Question 1041677
Assume the window is X wide and Y long.
So then the perimeter of the window would be.
{{{P=X+2Y+(pi/2)*X}}}
1.{{{(1+pi/2)X+2Y=40}}}
and the area of the window would be,
{{{A=XY+(pi/2)(X/2)^2}}}
2.{{{A=XY+(pi/8)X^2}}}
From eq. 1,
{{{2Y=40-(1+pi/2)X}}}
{{{Y=20-((1+pi/2)/2)X}}}
{{{Y=20-(1/2+pi/4)X}}}
Let's just call the constant in front of X, B, just for better readability.
{{{Y=20-BX}}}
So then substituting into 2,
{{{A=X(20-BX)+(pi/8)X^2}}}
{{{A=20X-BX^2+(pi/8)X^2}}}
{{{A=(pi/8-B)X^2+20X}}}
{{{A=(pi/8-1/2-pi/4)X^2+20X}}}
{{{A=(-pi/8-1/2)X^2+20X}}}
{{{A=-((4+pi)/8)X^2+20X}}}
Once again, let's use a constant, C.
{{{A=CX^2+20X}}}
Convert to vertex form to find the vertex and maximum area.
{{{A=C(X^2+(20/C)X)}}}
{{{A=C(X^2+(20/C)X+(10/C)^2)-C(10/C)^2}}}
{{{A=C(X+10/C)^2-100/C}}}
{{{A=-((4+pi)/8)(X-80/(4+pi))+800/(4+pi)}}}
So the maximum area of 
{{{A[max]=800/(4+pi)}}}
occurs when,
{{{highlight(X=80/(4+pi))}}}
and
{{{Y=20-(1/2+pi/4)(80/(4+pi))}}}

{{{Y=20-((2+pi)/4)(80/(4+pi))}}}
{{{Y=20-((2+pi)/(4+pi))*20}}}
{{{Y=20(1-(2+pi)/(4+pi))}}}
{{{Y=20((4+pi)/(4+pi)-(2+pi)/(4+pi))}}}
{{{Y=20((4+pi-2-pi)/(4+pi))}}}
{{{Y=20(2/(4+pi))}}}
{{{highlight(Y=40/(4+pi))}}}