SOLUTION: A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the win

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Question 399784: A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 21 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.


Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A Norman window has the shape of a rectangle surmounted by a semicircle.
(Thus the diameter of the semicircle is equal to the width of the rectangle.)
If the perimeter of the window is 21 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.
:
Let x = the width of the window and diameter of the semicircle
Let h = height of the rectangular portion of the window
:
Perimeter:
2h + x + .5x*pi = 21
2h + 2.57x = 21
2h = 21 - 2.57x
h = %28%2821-2.57x%29%29%2F2
h = (10.5-1.285x)
:
What would be the window with the greatest area;
Area = semicircle + rectangle
Radius = .5x
A = (.5*pi*(.5x)^2) + h*x
Replace h with (10.5-1.285x
A = (1.57*.25x^2) + x(10.5-1.285x)
A = .3927x^2 - 1.285x^2 + 10.5x
A = -.8923x^2 + 10.5x
Find the max area by finding the axis of symmetry; x = -b/(2a)
x = %28-10.5%29%2F%282%2A-.8923%29
x = 5.88 meter is the width with the greatest area
:
Find the max area
A = -.8923(5.88^2) + 10.5(5.88)
A = -.8923(5.88^2) + 10.5(5.88)
A = -30.85 + 61.74
A = 30.89 sq/ft is max area