Answer
Window height to width ratio: $\frac{4+\pi}{8}\approx0.9$
Work Step by Step
Step 1. Assume the semicircle has a radius of $R$ and the rectangle part has a height $h$. The width of the rectangle is then $2R$.
Step 2. The total perimeter is given by $P=\pi R+2h+2R)=(\pi+2)R+2h=C$ where $C$ is a constant. We then have $h=\frac{C-(\pi+2)R}{2}$
Step 3. The total light passing the window ($I$) will be proportional to the sum of half the semicircle area and the rectangle area, that is $I=k(\frac{A_1}{2}+A_2)$ where $k$ is a constant, $A_1=\frac{\pi R^2}{2}$ is the area of the semicircle and $A_2=2Rh$ is the area of the rectangle part.
Step 4. Combining the above results, we have $I=k(\frac{\pi R^2}{4}+2Rh)=k(\frac{\pi R^2}{4}+2R(\frac{C-(\pi+2)R}{2}))=k(\frac{\pi R^2}{4}+RC-(\pi+2)R^2)$
Step 4. Take the derivative of the above function to get $I'=k(\frac{\pi R}{2}+C-2(\pi+2)R)$. Let $I'=0$; we have $(3\pi/2+4)R=C$ or $R=\frac{2C}{(8+3\pi)}$. This gives $h=\frac{C-(\pi+2)R}{2}=\frac{C}{2}-\frac{\pi+2}{2}\times\frac{2C}{(8+3\pi)}=\frac{(4+\pi)C}{2(8+3\pi)}$
Step 5. Check $I''=-4-3\pi/2 \lt 0$, and the function is concave down which gives the maximum of $I$.
Step 6. The proportion of the window can be represented by the ratio of the two sides: $\frac{h}{2R}=\frac{(4+\pi)C}{2(8+3\pi)}\times\frac{(8+3\pi)}{4C}=\frac{4+\pi}{8}\approx0.9$
