If the perimeter of the window is 16 ft, the value of x is 8,4 ft * 4,2 ft, so that the greatest possible amount of light is admitted.
<h3>Explanation:
</h3>
Semicircle is a one-dimensional locus of points that forms half of a circle. The full arc of a semicircle always measured as 180°. Whereas a rectangle is a quadrilateral with four right angles.
The greatest possible amount of light will be admitted if the are of the window is maximum.
Let x denote as half the width of rectangle, therefore x is the radius of semicircle and let y as the height of the rectangle.
Window perimeter

Therefore


Window area is
A = area of rectangle + area of semicircle
Then by substitute the value of y we get
![A(x)=2x[15-(1+\frac{\pi}{2})x] + \frac{\pi x^2}{2} \\A(x)=30x-2x(1+\frac{\pi}{2})x + \frac{\pi x^2}{2} \\A(x)=30x-2x^2-\pi x^2+ \frac{\pi x^2}{2} \\A(x)=30x-(2+\frac{\pi }{2})x^2](https://tex.z-dn.net/?f=A%28x%29%3D2x%5B15-%281%2B%5Cfrac%7B%5Cpi%7D%7B2%7D%29x%5D%20%2B%20%5Cfrac%7B%5Cpi%20x%5E2%7D%7B2%7D%20%5C%5CA%28x%29%3D30x-2x%281%2B%5Cfrac%7B%5Cpi%7D%7B2%7D%29x%20%2B%20%5Cfrac%7B%5Cpi%20x%5E2%7D%7B2%7D%20%5C%5CA%28x%29%3D30x-2x%5E2-%5Cpi%20x%5E2%2B%20%5Cfrac%7B%5Cpi%20x%5E2%7D%7B2%7D%20%20%5C%5CA%28x%29%3D30x-%282%2B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%29x%5E2)
We need maximizing A(x) to find x value, for which it is maximum. Therefore we will equate the A'(x) is equal to zero.

The condition of domain are x>0, y>0, and A(x)>0
Therefore the domain is 
We will find the value of A at x=0,
and x = \frac{30}{4+\pi}
We find that A(0) = 0

Window area is maximum when 
When
, by using 
Therefore maximum light is admitted through the window.
When radius of semi-circular part is 
Therefore the dimensions of the rectangular part of the window are 8,4 ft * 4,2 ft
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