Question 327693
A Norman window allows in the maximum light when its area is greatest.
Let L=length of the rectangle and w=width of the rectangle.
{{{A=L*w + pi*((w/2)^2)/2}}}
=.3927w^2+Lw
As you can see from this formula, when w increases (and L decreases) the area becomes larger.
We want to find the largest w (and the smallest L) in order to make the perimeter of the window 24 ft. In other words, we want L to approach zero but not quite get there.
{{{P=2L+w+(pi*w)/2=24}}}
Without getting into calculus, we can make L=0.
{{{w+(pi*w)/2=24}}}
 2w+(pi*w)=48
w=9.3356... ft. L=very close to zero (.000001 ft for example) ANSWER
Now the area of the rectangular part of the window approaches zero and we are left with only the semicircle to let in virtually all the light.