Optimization problem
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If the radius of the semicircle is R, then the side of the rectangle is 2R by a, where a is the other side of the rectangle. The perimeter of the window is (pi)R+2R+2a=150, because the moulding is fixed around the perimeter. The rectangular part of the window has the most effective area when the rectangle is a square, so a=2R.

We can write (pi)R+6R=150, so R((pi)+6)=150, so R=150/((pi)+6)=150/9.142=16.41 inches.

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