Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 15.7 MAXIMUM AND MINIMUM VALUES ET SECTION 14.7 ¤ 205
43. x + y + z =100, so maximize f(x, y) =xy(100 − x − y). f x =100y − 2xy − y 2 , f y = 100x − x 2 − 2xy,
f xx = −2y, f yy = −2x, f xy =100− 2x − 2y. Thenf x =0implies y =0or y = 100 − 2x. Substituting y =0into
f y =0gives x =0or x = 100 and substituting y =100− 2x into f y =0gives 3x 2 − 100x =0so x =0or 100
Thus the critical points are (0, 0), (100, 0), (0, 100) and 100
, 100
3 3
D(0, 0) = D(100, 0) = D(0, 100) = −10,000 while D 100
, 100
3 3
(100, 0) and (0, 100) are saddle points whereas f 100
, 100
3 3
.
=
10,000
and f 100
3 xx , 100
3 3 = −
200
3 .
3
< 0. Thus (0, 0),
is a local maximum. Thus the numbers are x = y = z =
100
3 .
45. Center the sphere at the origin so that its equation is x 2 + y 2 + z 2 = r 2 , and orient the inscribed rectangular box so that its
edges are parallel to the coordinate axes. Any vertex of the box satisfies x 2 + y 2 + z 2 = r 2 ,sotake(x, y, z) to be the vertex
in the first octant. Then the box has length 2x, width2y, and height 2z =2 r 2 − x 2 − y 2 with volume given by
V (x, y) =(2x)(2y)
2
r 2 − x 2 − y 2 =8xy r 2 − x 2 − y 2 for 0 0. ThenV x = 1 4 cy − 2xy − y2 and V y = 1 4 cx − x2 − 2xy,
so V x =0=V y when 2x + y = 1 4 c and x +2y = 1 4 c. Solving, we get x = 1 12 c, y = 1
12 c and z = 1 4 c − x − y = 1
12 c.From
the geometrical nature of the problem, this critical point must give an absolute maximum. Thus the box is a cube with edge
length 1 12 c.
51. Let the dimensions be x, y and z, then minimize xy +2(xz + yz) if xyz =32,000 cm 3 .Then
f(x, y) =xy +[64,000(x + y)/xy] =xy +64,000(x −1 + y −1 ), f x = y − 64,000x −2 , f y = x − 64,000y −2 .
And f x =0implies y =64,000/x 2 ; substituting into f y =0implies x 3 =64,000 or x =40and then y =40.Now
D(x, y) = [(2)(64,000)] 2 x −3 y −3 − 1 > 0 for (40, 40) and f xx (40, 40) > 0 so this is indeed a minimum. Thus the
dimensions of the box are x = y =40cm, z =20cm.