Question 881026
The perimeter of the rectangle is,
{{{P=2(L+W)=50}}}
{{{L+W=25}}}
The area of the rectangle is,
{{{A=L*W}}}
Substitute from above,
{{{L=25-W}}}
{{{A=(25-W)W}}}
{{{A=25W-W^2}}}
Differentiate with respect to W and set the derivative equal to zero.
{{{dA/dW=25-2W=0}}}
{{{2W=25}}}
{{{W=25/2}}}
Then,
{{{L=25/2}}}
The maximum area for a given perimeter is a square.