Solução_Calculo_Stewart_6e

F.
206 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
53. Let x, y, z be the dimensions of the rectangular box. Then the volume of the box is xyz and
L = x 2 + y 2 + z 2 ⇒ L 2 = x 2 + y 2 + z 2 ⇒ z = L 2 − x 2 − y 2 .
Substituting, we have volume V (x, y) =xy L 2 − x 2 − y 2 , (x, y > 0).
V x = xy · 1
2 (L2 − x 2 − y 2 ) −1/2 (−2x)+y L 2 − x 2 − y 2 = y x 2 y
L 2 − x 2 − y 2 −
L2 − x 2 − y ,
2
V y = x L 2 − x 2 − y 2 −
xy 2
L2 − x 2 − y 2 . V x =0implies y(L 2 − x 2 − y 2 )=x 2 y ⇒ y(L 2 − 2x 2 − y 2 )=0 ⇒
2x 2 + y 2 = L 2 (since y>0), and V y =0implies x(L 2 − x 2 − y 2 )=xy 2 ⇒ x(L 2 − x 2 − 2y 2 )=0 ⇒
x 2 +2y 2 = L 2 (since x>0). Substituting y 2 = L 2 − 2x 2 into x 2 +2y 2 = L 2 gives x 2 +2L 2 − 4x 2 = L 2 ⇒
3x 2 = L 2 ⇒ x = L/ √
3 (since x>0)andtheny = L 2 − 2 L/ √ 3 2 √
= L/ 3. So the only critical point is
L/
√
3,L/
√
3
which, from the geometrical nature of the problem, must give an absolute maximum. Thus the maximum
volume is V L/ √ 3,L/ √ 3 = L/ √ 3 2 L 2 − L/ √ 3 2
−
L/
√
3
2
= L 3 / 3 √ 3 cubic units.
55. Note that here the variables are m and b,andf(m, b) = n [y i − (mx i + b)] 2
.Thenf m = n −2x i [y i − (mx i + b)] = 0
implies
n
xi y i − mx 2 i − bx i =0or
i =1
n
y i = m n x i +
i =1
Now f mm =
i =1
n
i =1
n
i =1
2x 2 i , f bb =
i =1
n
x i y i = m n x 2
i + b n x i and f b =
i =1
i =1
i =1
n
i =1
n
b = m x i + nb. Thus we have the two desired equations.
i =1
n
i =1
n
D(m, b) =4n x 2 n
i − 4
i =1
equations do indeed minimize
i =1
n
i =1
i =1
2=2n and f mb = n 2x i. And f mm(m, b) > 0 always and
2
n
x i =4 n
d 2 i .
i =1
x 2 i
i =1
−
−2[y i − (mx i + b)] = 0 implies
n
2
x i > 0 always so the solutions of these two
i =1
15.8 Lagrange Multipliers ET 14.8
1. At the extreme values of f, the level curves of f just touch the curve g(x, y) =8with a common tangent line. (See Figure 1
and the accompanying discussion.) We can observe several such occurrences on the contour map, but the level curve
f(x, y) =c with the largest value of c which still intersects the curve g(x, y) =8is approximately c =59, and the smallest
value of c corresponding to a level curve which intersects g(x, y) =8appears to be c =30.Thusweestimatethemaximum
value of f subject to the constraint g(x, y) =8to be about 59 and the minimum to be 30.
3. f(x, y) =x 2 + y 2 , g(x, y) =xy =1,and∇f = λ∇g ⇒ h2x, 2yi = hλy, λxi,so2x = λy, 2y = λx, andxy =1.
From the last equation, x 6= 0and y 6= 0,so2x = λy ⇒ λ =2x/y. Substituting, we have 2y =(2x/y) x ⇒
y 2 = x 2 ⇒ y = ±x. Butxy =1,sox = y = ±1 and the possible points for the extreme values of f are (1, 1) and
(−1, −1). Here there is no maximum value, since the constraint xy =1allows x or y to become arbitrarily large, and hence
f(x, y) =x 2 + y 2 canbemadearbitrarilylarge.Theminimumvalueisf(1, 1) = f(−1, −1) = 2.