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

242 0 CHAPTER 14 PARTIAL DERIVATIVES
57. f(x, y) = x 3 - 3x + y 4 - 2y 2 2
z 0 l\ "'' 0 and f xx( -1, 0) = - 6 0 and / u(1, ±1} = 6 > 0 indicating local minima / (1, ± 1) = - 3; and D ( -1, ± 1) = - 48 and
D(1, 0) = - 24, indicating saddle points.
59. f(x,y) = x 2 y, g(x,y) = x 2 +y 2 = 1 ~ 'ilf = (2xy,x 2 ) = )..'iJg = (2>.x,2>.y). Then 2xy = 2)..x impliesx = Oor
y = >.. Ifx = 0 then x +
2 y 2 = 1 gives y = ±1 and we have possible points (0, ±1) where f (0, ± 1) = 0. lfy =).. then
x 2 = 2)..y implies x 2 = 2y 2 and substitution into x 2 + y 2 = 1 gives 3y 2 = 1 ~ y = ± 7:J and x = ±jf. The
corresponding possible points are ( ±jf, ±7:J). The absolute maximum is f ( ± jf, -Ja) = ~ while the absolute
61. f(x, y, z) = xyz, g(x,y,z) = ~ 2 + y 2 + z 2 = 3. 'ilf = )..'iJg ~ (yz,xz,xy) = >.(2x, 2y,2z). lfany ofx, y, or z is
zero, then x = y = z = 0 which contradicts x 2 + y 2 + z 2 = 3. Then ).. = YZ ·=x 2 2 z = xy ~ 2y2 z = 2x 2 z ~
x y 2z
. .
y 2 = x 2 , and similarly 2yz 2 = 2x 2 y ~ z 2 = x 2 • Substituting into the constraint equation gives x 2 + x 2 + x 2 = 3 ~
I
x 2 = 1 = y 2 = z 2 . Thus the possible points are (1, 1, ± 1), (1, -1, ± 1), ( - 1, 1, ±1), ( - 1, -1, ±1). The absolute maximum
is / (1, 1, 1) = /(1, -1, - 1) = f( -1, 1, -1) = f( - 1, - 1, 1) = 1 and the absolute
minimum is /(1, 1, -1) = /(1, ....:.1, 1) = f( -1,1, 1) = f( - 1, - 1, - 1) = - 1.
63. f(x,y,z) = x 2 +y 2 +z 2 , .g(x,y,z) = xy 2 z 3 = 2 ~ 'ilf = (2x, 2y,2z) = X'ilg = (>.y 2 z 3 , 2>.xyz 3 ,3)..xy 2 z 2 ).
Since xy 2 z 3 = 2, x =I 0, y =I 0 and z =I 0, so 2x = )..y 2 z 3 (I), 1 = )..xz 3 (2), 2 = 3>.xy 2 z (3~ . Then (2) and (3) imply
~ = - 3
2
2
or y 2 = f z 2 soy = ±z q. Similarly (1) and (3) imply ~x 3
= - 3
2
? or 3x 2 = z 2 sox = ± -jsz. But
xz xy z V 3 y z xy-z
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