## pg. 827, #23

Use Lagrange Multipliers to prove that the
rectangle with maximum area that has a given perimeter p is a square.

### Solution:

Let x and y be the lengths of the sides of the rectangle. Then the
problem is one of maximizing xy subeject to the constraints
.
So, let
Then

Since x + y = p the maximum occurs when x = y = p/2. That is, the rectangle
is a square.
The fact that we have a maximum, and not a minimum follows from the fact that
the area function, A(x,y) = xy is non-negative, A(0,y) = 0 and A(x,0) = 0.
Thus A(x,y) must have a maximum on its domain.