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

222 D CHAPTER 14 PARTIAL DERIVATIVES
2 2) 2 2 ( ) 2 2 2 2 . 2 2
l" = ( x + y e'~~ - :r - 2x + 2xe'~~ -:r = 2x e'~~ - :r (1 - x - y ),
fzx = 2xe'fl 2 - o: 2 (- 2x) + (1- x 2 - y 2 ) ( 2x ( -2xe'~~ 2 -:r 2) + 2e'fl 2 - :r 2) = 2e'~~ 2 -:r 2 ((1- x 2 - y 2 )(1 - 2x 2 ) - 2x 2 ),
22 22 2 2 22
f xv = 2xe'11 - o: (- 2y) + 2x(2y)eY -:r (1 - x - y ) = ~ 4xyeY - x (x 2 + y 2 ),
f 11
v = 2yeV
2
- o:
2
(2y) + (1 + x
2
+ y
2
)
(
2y
(
2yeV
2 2)
-:r + 2eY
2 2 )
- :r = 2e 71 2 2
- :r ((1 + x 2 + y 2 )(1 + 2y 2 ) + 2y 2 ).
fv = 0 implies y = 0, and substituting into fx = 0 gives
2xe-"' 2 (1 - x 2 ) = 0 ~ x = o. or x = ± 1. Thus the critical points are
(0, 0) and (±1, 0). Now D(O, 0) = (2)(2) - 0 > 0 and f :r:x (O, 0) = 2 > 0,
so f(O, 0) = 0 is a local minimum. D(±l, 0) = ( - 4e- 1 )(4e- 1 )- 0 < 0
so (±1, 0) are saddle points.
17.f(x,y)=y 2 - 2ycosx ~
j ; =2ysinx,fv=2y -2cosx,
fxx = 2y cos x, f ., 11 = 2 sin x, f v 11 = 2. Then/:. = 0 implies y = 0 or
sinx = 0 ~
x = 0, 1r, or 211" for -1 :-::; x :-::; 7. Substituting y = 0 into
f v = 0 gives cosx = 0 ~ x = ~or 3 ;, substituting x = 0 or x = 21r
into f 71 = 0 gives y = 1, and substituting x = 1r into fv = 0 gives y = - 1.
Thus the critical points are (0, 1), ( ~. 0), (1r, -1), ( 3 ;, 0), and (21r, 1).
D (~ , 0) = D e 2
"", 0) = -4 < 0 so (~, 0) and e;, 0) are saddle points. D(O, 1) = D(1r, -1) = D(21r, 1) = 4 > 0 and
fa:x(O , 1) = fxx('Tr, - 1) = fxo:(27r, 1) = 2 > 0, so f(O, 1) = j(1r, -1) = f(27r, 1) = - 1 are local minima.
19. f(x, y) = x 2 + 4y 2 - 4xy + 2 ~ fx = 2x- 4y, fv =By - 4x, f x:r: = 2, f,v = -4, / 1171 = 8. Then fx = 0
and fv = 0 each implies y = ~x, so all points of the form (xo, ~xo) are critical points and for each ofth~se we have
D(xo, ~xo) = (2)(8) - (-4) 2 = 0. The Second Derivatives Test gives no information, but
f(x,y) = x 2 +4y 2 - 4xy + 2 = (x- 2y} 2 + 2 ;:: 2 with equality if and only ify = ~x. Thus f(x 0 , ~xo) = 2 are all local
(and absolute) minima.
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