Question 1093820
(Actually drawing the figure would help.)


Take 2x as rectangle length and y as height of just the rectangle.
Each side of the triangle is  also 2x (only two of them being sides for the window).


Perimeter,  {{{600=6x+2y}}}
{{{y=300-3x}}}
-
Total area, {{{2xy+x^2*sqrt(3)}}},<i>...adjusted...</i>
Total area, {{{A=2x(300-3x)+sqrt(3)x^2}}}, a quadratic function A
.

{{{A=sqrt(3)x^2+2x(300-3x)}}}
{{{A=sqrt(3)x^2+600x-6x^2}}}
{{{A=(sqrt(3)-6)x^2+600x}}}
.
.
Maximum area will occur for {{{2(sqrt(3)-6)x+600=0}}}
or for 
{{{highlight_green((sqrt(3)-6)x+300=0)}}}