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

228 0 CHAPTER 14 PARTIAL DERIVATIVES
implies x 2 + 2y 2 = r 2 . Substituting, we have 2x 2 + y 2 = x 2 + 2y 2 ::::> x 2 = y 2 ::::> y = x. Then x 2 + 2y 2 = r 2 '*
3x 2 = r 2 ::::> x = .jr"i]3 = r I vf3 = y. Thus the oni,Y critical point is (r I vf3, r I ,/3). There must be a maximum
volume and here it must occur at a critical point, so the maximum volume occurs when x = y = r 1 V3 and the maximum
.47. Maximize f(x, y) = x; (6-x- 2y), then the maximum volume is V = xyz.
f., = i(6y ~ 2xy- y 2 ) = h(6- 2x- 2y) and fv = ~x (6 - x - 4y). Setting fx =:= 0 and j 11 = 0 gives the critical point
(2, 1} which geometrically must give a maximum. Thus the volume of the largest such box is V = (2)(1) ( ~) = t·
49. Let the dimensions be x, y, and z; then 4x + 4y + 4z = c and the volume is
V = xyz = xv(ic- x - y) = ic:cy- x 2 y - xy 2 , x > 0, y > 0. Then V, = ~cy- 2xy- y 2 and Vy = ~ex- x 2 -
2xy,
so Vx = 0 = Vy when 2x + y = ic and x + 2y =~c. Solving, we get x =f-ie, y = 1
12
c and z = ic- x- y = f-ie. From
the geometrical nature of the ·problem, this critical point must give an absolute maximum. Thus the box is a cube with edge
length -f2c.
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)fxy] = xy + 64,000(x- 1 + y- 1 ), fx = y- 64,000x- 2 , fv = x- 64,000y- 2 .
And fx = 0 implies y = 64,000/x 2 ; substituting into /y = 0 implies x 3 = 64,000 or x = 40 and then y = 40. Now
D(x, y) = [(2)(64,000Wx - 3 y- 3 -
dimensions of the box are x = y = 40 em, z = 20 em.
1 > 0 for ( 40, 40) and 'f.,, ( 40, 40) > 0 so this is indeed a minimum. Thus the
53. Let x, y, z be the dimensions of the rectangular box. Then the volume of the box is xyz and'
L = J xz + y2 + z2 ::::> L 2 = x 2 + y 2 + z 2 ::::> z = J L 2 - x 2 - y 2 .
Substituting, we have volume V(x,y) = xy JL2- x 2 - y 2 (x,y > 0).
2
Vy = xJL2 -x 2 -y 2 - xy . V., =Oimpliesy(L 2 - x 2 -y 2 )=x 2 y
JP - x2 - y2
::::> y(L 2 -2x 2 -y 2 ) =0 ::::>
2x 2 + y 2 = L 2 (since y > 0), and v; 1
= o ·implies 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/vf3 (since x > 0) f!nd then y = J L2- 2(L/vf3) 2 = L/vf3.
So the only critical point is (Lf,/3, Llvf3 ) which, from th~ geometrical nature of the problem, must give an absolute
maximum. Thus th~ maximum volume is V(LI,/3, Ll,/3) = (LI,/3) 2 j L2- (Livf3) 2 - (LI ,/3) 2 = L 3 I (3 ,/3)
cubic units.
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