We want to put a Norman window inside a 1 m x 3 m frame. The window is to have a rectangular base with an equilateral triangle on top. The frame will be made from a board 6 m in length. We want to maximize the amount of light to be admitted.

We want to maximize the area of a rectangle joined to an equilateral triangle. If we denote the triangle as ABC with A the apex, BC the base, and the rectangle with width w and height h we see that BC=w.

(See attachment.)

We want to write a formula for the area of the window in terms of a single variable.

6=2h+3w so h=3-2w. (This is the length of the board we are using.)

`A_("Total")=A_("rectangle")+A_("triangle")`

`A=wh+1/2(w)(sqrt(3)/2 w)`

Substituting for h we get:

`A=w(3-2w)+sqrt(3)/4 w^2=3w-2w^2+sqrt(3)/4 w^2`

To maximize the area, we differentiate with respect to w and evaluate at the critical point(s).

`(dA)/(dw)=3-4w+sqrt(3)/2 w` The critical point(s) will occur when the derivative is equal to zero (note that this function is linear in w so exists everywhere.)

`w(sqrt(3)/2 -4)=-3 => w=(-3)/(sqrt(3)/2 -4)~~.9573`

Then the height would be `h~~1.0855`

The total area is `A~~1.436 "m"^2`

If the carpenter only built an equilateral triangle:

i. if she used the 1 m base as the base of the triangle then the area would be `A~~.433"m"^2`

ii. But this is not the largest equilateral triangle that can be included in a 1 x 3 rectangle. If one vertex of the triangle is on a side of the rectangle and the base also lies along a side, then the area is `A~~.5774"m"^2`

Note that either the height or the base of the triangle is limited to one.

**Further Reading**