A norman window is in the shape of a rectangle surmounted by a semicircle. What is the ratio of the width of the rectangle to the total height so that it will yield a window admitting the most light for a given perimeter?
+130548 Let the area of the semi-circle = (1/2) pi*r^2 where r is the radius
Let the height of the semi-circle = r
Let the perimeter of the semi-circle = pi*r
Let the width of the rectangular part of the window = 2r
The the length of one side of the rectangle = [ P - pi*r - 2r] / 2 where P is the perimeter
The.....the total area, A =
A = (1/2)pi *r^2 + 2r * [ P - pi*r - 2r] / 2
A = (1/2) pi*r^2 + r [ P - pi*r - 2r]
A = (1/2)pi*r^2 + Pr - pi*r^2 - 2r^2
Take the derivative
A' = pi*r + P - 2pi*r - 4r
A' = P - r [ 4 + pi]
Set the derivative = 0
P - r [4 + pi] = 0
P = r [ 4 + pi ]
Solve for r
r = P / [ 4 + pi ]
So......the ratio of the width of the rectangular part of the window to the total height =
2r / [ r + [ P - pi*r - 2r] / 2 ] =
2r / [ 2r + P - pi*r - 2r] / 2 =
4r / [ P - pi*r ]
Substitute P / [ 4 + pi ] for r
(4 P / [ 4 + pi ] ) / ( P - pi *P / [ 4 + pi] )
(4P) / ( 4P + pi*P - pi*P) =
(4P) / (4P) =
1 / 1
So.....the width of the window = the total height of the window to maximize the area for any given perimeter
