Calculus Help~~~~~

anonymous

2008-09-07 6:17 pm

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 30ft, find the dimension of the window so that the greatest possible amount of the light is admitted.

回答 (1)

用戶10012

2008-09-08 9:30 pm

✔ 最佳答案

Let width of window = 2y, height of rectangular part of window = x.
Therefore, perimeter of window = 2py + 2y + 2x = 30..........(1).
Area of window, A = 2xy + py^2/2 where p = pi ................(2)
from (1) x = 15 - (1+p)y. Sub into (2), we get
A = 2y[15 -(1+p)y] + py^2 = 15y - 2py^2 - 2y^2 + py^2 =15y - 2y^2 -py^2.
dA/dy = 15 - 4y - 2py = 15 - 2(2 + p)y.
Put it to zero, we get y = 15/{2(2 +p)}, so width = 2y = 15/(2+p).
So x = 15 -15 (1+p)/[2(2 +p)] = (60 + 30p - 15 - 15p)/[2(1+p)] = (45+ 15p)/[2(2 +p)] = 15(3 + p)/[2(2 +p)].

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