# Section 12.8 Lagrange Multipliers subject to the

```Section 12.8 Lagrange Multipliers
The problem is to find the extreme values of a given function f ( x , y , z ) subject to the
constraint g ( x , y , z )  k .
The method of Lagrange multipliers is to solve  f    g for some constant  .
Intuitive explanation: g ( x , y , z )  k is a level surface of g. Suppose that f ( x , y , z ) has an
extreme value at some point (a, b, c) on this surface. Then for any curve on this surface


with vector equation r (t ) with r ( t 0 )  a , b , c , the function f ( x ( t ), y ( t ), z ( t )) has an
extreme value at t 0 . So
But
d
d
dt
f ( x ( t ), y ( t ), z ( t ))  0 at t 0 .

f ( x ( t ), y ( t ), z ( t ))   f  r  ( t )
dt

. So  f ( a , b , c )  r ( t 0 )  0 . We have shown that the
gradient of f at (a, b, c) is orthogonal to every curve on g ( x , y , z )  k that passes through
(a, b, c). So the gradient of f at (a, b, c) is orthogonal to the tangent plane to the surface
at this point and must be parallel to the gradient of g at this point.
Examples: Find the extreme values of the given function subject to the given constraint.
1. Pg 788 problem 1 in Stewart
2
f ( x, y )  x  y
2
subject to
2
x  y
2
 1.
2. pg 788 problem 2
f ( x, y )  2 x  y
subject to
2
x  4y
2
1
3. pg 788 problem 8
2
2 2
f ( x, y, z)  x y z
subject to
2
2
2
x  y  z 1
```