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

232 D CHAPTER 14 PARTIAL DERIVATIVES
25. P(L,K)=bLo.Kl-o., g(L,K)=mL+nK=p '* 'VP=(abLo:- 1 K 1 - o.,(1-a)bL'""JC'""), >..'Vg=(>..m, >..n).
Then ab(K/ L) 1 -o. = >..m and (1- a)b(L/ K)"' =>..nand mL + nK = p, so ab(K/ L?-"'/m = (1 - a)b(L/ K) 0 /n or
naj[m(1-a)]= (L/ K)"'(L/ K) 1 -"'. or L = Kno/[m{1- a)]. Substituting into mL + nK = p gives K = (1- a)pjn
and L = apjm for the maximum production.
27. Let the sides of the rectangle be x andy. Then f(x, y) = xy, g(x, y) = 2x + 2y = p '* 'V f(x, y) = (y, x)·,
>.. 'Vg = (2>.., 2>..). Then A= ~y = ~x i~pli es x = y and the rectangle with maximum area is a square with side length iP·
29. The distance from {2,0, - 3) to a point (x,y,z) on the plane is d = .j(x- 2)2 + y 2 + (z + 3) 2 , so we seek to minimize
d 2 = f(x, y, z) = (x- 2) 2 + y 2 + (z + 3) 2 subject to the constraint that (x, y, z) lies on the plane x + y + z = 1, that is,
that g(x, y, z) = x + y + z = 1. Then 'V f =A 'Vg =* (2(x- 2), 2y, 2(z + 3)) = (A, .A, .A), sox= (A+ 4)/2,
y = >./2, z = (>.. - 6)/2. Substituting into the constraint equation gives .A~ 4 + i + A; 6 = 1
>.. = ~ . sox = ~. y = ~.and z = -i· This must correspond to a minimum, so the shortest distance is
d = J ( ~ - 2) ~ · + ( ~ )2 + (- i + 3) 2 = ..fi = ~ -
'* 3A- 2 = 2 '*
31. Let f(x, y , z) = d 2 = (x- 4) 2 + (y- 2) 2 + z 2 • Then we want to minimize f subject to the constraint
g (x,y,z) = x 2 +y 2 - z 2 = 0. 'Vf = )..'\Jg '* (2 (x --; 4) ,2(y- 2) ,2z) = (2>..x,2>..y, -2-Xz), sox- 4 = J..x,
y - 2 = .Ay, and z = -J..z. From the last equation we have z + >..z = 0 '* z (1 +.A) = 0, so either z = 0 or>..= -1.
But from the constraint equation we have z = 0 '* x 2 + y 2 = 0 '* x = y = 0 which is not possible from the first
two equations. So ).. = - 1 and x - 4 = >..x '* x = 2, y - 2 = >..y '* y = 1, and x·2 + y 2 - z 2 = 0 '*
4 + 1 - z 2 = 0 '* z = ±v'S. This must correspond to a minimum, so the points on the cone closest to ( 4, 2, 0)
are (2, 1, ±v'S).'
33. f(x,y,z) = xyz,g(x,y,z) = x +y + z = 100 =* 'Vf = (yz,xz, xy) = )..'\Jg =(>.., >..,>..). Then>.. = yz = xz = xy
implies x = y = z = 1 ~ 0 .
35. If the dimensions are 2x, 2y, and 2z, then maximize f(x, y, z) = (2x)(2y)(2z) = 8xyz subject to
g(x, y, z) := x 2 + y 2 + z 2 = r 2 (x > 0, y > 0, z > 0). Then 'V f = .A 'V g '* (8yz, Bxz, 8xy) = .A (2x, 2y, 2z) =*
41JZ 4xz 4xy . . 2 2 2 2
8yz = 2-Xx, Bxz = 2>.y, and 8xy = 2.Xz, so>.= - ·- = - = - .This gives x z = y z '* x = y (since z t= D)
X Y Z
'
and xy 2 = xz 2 '* z 2 = y 2 , so x 2 = y 2 = z 2 '* x = y = z, and substituting into the constraint
equation gives 3x 2 = r 2
'* x = r/../3 = y = z. Thus the largest volume of such a box is
. f ( ~, ~, ~ ) = 8 ( ~) ( 7J) ( ~) = 3
~ r 3 .
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