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

230 0 CHAPTER 14 PARTIAL DERIVATIVES
the points (2, 2, 1) ~nd ( - 2, - 2, - 1). The maximum value off on x 2 + y 2 + z 2 = 9 is f(2, 2, 1) = 9, and the minimum is
J( -2, -2, - 1) = - 9.
9. f(x, y, z) = xyz, g(x, y, z) = x 2 + 2y 2 '+ 3z 2 = 6. V f = >. Vg =:. (yz, xz, x y) = .>. {2x, 4y, Gz). If any ofx, y, or z is
zero then x = y = z = 0 which contradicts x 2 + 2y 2 + 3z 2 = 6. Then.>.= (yz) /(,2x) = '(xz)/ (4y) = (xy)/(6z) or
x 2 = 2y 2 and z 2 = ~y 2 .
Thus x 2 + 2y 2 + 3z 2 = 6 implies 6y 2 = 6 or y = ±1. Then the possible points are
( .,/2, ±1, /j), (.,/2, ±1, -/j), (-.,/2, ± 1, /j), (- .,/2,±1,-/j). The maximum valu~ off on the ellipsoid is
-J:i, occurring when all coordinates are positive or exactly two are negative and the minimum is -7a occurring when 1 or 3 of
the coordinates are negative.
11. f(x,y,z) = x 2 +y 2 +z 2 , g(x,y, z) = x~ +y 4 + z 4 = 1 =? Vf = (2x,2y,2z)~ >.Vg = (4>.x 3 , 4>.y 3 ,4>.z 3 ).
Case 1: lf x f. 0, y f. 0 and z f. 0, then V f = >. Vg implies.>. = l /(2x 2 ) = 1/(2y 2 ) = l /(2z 2 ) or x 2 = y 2 = z 2 and
giving the points (±-L -L -L) (±-L _ _!_ -L) (±-L -L _ _!_) (±-L _ _!_ _ _!_)
3x 4 = 1 or x = ± - 1 -
V3 VJ' VS' V3 ' · VJ' VS' V3 ' VS' VJ ' V3 ' VJ' W' V3
all with an !-value of v'3.
Case 2: If one of the variables equals zero and the other two are not zero, then the squares of the two nonzero coordinates are
equal with common value ~ and corresponding f value of .,/2.
Case 3: If exactly two of the variables are zero, then the third variable has value ± 1 with the corresponding f value ofl. Thus
on x 4 + y' 1 + z 4 = 1, the maximum value off is y'3 and the minimum value is 1.
13. f(x, y, z, t) = X -1- y -1- Z -1- t, g(x, y, z, t) = x 2 -1- y 2 -1- z 2 -1- e = 1 => (1, 1, 1, 1) = (2>.x, 2).y, 2).z, 2>.t), SO
). = 1/(2x) = 1/(2y) = 1/(2z) = 1/(2t) and x = y ~ z = t. But x 2 -1- y 2 -1- z 2 -1- t2 = 1, so the possible points are
(±~ , ±~, ±~, ±~). Thus the maximum value off is f(~, ~~ ~~ t) = 2 and the minimum value is
15. f(x,y, z) = x + 2y, g(x,y,z) = x +y + z = 1, h(x,y, z ) = y 2 + z 2 = 4 =:. Vf = (1,2,0), .>.Vg = (>.,>., >.)
and p.Vh = (0, 2p.y, 2p.z). Then 1 = .>., 2 = .>. + 2p.y and 0 = >. + 2~tz so p.y = ~ = - p.z or y = 1/ (2p.), z = -1/ (2p.).
Thus x + y + z = 1 implies x = 1 and y 2 + z 2 = 4 implies p. =±~.Then the possible points are (1, ±.J2, =FV2)
and the'mfJ.Ximum value is f (1, .J2, -.J2) = 1 + 2 .J2 and the minimum value is f (1, - .,/2, .J2) = 1 - 2 .J2.
17. f(x, y, z ) = yz + xy, g(x, y , z ) = xy = 1, h(x, y, z) = y 2 + z 2 = 1 =:. V f = (y, x + z, y), >.Vg = (.>.y, >.x, 0),
p.Vh = (0, 2p.y, 2p.z). Then y = >.y implies>. = 1 [y f. 0 since g(x, y, z) = 1], x + z = .>.x + 2p.y andy= 2p.z. Thus
p. = z/(2y) = y/(2y) or y 2 = z 2 , and so y 2 + z 2 = 1 implies y = ±72, z = ±72. Then xy = 1 implies x = ±.J2 and
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