Question 761289
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First thing is to make the rather reasonable assumption that the maximum amount of light would be admitted by the maximum area window.


The perimeter of the overall window is *[tex \LARGE \pi r\ +\ 2r\ +\ 2h\ =\ 24]


Where *[tex \LARGE h] is the height of the rectangular part of the window and *[tex \LARGE r] is the radius of the semi-circular piece.


From this relationship we can determine that *[tex \LARGE h\ =\ 12\ -\ \left(1\ -\frac{\pi}{2}\right)r] 


The area of the rectangular portion of the window is then *[tex \LARGE A_r\ =\  24r\ -\ \left(2\ +\ \pi\right)r^2]


The area of the semi-circular portion of the window is  *[tex \LARGE A_s\ =\ \frac{\pi}{2}r^2]


Add 'em up:  *[tex \LARGE A_w(r)\ =\ 24r\ -\ \left(2\ +\ \pi\right)r^2\ +\ \frac{\pi}{2}r^2]


Take the first derivative: *[tex \LARGE \frac{d}{dr}A_w(r)\ =\ 24\ -\ \left(4\ +\ 2\pi\right)r\ +\ \pi r\ =\ 24\ -\ (4\ +\ \pi)r]


Set the derivative equal to zero: *[tex \LARGE 24\ -\ (4\ +\ \pi)r\ =\ 0]


So *[tex \LARGE r\ =\ \frac{24}{4\ +\ \pi}]


Take the second derivative:  *[tex \LARGE \frac{d^2}{dr^2}A_w(r)\ =\ -4\ -\ \pi] which is negative for all values of *[tex \LARGE r].  So the function is concave down in the region of the extremum and therefore the value of the function at that point is a maximum.


So the rectangular part of the window must have a width of twice the radius of the semi-circle, or *[tex \LARGE w\ = \frac{48}{4\ +\ \pi}], and the height must be *[tex \LARGE h\ =\ 12\ -\ \left(1\ -\frac{\pi}{2}\right)\left(\frac{24}{4\ +\ \pi}\right\) =\ \frac{24}{4\ +\ \pi}]


I'll leave the verification of that simplification to you.


That's all you need to answer your question.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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