Norman Window Optimization
Let us explore the consequences of making the radius of the semicircle equal to $x$. Let the height of the rectangular part be $h$.
Then the perimeter is equal to $2h+ \pi x+2x$. This is equal to $24$, and therefore $$2h+\pi x+2x=24,\quad\text{or equivalently}\quad h=\frac{1}{2}[24-x(\pi+2)].\tag{1}$$
Note that the area of the window is equal to $2xh+\frac{\pi}{2}x^2$.
Substitute the expression for $h$ that we got in (1) into the formula for the area of the window. We will get a quadratic in $x$, with somewhat messy coefficients. Now maximize in one of the usual ways.