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November 27, 2014

November 27, 2014

Posted by **Yadira** on Sunday, April 22, 2012 at 10:42pm.

- Algebra -
**Steve**, Monday, April 23, 2012 at 10:12amno figure, but from the context it appears to be a rectangle and something else. Since it's a church, and the same kind of problem has appeared here before, I assume the top portion is a semi-circle.

So, with width w and height h, the perimeter is w+2h+pi*w/2. The trim is not used between the top and bottom portions of the window.

So, we have

w+2h+pi*w/2 = 126

so,

h = 63 - (pi+2)w/4

window area is rectangle + half-circle

a = wh + pi/8 w^2

= w(63 - (pi+2)w/4) + pi/8 w^2

= 63w - pi/8 (2pi+3)w^2

da/dw = 63 - pi/4 (2pi+3)w

max/min area is where da/dw = 0

w = 252/(3pi + 2pi^2) = 8.64

Now you can figure the areas.

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