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

SECTION 14.7 MAXIMUM AND MINIMUM VALUES 0 225
local minima, !(0.170, - 1.215) ::::: 3.197 is a local maximum, and ( -1.301, -1.215), (0.170, 0.549), and (1.131, -1.215)
are saddle points. There is no highest or lowest point on the graph.
3
z
29. Since f is a polynomial it is contint,Jous on D, so an absolute maximum and minimum exist. Here f x = 2x- 2, jy = 2y, and
setting fx = jy = 0 gives (1, 0) as the only critical point (which is inside D), where f (1, 0) = - 1. Along £ 1 : x = 0 and
f(O, y) = y 2 for -2 ~ y ::=:; 2, a quadratic function which attains its minimum at y = b, where f(O, 0) = 0, and its maximum
·aty = ±2, wheref(0,±2) = 4. Along £2: y = x- 2 forO ::=:; x ~ 2, and f(x,x- 2) = 2x 2 - 6x + 4 = 2 (x _ ~) 2 _ ~ .
a quadratic which attains its minimum at x = £, where f ( ~, - ~) = - ~, and its maximum at x = 0, where f ( 0, - 2) = 4.
Along £3: y = 2 - x for 0 ~ x ::::; 2, and
f(x, 2- x) = 2x 2 - 6x + 4 = 2(x- ~) 2 - ~.a quadratic which attains
its minimum at x =~~where f(~, ~) =-~,and its maximum at x = 0,
where f (O, 2) = 4. Thus the absolute maximum off on D is f(O, ±2) = 4
and the absolute minimum is !(1, 0) = -1.
y
(0,2)
L,
(0, - 2)
(2,0)
X
31. f x(x,y) = 2x +2xy, jy(x,y) = 2y +x 2 , and setting f.,= f" = 0
gives (0, 0) as the only critical point in D , with .f(O, 0) = 4.
On £ 1: y = -1, j(x, - 1) = 5, a constant.
On £2: x = 1, /(1, y) = y 2 + 11 + 5, a quadratic in y which attains its
maximum at (1, 1), /(1, 1) = 7 and its minimum at (1, - ~) ,
f(1, - t) =.!f.
On £ 3: f(x, 1) = 2x 2 + 5 which attains its maximum at (-1, 1) and (1, 1)
with f(±l , 1) = 7 and its minimum at (0, 1), f(O, 1) = 5.
(-1, 1)
L•
(-1,-1)
y
L, (I, I)
L2
0
L, (1,-1)
X
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