A Norman window is a window with a semi-circle on top of regular rectangular window. What should be the dimensions of the window to allow in as much light as possible, if there are only 12 ft of the frame material available?

Side lengths = 1.68 ft and width = 3.36 ft.

Step-by-step explanation:

Let the side lengths of the window be L and the width = 2r (r is also the radius of the semi-circle).

So we have

Perimeter = 2L + 2r + πr = 12

Area = 2rL + 0.5πr^2

From the first equation

2L = 12 - 2r - πr

Substitute for 2L in the equation for the area:

A = r (12 - 2r - πr) + 0.5πr^2

A = 12r - 2r^2 - πr^2 + 0.5πr^2

A = 12r - 2r^2 - 0.5πr^2

We need to find r for the maximum area:

Finding the derivative and equating to zero:

A' = 12 - 4r - πr = 0=

4r + πr = 12

r = 12 / (4 + π)

r = 1.68 ft.

So the width of the window = 2 * 1.68 = 3.36 ft.

Now 2L = 12 - 2r - πr

= 12 - 2*1.68 - 1.68π

= 3.36

L = 1.68.

Dimensions: 1.68 ft * 3.36 ft

Step-by-step explanation:

Let x be the radius

Arc length = 3.14x

Width = 2x

Height = [12 - (2x) - 3.14x]/2

= 6 - 2.57x

Area = 3.14x²/2 + 2x (6 - 2.57x)

Area = 12x - 3.57x²

dA/dx = 12 - 7.14x = 0

x = 1.6806722689

Width = 2x

= 3.361344538

Height = 6 - 2.57x

= 1.6806722689