SOLUTION: A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. The goal is to find the area of the la

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Question 373688: A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. The goal is to find the area of the largest possible Norman window with a perimeter of 35 feet?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A Norman window has the shape of a semicircle atop a rectangle so that the
diameter of the semicircle is equal to the width of the rectangle.
The goal is to find the area of the largest possible Norman window with a
perimeter of 35 feet?
:
Let x = the width and diameter of the window
Let L = the rectangle portion length of the window
:
radius of the semicircle = .5x
half circumference of the semi circle = pi%2A.5x = 1.5708x
:
Perimeter:
1.5708x + x + 2L = 35
2.5708x + 2L = 35
2L = (35-2.5708x)
L = 1%2F2(35-2.5708x)
L = 17.5 - 1.2854x
:
The area:
A = x * L
Replace L with (17.5-1.2854x)
A = x(17.5-1.2854x)
A = -1.2854x^2 - 17.5x
Max area occurs at the axis of symmetry of this quadratic equation
x = %28-17.5%29%2F%282%2A-1.2854%29
x = %28-17.5%29%2F%28-2.5708%29
x = 6.8072 ft the width that gives max area
:
Find the area, replace x with 6.8072
A = -1.2854(6.8072^2) + 17.5(6.8072)
A = -1.2854(46.3382) + 17.5(6.8072)
A = -59.556 + 119.126
A = 59.57 sq/ft, max area for 35 ft perimeter