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

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SECTION 14.8 LAGRANGE MULTIPLIERS 0 229
55. Note that here the variables are m and b, and f{m, b) = :L: [y,- (mxi +b)] . Then /m. = :L: - 2xi [Vi - (mxi + b)J = 0
. i =l i=l
n. · n n n · n
implies :L: (x•Y•- mx; - bxi) = 0 or .:L: XiYi = m .:L: x~ + b .:L: Xi and fb = :L: -2[yi- (mxi +b)] = 0 implies
i= l t. = l \=1 •=1 1.=1
n n n (n )
.~ Yi = m i~l Xi + i~l b = m i~t Xi + nb. Thus we have the two desired equations.
1
n n n .
Now fmm = :L: 2x~, fbb = :L: 2 = 2n and fmb = :L: 2Xi. And fmm{m, b) > 0 always and
i =l i = l i= l
2
n
equations do indeed minimize f= d;.
i = 1
14.8 Lagrange Multipliers
1. At the extreme val';les off, the level curves off just touch the curve g(x, y) = 8 with a common tangent line. (See Figure I
and the accompanying discussion.) We can observe several such occurrenc~ 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) = 8 is approximately c =59, and the smallest
value of c corresponding to a level curve which intersects g(x, y) = 8 appears to be c = 30. Thus we estimate the maximum
value off subject to the constraint g(xj y) = 8 to be about 59 and the minimum to be 30.
3. f(x, y) = x 2 + y 2 , g(x, y) = xy = 1, and \1 f = >.. \lg => (2x, 2y) = (>..y, >..x), so 2x = >..y, 2y = >..x, and xy = 1.
From the last equation, x =I= 0 andy =I= 0, so 2x = >.y => >.. = 2xjy. Substituting, we have 2y = (2xjy) x =>
y 2 = x 2 => y = ±x. But xy = 1, so X= y = ± 1 and the possible points for the extreme values off are {1, 1) and
( -1, - 1). Here there is no maximum value, since the constraint xy = 1 allows x or y to become arbil:(arily large, and hence
f(x, y) = x 2 + y 2 can be made arbitrarily large. The mi~imum value is !{1, 1) = f( - 1, -1) = 2.
5. f(x, y) = y 2 - x 2 , g(x, y) = tx +
2 y 2 = 1, and \1 f = >.\lg =>
{-2x, 2y) = (t>..x, 2>..y), so -2x = t>.x, 2y = 2>..y,
and ix 2 + y 2 = 1. From the first equation we have x{ 4 + >..) = 0 => x = 0 or>. = - 4. If x = 0 then the third equation
gives y = ±1. If), = - 4 then the second equation gives 2y = -8y => y = 0, and substituting into the third equation,
we have x = ±2. Thus the possible extreme values off occur at the points (0, ± 1) and (± 2, 0). Evaluating fat these points,
we see that the maximum value is f{O, ± 1) = 1 and the minimum is !{±2, 0) = -4.
7. f(x, y, z) = 2x + 2y + z, g(x, y, z) = x 2 + y 2 + z 2 = 9, and \1 f = >.. \1 g => (2, 2, 1) = (2>..x, 2>..y, 2>.z), so 2>.x = 2,
2>.y = 2, 2)..z = 1, and x 2 + y 2 + z 2 = 9. The first three equations imply x = l• y = ·l· and z = 2
\ . ~ut substitution into
2 2
the fourth equation gives (l)2
+ ( l) + ( \ ) = 9 2
=>
9
-
4)..
2 = 9 => ).. = ±t, so f has possible extreme values at
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