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

SECTION 14.7 MAXIMUM AND MINIMUM VALUES 0 223
8 ·
7
8
6
4
2
6
z 5 -2.5
4
3
2 - 2.5
- 2 - I 0 2
y
2.5
From the graphs, there appear to be local minimaofabout/(1, ±1) = f(- 1,±1) ~ 3 (and no local maxima or saddle
points). f, = 2x - 2x- 3 y- 2 , fv = 2y - 2x- 2 y- 3 , f xx = 2 + 6x- 4 y- 2 , fx11 = 4x- 3 y- 3 , / 1111 = 2 + 6x- 2 y- 4 • Then
f, = 0 implies 2x 4 y 2 - 2 = 0 or x 4 y 2 = 1 or y 2 = x - 4 • Note that neither x nor y can be zero. Now f v = 0 implies
2x 2 y 4 - 2 = 0, and with y 2 = x- 4 this implies 2x- 6 - 2 = 0 or x 6 = 1. Thus x = ±1 and if x = 1, y = ± 1; if x = - 1, ·
y = ±1. So the critical points are (1, 1), (1, - 1),( - 1, 1) and (- 1, -1). Now D(1, ± 1) = D( -1, ±1) = 64 - 16 > 0 and
fxx > 0 always, so f(1, ± 1) = f (-1, ± 1) = 3 are local minima.
23. f(x,y) = sin x +siny+sin(x +y), 0 :S: x :S: 2rr, 0 :S: y :S: 2rr
y
From the graphs it appears that f has a local maximum at about (1, 1) with value approximately 2.6, a local minimum
at about (5, 5) with value approximately -2.6, and a saddle point at about (3, 3).
f :z: = cos x + cos(x + y), fv = co:;y + cos(x + y), f xx = - sinx- sin(x + y ), / 11 y = - siny- sin(x + y),
f x y = - sin(x + 1J). Setting fx = 0 and / 11 = 0 and subtracting gives cos x - cos y = 0 or cos i = cos y. Thus x = y
or x = 2rr - y. lfx = y, f, = 0 becomes cosx + cos 2x = 0 or 2 cos 2 x + cosx - 1 = 0, a quadratic in cosx. Thus
cos x = - 1 or ~ and x = rr, i , or r; , giving the critical points ( rr, rr ), ( i, i) and ( 5 ;, 5 ;). Similarly if
x = 2rr - y, f :z: = 0 becomes (cos x) + 1 = 0 and the resulting critical point is (rr, rr). Now
D(x, y) = sin x sin y + sinx sin(x + y) +sin y sin(x + y). So D(1r, 1r) = 0 and the Second Derivatives Test doesn't apply.
However, along the line y = x we have f(x, x) = 2sin x + sin2x = 2 sinx + 2sinx cos x = 2sinx (1 + cosx), and
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