Question 1099681: A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. If the perimeter of the window is 860 cm, find maximum area of the window in order to admit the most light and the dimensions of the window.
(a) max area=(in cm^2)
(b) Length of the base of the window (or the length of the diameter of the semicircle) =(in cm)
(c) Total height of the window including the rectanle and the semicircle = (in cm)
Any help would be greatly appreciated! thank you!
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The shape of the window is a rectangle with a semicircle sitting on top.
Let x = the width of the window, and h = the height of the rectangular portion
Then the perimeter of the window, P, is given by:
P = x + 2h + pi*x/2 = 2h + x(1 + pi/2)
The semicircle has a radius of x/2, so the total area, A, of the window is:
A = x*h + pi/2*(x/2)^2
Solve for h in terms of x:
h = (P - x(1 + pi/2))/2
Substitute the value for h into the expression for the area:
A = (P/2)x - x^2/2*(1 + pi/4)
The area is a maximum when dA/dx = 0
dA/dx = P/2 - x(1 + pi/4) = 0
Solve for x:
x = P/(2 + pi/2)
Solving for h in terms of P and x, and simplifying gives:
h = (1/2)*P/(2 + pi/2) = x/2
Using the value of P gives:
x = 240.84 cm
h = 120.42 cm
A = 51781.2 cm^2
The total height of the window is h + x/2 = 240.84 cm
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