Solução_Calculo_Stewart_6e

F.
208 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
15. f(x, y, z) =x +2y, g(x, y, z) =x + y + z =1, h(x, y, z) =y 2 + z 2 =4 ⇒ ∇f = h1, 2, 0i, λ∇g = hλ, λ, λi
and μ∇h = h0, 2μy, 2μzi. Then1=λ, 2=λ +2μy and 0=λ +2μz so μy = 1 = −μz or y =1/(2μ), z = −1/(2μ).
2
Thus x + y + z =1implies x =1and y 2 + z 2 =4implies μ = ± 1
2 √ . Then the possible points are 1, ± √ 2, ∓ √ 2
2
and the maximum value is f 1, √ 2, − √ 2 =1+2 √ 2 and the minimum value is f 1, − √ 2, √ 2 =1− 2 √ 2.
17. f(x, y, z) =yz + xy, g(x, y, z) =xy =1, h(x, y, z) =y 2 + z 2 =1 ⇒ ∇f = hy, x + z, yi, λ∇g = hλy, λx, 0i,
μ∇h = h0, 2μy, 2μzi. Theny = λy implies λ =1[y 6= 0since g(x, y, z) =1], x + z = λx +2μy and y =2μz. Thus
μ = z/(2y) =y/(2y) or y 2 = z 2 ,andsoy 2 + z 2 =1implies y = ± 1 √
2
, z = ± 1 √
2
.Thenxy =1implies x = ± √ 2 and
the possible points are ± √
2, ± √ 1 1
2
, √
2
, ± √ 2, ± √ 1
2
, − √ 1
2
. Hence the maximum of f subject to the constraints is
f ± √
2, ± √ 1
2
, ± √ 1
2
= 3 and the minimum is f ± √
2, ± √ 1
2 2
, ∓√ 1
2
= 1 . 2
Note: Since xy =1is one of the constraints we could have solved the problem by solving f(y, z) =yz +1subject to
y 2 + z 2 =1.
19. f(x, y) =e −xy . For the interior of the region, we find the critical points: f x = −ye −xy , f y = −xe −xy , so the only
critical point is (0, 0), andf(0, 0) = 1. For the boundary, we use Lagrange multipliers. g(x, y) =x 2 +4y 2 =1
⇒
λ∇g = h2λx, 8λyi, sosetting∇f = λ∇g we get −ye −xy =2λx and −xe −xy =8λy. Thefirst of these gives
e −xy = −2λx/y, and then the second gives −x(−2λx/y) =8λy ⇒ x 2 =4y 2 . Solving this last equation with the
constraint x 2 +4y 2 =1gives x = ± √ 1
2
and y = ± 1
2 √ .Nowf ± √ 1
2 2
, ∓ 1
2 √ = e 1/4 ≈ 1.284 and
2
f ± √ 1
2
, ± 1
2 √ = e −1/4 ≈ 0.779. The former are the maxima on the region and the latter are the minima.
2
21. (a) f(x, y) =x, g(x, y) =y 2 + x 4 − x 3 =0 ⇒ ∇f = h1, 0i = λ∇g = λ 4x 3 − 3x 2 , 2y .Then
1=λ(4x 3 − 3x 2 ) (1) and 0=2λy (2). Wehaveλ 6= 0from (1),so(2) gives y =0. Then, from the constraint equation,
x 4 − x 3 =0 ⇒ x 3 (x − 1) = 0 ⇒ x =0or x =1.Butx =0contradicts (1), so the only possible extreme value
subject to the constraint is f(1, 0) = 1. (The question remains whether this is indeed the minimum of f.)
(b) The constraint is y 2 + x 4 − x 3 =0 ⇔ y 2 = x 3 − x 4 . The left side is non-negative, so we must have x 3 − x 4 ≥ 0
which is true only for 0 ≤ x ≤ 1. Therefore the minimum possible value for f(x, y) =x is 0 which occurs for x = y =0.
However, λ∇g(0, 0) = λ h0 − 0, 0i = h0, 0i and ∇f(0, 0) = h1, 0i,so∇f(0, 0) 6=λ∇g(0, 0) for all values of λ.
(c) Here ∇g(0, 0) = 0 but the method of Lagrange multipliers requires that ∇g 6=0 everywhere on the constraint curve.
23. P (L, K) =bL α K 1−α , g(L, K) =mL + nK = p ⇒ ∇P = αbL α−1 K 1−α , (1 − α)bL α K −α , λ∇g = hλm, λni.
Then αb(K/L) 1−α = λm and (1 − α)b(L/K) α = λn and mL + nK = p,soαb(K/L) 1−α /m =(1− α)b(L/K) α /n or
nα/[m(1 − α)] = (L/K) α (L/K) 1−α or L = Knα/[m(1 − α)]. Substituting into mL + nK = p gives K =(1− α)p/n
and L = αp/m for the maximum production.