Solução_Calculo_Stewart_6e

F.
204 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
35. f x(x, y) =6x 2 and f y(x, y) =4y 3 .Andsof x =0and f y =0only occur when x = y =0. Hence, the only critical point
inside the disk is at x = y =0where f(0, 0) = 0. Now on the circle x 2 + y 2 =1, y 2 =1− x 2 so let
g(x) =f(x, y) =2x 3 +(1− x 2 ) 2 = x 4 +2x 3 − 2x 2 +1,−1 ≤ x ≤ 1. Theng 0 (x) =4x 3 +6x 2 − 4x =0 ⇒ x =0,
−2, or 1 . f(0, ±1) = g (0) = 1, f 1
, ± √
3
= g
1
2 2 2
2 =
13
,and(−2, −3) is not in D. Checking the endpoints, we get
16
f(−1, 0) = g(−1) = −2 and f(1, 0) = g(1) = 2. Thus the absolute maximum and minimum of f on D are f(1, 0) = 2 and
f(−1, 0) = −2.
Another method: On the boundary x 2 + y 2 =1we can write x =cosθ, y =sinθ,sof(cos θ, sin θ) =2cos 3 θ +sin 4 θ,
0 ≤ θ ≤ 2π.
37. f(x, y) =−(x 2 − 1) 2 − (x 2 y − x − 1) 2 ⇒ f x(x, y) =−2(x 2 − 1)(2x) − 2(x 2 y − x − 1)(2xy − 1) and
f y (x, y) =−2(x 2 y − x − 1)x 2 .Settingf y (x, y) =0gives either x =0or x 2 y − x − 1=0.
There are no critical points for x =0,sincef x (0,y)=−2,sowesetx 2 y − x − 1=0 ⇔ y = x +1 [x 6= 0],
x 2
so f x x, x +1
= −2(x 2 − 1)(2x) − 2 x 2 x +1
− x − 1 2x x +1
− 1 = −4x(x 2 − 1). Therefore
x 2
x 2 x 2
f x (x, y) =f y (x, y) =0at the points (1, 2) and (−1, 0). To classify these critical points, we calculate
f xx(x, y) =−12x 2 − 12x 2 y 2 +12xy +4y +2, f yy(x, y) =−2x 4 ,
and f xy (x, y) =−8x 3 y +6x 2 +4x. In order to use the Second
Derivatives Test we calculate
D(−1, 0) = f xx(−1, 0) f yy(−1, 0) − [f xy(−1, 0)] 2 =16> 0,
f xx(−1, 0) = −10 < 0, D(1, 2) = 16 > 0,andf xx(1, 2) = −26 < 0,so
both (−1, 0) and (1, 2) give local maxima.
39. Let d be the distance from (2, 1, −1) to any point (x, y, z) on the plane x + y − z =1,so
d = (x − 2) 2 +(y − 1) 2 +(z +1) 2 where z = x + y − 1, and we minimize
d 2 = f(x, y) =(x − 2) 2 +(y − 1) 2 +(x + y) 2 .Thenf x (x, y) =2(x − 2) + 2(x + y) =4x +2y − 4,
f y(x, y) =2(y − 1) + 2(x + y) =2x +4y − 2. Solving 4x +2y − 4=0and 2x +4y − 2=0simultaneously gives x =1,
y =0. An absolute minimum exists (since there is a minimum distance from the point to the plane) and it must occur at a
critical point, so the shortest distance occurs for x =1, y =0for which d = (1 − 2) 2 +(0− 1) 2 +(0+1) 2 = √ 3.
41. Let d be the distance from the point (4, 2, 0) to any point (x, y, z) on the cone, so d = (x − 4) 2 +(y − 2) 2 + z 2 where
z 2 = x 2 + y 2 , and we minimize d 2 =(x − 4) 2 +(y − 2) 2 + x 2 + y 2 = f (x, y). Then
f x (x, y) =2(x − 4) + 2x =4x − 8, f y (x, y) =2(y − 2) + 2y =4y − 4, and the critical points occur when f x =0
⇒
x =2, f y =0 ⇒ y =1. Thus the only critical point is (2, 1). An absolute minimum exists (since there is a minimum
distance from the cone to the point) which must occur at a critical point, so the points on the cone closest
to (4, 2, 0) are 2, 1, ± √ 5 .