Answer
$l=\frac{24}{\pi +4}\text{ft}$
Work Step by Step
Let $l$ be the length of the rectangle and $w$ be the width of the rectangle.
Hence, the area of the rectangle is $l\cdot w$.
The area of the semi-circle is $\frac{\pi {{r}^{2}}}{2}$.
The radius of the semi circle is half of the width of the rectangle.
$r=\frac{w}{2}$
Total area of the window is,
$\begin{align}
& A=\left( l\cdot w \right)+\left( \frac{\pi {{r}^{2}}}{2} \right) \\
& =\left( l\cdot w \right)+\left( \frac{\pi {{\left( \frac{w}{2} \right)}^{2}}}{2} \right)
\end{align}$
The perimeter of the window is,
$2l+w+\pi \left( \frac{w}{2} \right)=24$
Solve for $l$,
$\begin{align}
& 2l+\left( 1+\frac{\pi }{2} \right)w=24 \\
& 2l+\left( \frac{\pi +2}{2} \right)w=24 \\
& 2l=24-\left( \frac{\pi +2}{2} \right)w
\end{align}$
Further simplify,
$l=12-\left( \frac{\pi +2}{4} \right)w$
Substitute the value of $l$ in the equation of area,
$\begin{align}
& A=\left( 12-\left( \frac{\pi +2}{4} \right)w \right)\left( w \right)+\frac{\pi }{8}{{w}^{2}} \\
& =12w-\left( \frac{\pi +2}{4} \right){{w}^{2}}+\frac{\pi }{8}{{w}^{2}} \\
& =12w-\left( \frac{2\pi +4-\pi }{8} \right){{w}^{2}} \\
& =12w-\left( \frac{\pi +4}{8} \right){{w}^{2}}
\end{align}$
Compare the equation with the equation $A=a{{w}^{2}}+bw+c$
$a=-\left( \frac{\pi +4}{8} \right),b=12$.
The vertex of the equation is,
$\begin{align}
& w=\frac{-b}{2a} \\
& =\frac{-12}{2\cdot \left( -\left( \frac{\pi +4}{8} \right) \right)} \\
& =\frac{48}{\pi +4}
\end{align}$
The value of $l$ is,
$\begin{align}
& l=12-\left( \frac{\pi +2}{4} \right)\left( \frac{48}{\pi +4} \right) \\
& =12-12\left( \frac{\pi +2}{\pi +4} \right) \\
& =\frac{24}{\pi +4}
\end{align}$
The value of $r$ is,
$\begin{align}
& r=\frac{w}{2} \\
& =\frac{\frac{48}{\pi +4}}{2} \\
& =\frac{24}{\pi +4}
\end{align}$
