Question 194404
A norman window is one that consists of a rectangle with a curved top to it. 
the curved top is actually a semicircle that sits on top of the rectangle.
 if the perimeter around the norman window is 20 feet, what radius of the
 semicircle will provide the greatest window area?
:
Let H = the height of the rectangular portion of the window
:
Let x = the radius of the semi circle on top
:
Half the circumference = pi*x
:
Width of the window = 2x
:
The perimeter:
2H + 2x + pix = 20
Solve for H
2H = 20 - 2x - pix 
H = {{{(20-2x-pix)/2}}}
:
Area formula: A = (H*2x) + .5pi*x^2
Replace H with {{{(20-2x-pix)/2}}}
A = 2x({{{(20-2x-pi*x)/2}}})+ {{{.5pi*x^2}}}
Cancel the 2
A = {{{x(20-2x-pi*x)}}}+ {{{.5pi*x^2}}}
A = {{{20x - 2x^2 - (pix^2) + (.5pix^2)}}}
Leaving us with
A = {{{-2x^2 + 20x - .5pix^2}}}
.5*pi = 1.57
A = {{{-2x^2 - 1.57x^2 + 20x}}}
A = {{{-3.57x^2 + 20x}}}
A quadratic equation, find the vertex for max area; a=-3.57, b=20
x = {{{(-20)/(2*-3.57)}}}
x = {{{(-20)/(-7.14)}}}
x = 2.8 ft radius for max area
:
Check solution by finding the perimeter with this radius
Find 2H
2H = 20 - 2(2.8) - 2.8pi
2H = 5.6 ft
Find perimeter: 
2H + 2x + pix = 20
5.6 + 2(2.8) + 2.8*pi = 
5.6 + 5.6 + 8.8 = 20
;
note that max area is when the rectangular portion, is a square, 5.6 by 5.6