Question 900801
Let L and W be the length and width of the rectangle.
The perimeter is,
{{{P=2L+2W=36}}}
So,
{{{L+W=36}}}
The area of the rectangle is,
{{{A=L*W}}}
From the perimeter,
{{{L=36-W}}}
Substitute,
{{{A=(36-W)W=36W-W^2}}}
To find the extremum, set the derivative of A equal to zero.
{{{dA/dW=36-2W}}}
{{{36-2W=0}}}
{{{2W=36}}}
{{{W=18}}}
So then,
{{{L+18=36}}}
{{{L=18}}}
The maximum area for a given perimeter is actually a square. 
{{{A[max]=18^2=324}}}{{{m^2}}}