Posted by **Daniel** on Sunday, November 20, 2011 at 4:12am.

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30ft, find the dimensions of the window so that the greatest possible amount of light is admitted.

I keep screwing up in creating my equation.

p = perimeter of window

l = length of rectangle

d = w = width of rectangle

circumference of a semi circle = 2(PI*r)/2 = rPi

I get p = rPi + 2L + 2W

I don't know what I am doing wrong since I am unable to get a proper derivative.

- calculus -
**Reiny**, Sunday, November 20, 2011 at 9:09am
How about starting with simpler definitions.

Let the radius of the semicircle be r

then the length of the rectangle is 2r.

let the width be x.

then:

2x + 2r + (1/2)2πr = 30

2x + 2r + πr = 30

x = (30-2r-πr)/2

area = (1/2)πr^2 + 2xr

= (1/2)πr^2 + 2r(30-2r-πr)

=(1/2)πr^2 + 30r - 2r^2 - πr^2

d(area)/dr = πr + 30 - 4r - 2πr

= 0 for a max of area

r(π-4-2π) = -30

r = 30/(π+4) = appr. 4.2

then x = 4.2 also

so the rectangle is 8.4 long and 4.2 high and the semicircle sitting on top has a diameter of 8.4

- calculus -
**Daniel**, Sunday, November 20, 2011 at 11:16pm
Wouldn't the equation for the perimeter be

2x + 4r + πr = 30?

Since there are two lengths?

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