### 13.5 Lagrange Multipliers

Lagrange multipliers are used to solve extremum problems for a function defined on a level set of another
function. For example, suppose you want to maximize xy given that x + y = 4. This is not too hard to do
using methods developed earlier. Solve for one of the variables, say y, in the constraint equation
x + y = 4 to find y = 4 − x. Then the function to maximize is f

=

x and the answer is
clearly

x = 2. Thus the two numbers are

x =

y = 2. This was easy because you could easily
solve the constraint equation for one of the variables in terms of the other. Now what if you
wanted to maximize

f =

xyz subject to the constraint that

x^{2} +

y^{2} +

z^{2} = 4? It is still
possible to do this using similar techniques. Solve for one of the variables in the constraint
equation, say

z, substitute it into

f, and then find where the partial derivatives equal zero to
find candidates for the extremum. However, it seems you might encounter many cases and it
does look a little fussy. However, sometimes you can’t solve the constraint equation for one
variable in terms of the others. Also, what if you had many constraints? What if you wanted
to maximize

f subject to the constraints

x^{2} +

y^{2} = 4 and

z = 2

x + 3

y^{2}. Things are
clearly getting more involved and messy. It turns out that at an extremum, there is a simple
relationship between the gradient of the function to be maximized and the gradient of the constraint
function.

This relation can be seen geometrically in the following picture.