Answer
$$x = \frac{{32}}{{\pi + 4}}{\text{ft and }}y = \frac{{16}}{{\pi + 4}}{\text{ft}}$$
Work Step by Step
$$\eqalign{
& {\text{From the image shown below:}} \cr
& {\text{Let }}A{\text{be the area to be maximized}} \cr
& A = {\text{area of the rectangle}} + {\text{ area of semicircle}} \cr
& A = xy + \frac{1}{2}\pi {\left( {\frac{x}{2}} \right)^2} \cr
& A = xy + \frac{1}{8}\pi {x^2}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{We know that the perimeter of the windows is 16ft}},{\text{ then}} \cr
& x + 2y + \pi \left( {\frac{x}{2}} \right) = 16 \cr
& x + 2y + \frac{1}{2}\pi x = 16 \cr
& {\text{Solving the previous equation for }}y \cr
& 2x + 4y + \pi x = 32 \cr
& y = \frac{{32 - 2x - \pi x}}{4} \cr
& {\text{Substitute }}\frac{{32 - 2x - \pi x}}{4}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& A = x\left( {\frac{{32 - 2x - \pi x}}{4}} \right) + \frac{1}{8}\pi {x^2}{\text{ }} \cr
& A = 8x - \frac{1}{2}{x^2} - \frac{1}{4}\pi {x^2} + \frac{1}{8}\pi {x^2}{\text{ }} \cr
& A = 8x - \frac{1}{2}{x^2} - \frac{1}{8}\pi {x^2}{\text{, The domain is }}x > 0 \cr
& {\text{Differentiate}} \cr
& \frac{{dA}}{{dx}} = 8 - x - \frac{1}{4}\pi x \cr
& {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr
& 8 - x - \frac{1}{4}\pi x = 0 \cr
& x\left( {1 + \frac{1}{4}\pi } \right) = 8 \cr
& x\left( {\frac{{4 + \pi }}{4}} \right) = 8 \cr
& x = \frac{{32}}{{\pi + 4}} \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {8 - x - \frac{1}{4}\pi x} \right] \cr
& \frac{{{d^2}A}}{{d{x^2}}} = - 1 - \frac{1}{4}\pi \cr
& {\text{By the second derivative test:}} \cr
& {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = \frac{{32}}{{\pi + 4}}}} = - 1 - \frac{1}{4}\pi < 0{\text{ Relative maximum}} \cr
& {\text{Calculate }}y \cr
& y = \frac{{32 - 2x - \pi x}}{4} \to y = \frac{{32 - 2\left( {\frac{{32}}{{\pi + 4}}} \right) - \pi \left( {\frac{{32}}{{\pi + 4}}} \right)}}{4} \cr
& {\text{Simplifying}} \cr
& y = \frac{{16}}{{\pi + 4}} \cr
& {\text{Therefore, the dimensions are:}} \cr
& x = \frac{{32}}{{\pi + 4}}{\text{ft and }}y = \frac{{16}}{{\pi + 4}}{\text{ft}} \cr} $$
