Multivariate Calculus - Bruce E. Shapiro

LECTURE 15. CONSTRAINED OPTIMIZATION: LAGRANGE
MULTIPLIERS 121
Hence
and
f
x = 6 7 , , y = 18
7 , z = −12 7
( 6
7 , 18 )
7 , −12 36 + 324 + 144
= = 504
7
49 49 ≈ 10.28
To determine if this is a maximum or minimum we need to compare with some
other point on the plane x + 3y − 2z = 12. Since there are two free variables and
one dependent variable we can pick x = 0 and y = 0. Then z = −6, and
f(0, 0, −6) = 36 > 10.28
Hence the point we found is a minimum, as desired.
Example 15.5 Find the point on the plane 2x − 3y + 5z = 19 that is nearest to the
origin.
Solution.The distance from the point (x, y, z) to the origin is
d(x, y, z) = √ x 2 + y 2 + z 2
We want to minimize this distance. Because differentiating square roots is messy,
we observe that d is minimized if and only if d 2 is also minimized. So we choose to
minimize
f(x, y, z) = x 2 + y 2 + z 2
subject to the constraint
g(x, y, z) = 2z − 3y + 5z − 19 = 0
Using Lagrange’s method ∇f = λ∇g so that
(2x, 2y, 2z) = λ (2, −3, 5)
Hence we have four equations in four unknowns,
x = λ
y = − −3λ
2
From the fourth equation
Using λ = 1 gives
z = 5λ 2
19 = 2x − 3y + 5z
19 = 2λ + 9λ 2 + 25λ
2 = 19λ
x = 1, y = 3/2, z = 5/2
Math 250, Fall 2006 Revised December 6, 2006.