Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 15.7 MAXIMUM AND MINIMUM VALUES ET SECTION 14.7 ¤ 199
7. f(x, y) =x 4 + y 4 − 4xy +2 ⇒ f x =4x 3 − 4y, f y =4y 3 − 4x,
f xx =12x 2 , f xy = −4, f yy =12y 2 .Thenf x =0implies y = x 3 ,
and substitution into f y =0 ⇒ x = y 3 gives x 9 − x =0 ⇒
x(x 8 − 1) = 0 ⇒ x =0or x = ±1. Thus the critical points are (0, 0),
(1, 1),and(−1, −1). NowD(0, 0) = 0 · 0 − (−4) 2 = −16 < 0,
so (0, 0) is a saddle point. D(1, 1) = (12)(12) − (−4) 2 > 0 and
f xx (1, 1) = 12 > 0, sof(1, 1) = 0 is a local minimum.
D(−1, −1) = (12)(12) − (−4) 2 > 0 and
f xx =(−1, −1) = 12 > 0,sof(−1, −1) = 0 is also a local minimum.
9. f(x, y) =(1+xy)(x + y) =x + y + x 2 y + xy 2 ⇒
f x =1+2xy + y 2 , f y =1+x 2 +2xy, f xx =2y, f xy =2x +2y,
f yy =2x. Thenf x =0implies 1+2xy + y 2 =0and f y =0implies
1+x 2 +2xy =0. Subtracting the second equation from the first gives
y 2 − x 2 =0 ⇒ y = ±x,butify = x then 1+2xy + y 2 =0 ⇒
1+3x 2 =0which has no real solution. If y = −x then
1+2xy + y 2 =0 ⇒ 1 − x 2 =0 ⇒ x = ±1, so critical points are (1, −1) and (−1, 1).
D(1, −1) = (−2)(2) − 0 < 0 and D(−1, 1) = (2)(−2) − 0 < 0,so(−1, 1) and (1, −1) are saddle points.
11. f(x, y) =x 3 − 12xy +8y 3 ⇒ f x =3x 2 − 12y, f y = −12x +24y 2 ,
f xx =6x, f xy = −12, f yy =48y. Thenf x =0implies x 2 =4y and
f y =0implies x =2y 2 . Substituting the second equation into the first
gives (2y 2 ) 2 =4y ⇒ 4y 4 =4y ⇒ 4y(y 3 − 1) = 0 ⇒ y =0or
y =1.Ify =0then x =0and if y =1then x =2, so the critical points
are (0, 0) and (2, 1).
D(0, 0) = (0)(0) − (−12) 2 = −144 < 0, so(0, 0) is a saddle point.
D(2, 1) = (12)(48) − (−12) 2 =432> 0 and f xx(2, 1) = 12 > 0 so f(2, 1) = −8 is a local minimum.
13. f(x, y) =e x cos y ⇒ f x = e x cos y, f y = −e x sin y.
Now f x =0implies cos y =0or y = π + nπ for n an integer.
2
But sin π
2 + nπ 6=0,sotherearenocriticalpoints.