Multivariate Calculus - Bruce E. Shapiro

LECTURE 15. CONSTRAINED OPTIMIZATION: LAGRANGE
MULTIPLIERS 125
Applying our usual technique, ∇f = λ∇g + µ∇h, so that
or
(2x, 2y, 2z) = λ (−2x, −2y, 2z) + µ (−1, −1, 1) (15.26)
2x = −2xλ − µ (15.27)
2y = −2yλ − µ (15.28)
2z = 2zλ + µ (15.29)
We can elminate µ by adding equation 15.29 to each of equations 15.27 and 15.28
to give
2x + 2z = 2λ(z − x) (15.30)
or
Equating the two expressions for λ,
Hence y = x or z = 0.
Case 1: z=0
2y + 2z = 2λ(z − y) (15.31)
λ = z + x
z − x
λ = z + y
z − y
z + x
z − x = z + y
z − y
(z + x)(z − y) = (z + y)(z − x)
z 2 + xz − yz − xy = z 2 + yz − xz − xy
xz − yz = yz − xz
yz = xz
(15.32)
(15.33)
From the constraint g we have x 2 + y 2 = 0 which means x = 0 and y = 0. The
origin is the only point on the cone that satisfies z = 0. But this point is not on the
plane z = 1 + x + y. So there is no solution here.
Case 2: x=y
The first constraint (g) gives z 2 = 2x 2 . The second constraint (h) gives z = 1 + 2x,
hence
(1 + 2x) 2 = 2x 2 ⇒ 1 + 4x + 4x 2 = 2x 2 ⇒ 2x 2 + 4x + 1 = 0
hence
y = x = −4 ± √ 8
4
= −1 ± 1 √
2
(
z = 1 + 2 −1 ± √ 1 )
= −1 ± √ 2
2
Math 250, Fall 2006 Revised December 6, 2006.