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
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.
3 answers
Wouldn't the equation for the perimeter be
2x + 4r + πr = 30?
Since there are two lengths?
2x + 4r + πr = 30?
Since there are two lengths?
the answer for this questions is x and y = 4.2