A Norman window has the shape of a rectangle surmounted by a semicircle.

To admit the most light, the area must be largest. For maximum area:
$y = \frac{1}{2}x$
 

 

$x + 2y + \pi r = 30$

$x + 2(\frac{1}{2}x) + \pi (\frac{1}{2}x) = 30$

$(2 + \frac{1}{2}\pi) x = 30$

$x = 8.4 ~ \text{ft}$

Perimeter of the window:
$x + 2y + \pi r = 30$

$x + 2y + \pi \left( \frac{1}{2}x \right) = 30$

$2y = 30 - x - \frac{1}{2}\pi x$

$y = 15 - \frac{1}{2}x - \frac{1}{4}\pi x$
 

 

Area of the window:
$A = xy + \frac{1}{2}\pi r^2$

$A = x \left( 15 - \frac{1}{2}x - \frac{1}{4}\pi x \right) + \frac{1}{2}\pi \left( \frac{1}{2}x \right)^2$

$A = \left( 15x - \frac{1}{2}x^2 - \frac{1}{4}\pi x^2 \right) + \frac{1}{8}\pi x^2$

$A = 15x - \frac{1}{2}x^2 - \frac{1}{8}\pi x^2$
 

For the window to admit the greatest possible amount of light:
$\dfrac{dA}{dx} = 15 - x - \frac{1}{4}\pi x = 0$

$15 = x + \frac{1}{4}\pi x$

$15 = \left( 1 + \frac{1}{4}\pi \right) x$

$x = \dfrac{15}{1 + \frac{1}{4}\pi}$

$x = 8.4 ~ \text{ft}$