7:i, 72), ( ±V-2, ±~, - );) . Hence the maximum off subject to the constraints is
f(±J2,±~,±jz ) = ~ andtheminimumisf(±J2,±~, 'f72) = ~·
Note: Since xy = 1 is one of the constraints we could have solved the problem by solving f (y, z) = yz + 1 subject to
y2 + z2 = 1.
19. f(x, y) = x 2 + y 2 + 4x - 4y. For the interior of the region, we find the critical points: fx = 2x + 4, f u = 2y - 4, so the
only-critical point is ( - 2, 2) (which is inside the region) and f ( - 2, 2) = -8. For the boundary, we use Lagrange multipliers.
g(x, y) = x 2 + y 2 = 9, so "iJ f = >. "ilg =;. (2x + 4, 2y - 4) = (2>.x, 2>.y). Thus 2x + 4 = 2>.x and 2y - 4 = 2>.y.
Adding the two equations gives 2x + 2y = 2>.x + 2>.y =:. x + y = >.(x + y) =:. (x + y)(>.- 1) = 0, so
x + y = 0 =;. y = -x or>. - 1 = 0 =;. >. = 1. But >. = 1 leads to a contradition in 2x + 4 = 2>.x, soy = - x and
x 2 + y 2 = 9 implies 2y 2 = 9 => y = ±12. We have f ( ~· -72) = 9 + 12-/2 ~ 25.97 and
r( - 12, ~ ) = 9- 12-/2 ~ - 7.97, so the maximum value off on the disk x 2 ~ y 2 :s: 9 is f ( 72· -12) = 9 + 12-/2
and the minimum is f( - 2, 2) = -8.
21. f(x, y) = e- xy. For the interior ofthe region, we find the critical points: f x = - ye-xu, / 11
= -xe-"'Y, so the only
critical point is (0, 0), and f(O, 0) = 1. For the boundary, we use Lagrange multipliers. g(x, y) = x 2 + 4y 2 = 1 =;.
>. "iJ g = (2>.x, 8>.y), so setting "iJ f = >. "iJ g we get -ye-"' 11 = 2>.x and -xe-"' 11 = 8>.y. The first of these gives
e- "' 11 = -2>.xjy, and then .the second gives - x( - 2>.xjy) = 8>.y => x 2 = 4y 2 • Solving this last equation with.the
constraint x 2 + 4y 2 = 1 gives X= ±72 and y=±~ . Now t( ±~, 'f 2~) = e 1 1 4 ~ 1.284 and
f ( ±~ , ±;~ ) = e- 1 / 4 ~ 0.779. The former arc the maxima on the region and the latter are the min.ima.
23. (a) f(x, y) = x , g(x,y) = y 2 + x 4 - x 3 == 0 =;. "ilf = (1, 0) = >.Vg = >.(4x 3 - 3x 2 ,2y). Then
1 = >.( 4x 3 - 3x 2 ) (1) and 0 = 2>.y (2). We have >. i= 0 from (1), so (2) gives y = 0. Then, from the constraint equation,
x 4 - x 3 =. 0 =;. x 3 (3; - 1) = 0 =;. x = 0 or x = 1. But x = 0 contradicts (1), so the only possible extreme value
subject to the constraint is /(1, 0) = 1. (The question remains whether this is indeed the minimum of f.)
(b) The constraint is y 2 + x 4 - x 3 = 0 ¢} y 2 = x 3 - x 4 . The left side is non-negative, so we must have x 3 - x 4 :2: 0
which is true only for 0 :S: x :S: 1. Therefore the minimum possible value for f(x, y) =x is 0 which occurs for x = y = 0.
However,>. "iJ g(O, 0) = >. (0- 0, 0) = (0, 0) and "iJ f(O, 0) = (1, 0), so "iJ f(O, 0) # >. "iJ g(O, 0) for all values of>..
(c) Here "iJ g(O, 0) = 0 but the method of Lagrange multipliers requires that "iJ g # 0 everywhere on the constraint curve.
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