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

CHAPTER 14 REVIEW D 241
45. j(x, y) = x 2 e- 11 => 'V f = (2xe- 11 , - x 2 e- 11 ), 'V f( -2, 0) = (-4, -4). The direction is given by (4, - 3), so
u =
1
yf42+(-3)2 ... ...
(4,-3} = { (4,-3) and Duf(-2,0} = 'Vf{-2,0} · u = (-4, -4) · t (4, -3) = t(-16 + 12} = -~.
47. 'V f = (2xy,x 2 + 1/(2JY)), I'Vf(2, 1)1 = 1(4, ~) I · Thus the maximum rate of change off at (2, 1} is o/ in the
direction ( 4·, !).
49. First we draw a line passing through Homestea'd and the eye of the hurricane. We can approximate the directional derivative at
Homestead in the direction of the eye of the hurricane by the average rate of change of wind speed between the points where
this line intersects the contour lines closest to Homestead. In the direction of the eye of the hurricane, the wind speed changes
from 45 to 50 knots. We estimate the distance between these two points to be approximately 8 miles, so the rate of change of
wind speed in the direction given is approximately 50 8 45 = i = 0.625 knot/ mi.
51 . f(x,y)=x 2 - xy+y 2 + 9x -6y+10 => f, = ·2x-y+9,
j 11 = - x + 2y- 6, f xx = 2 = / 1111 ,
f xv = - 1. Then f x = 0 and ! 11 = 0 imply
y = 1, x = -4. Thus the only critical point is ( - 4, 1} and fx:r:( -4, 1) > 0,
D(- 4, 1) = 3 > 0, so f( - 4, 1) = -11 is a local minimum.
53. f(x, y) = 3xy - x 2 y- xy 2 => fx = 3y - 2xy - y 2 , / 11 = 3x- x 2 - 2xy,
f,, = - 2y, fv·u = - 2x, f xv = 3-2x- 2y. Then f, = 0 implies
y(3- 2x- y) = 0 soy= 0 or y = 3 - 2x. Substituting into / 11 = 0 implies
x(3 - x) = 0 or 3x( - 1 + x )' = 0. Hence the critical points are (0, 0), (3, 0},
(0, 3) and (1, 1). D(O, 0} = D(3, 0} = D(O, 3) = ·-9 < 0 so (0, 0), (3, 0), and
(0, 3) are saddle points. D(1, 1) = 3 > 0 and f xx (1, 1} = - 2 < 0, so
!(1, 1) = 1 is a local maximum.
55. First solve inside D. Here f ., = 4y 2 - 2xy 2 - y 3 , f 11 = 8xy - 2x 2 y - 3xy 2 • y
Then f, =·o implies y = 0 or y = 4- 2x, but y = 0 isn't inside D. Substituting (O, 6 )
y = 4- 2x into / 11 = 0 implies x = 0, x = 2 or x = 1, but x = 0 isn't inside -D,
and when x = 2, y = 0 but (2, 0} isn't inside D. Thus the only critical point inside
D is (1, 2) and f (l , 2) = 4. Secondly we consider the boundary of D.
On L1: f(x, 0) = 0 and so f = 0 on L 1. On L2: x = -y + 6 and
(6,0)
.t
f( - y + 6, y) = y 2 (6 - y)( - 2) = -2{6y 2 - y 3 ) which has critical points
at y = 0 and y = 4. Then !{6, 0) = 0 while !(2, 4) = -64. On La: f(O , y) = 0, so f = 0 on £ 3. Thus on D the absolute
maximum off is !{1, 2} = 4 while the absolute minimum is /{2, 4) = - 64.
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