Maximizing the area of a Norman window with fixed perimeter

The perimeter consists of two vertical sides of length $h$, the bottom of length $x$, and the semicircular arc of length $\frac{\pi x}{2}$, so $2h + x + \frac{\pi x}{2} = 20$. Solving for $h$ gives $h = \frac{20 - x - \frac{\pi x}{2}}{2} = 10 - \frac{x}{2} - \frac{\pi x}{4}$. This must be positive, which restricts $x$ to $0 < x < \frac{40}{\pi + 2}$.

The total area is the rectangle plus the semicircle: $A = xh + \frac{\pi x^2}{8}$. Substituting $h$ gives $A(x) = x\left(10 - \frac{x}{2} - \frac{\pi x}{4}\right) + \frac{\pi x^2}{8} = 10x - \frac{x^2}{2} - \frac{\pi x^2}{4} + \frac{\pi x^2}{8} = 10x - \left(\frac{1}{2} + \frac{\pi}{8}\right)x^2$.

Differentiating, $A'(x) = 10 - \left(1 + \frac{\pi}{4}\right)x$. Setting $A'(x) = 0$ gives $x = \frac{10}{1 + \frac{\pi}{4}} = \frac{40}{4+\pi}$. Since $A''(x) = -\left(1+\frac{\pi}{4}\right) < 0$, this critical point is a maximum, and it lies within the valid domain.

Plugging $x = \frac{40}{4+\pi}$ back into $A(x) = 10x - \left(\frac{1}{2}+\frac{\pi}{8}\right)x^2$ gives the maximum area, approximately $A \approx 28.0$ square feet, achieved when $x \approx 5.6$ ft.