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

SECTION 14.7 MAXIMUM AND MINIMUM VALUES 0 221
1 '+ 2x 2 + x 2 = 0 =} 3x 2 = -1 which has no real solution. lfy = x
then substitution into f, = 0 gives 1 - 2x 2 + x 2 = 0 =} x 2 = 1 =}
x = ±1, so the critical points are (1, 1) and (-1, -1). Now
D (1, 1) = (- 2)(2) - 0 2 = - 4 < 0 and
D( - 1, - 1) = (2)(-2)- 0 2 = -4 < 0, so (1, 1) and (-1, - 1) are
saddle points.
9. f(x, y) = y 3 + 3x 2 y - 6x 2 - 6y 2 + 2 =} f ,. = 6xy - 12x, ! 11 = 3y 2 + 3x 2 - 12y, f x:t = 6y- 12, f xu = 6x,
fw = 6y - 12. Then f z = 0 implies 6x(y- 2) = 0, sox = 0 or y = 2. If x = 0 then substitution into j y = 0 gives
3y 2 - 12y = 0 =} 3y(y- 4) = 0 =} y = 0 or y = 4, so we have critical points (0, 0) and (0, 4). If y = 2,
substitution into jy = 0 gives 12 + 3x 2 - 24 = 0 =} x 2 = 4 =}
x = ±2, so we have critical points (±2, 2).
D(O, 0) = (- 12)( -12) - 0 2 = 144 > 0 and fu(O, 0) = - 12 < o •. so
f(O, 0) = 2 is a local maximum. D{O, 4) = (12){12) - 0 2 = 144 > 0
and f xx(O, 4) = 12 > 0, so f(O , 4) = -30 is a local minimum.
D(±2, 2) = {0)(0)- {±12) 2 = -144 < 0, so (±2, 2) are saddle points.
y
11. f(x, y) = x 3 - 12xy + _8y 3 =} fx = 3x 2 - 12y, f u = - 12x + 24y 2 , fxx = 6x, fx 11 = - 12,/ 1111 = 48 y. Then fx = 0
implies x 2 ·= 4y and / 11 = 0 implies x = 2y 2 • Substituting the second equation into the first gives {2y 2 } 2 = 4y =}
4y 4 = 4y =} 4y(y 3 - 1) = 0 =} y = 0 or y = 1. lfy = 0 then
x = 0 and if y = 1 then x = 2, so the critical points are (0, 0) and {2, 1).
D(O, 0) = {0){0) - ( - 12) 2 = - 144 < 0, so (0, 0) is a saddle point.·
D(2, 1) = (12)(48)- ( -12? = 432 > 0 and fxx(2, 1) = 12 > 0 so
!(2, 1) = -8 is a local .minimum.
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13. f(x,y)=e"'cosy =} f x=e"'cosy, / 11 =-e"'siny.
Now f a: = 0 implies cosy = 0 or y = f + mr for nan integer.
But sin ( f + mr) =1- 0, so there are no critical points.
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