The base length that will maximize the area for such a window is 168.03 cm. The exact largest value of x when this occurs is 233.39 cm
Suppose we make an assumption that:
- (x) should be the width of the rectangle base;
- (h) should be the height of the rectangle
Also, provided that the diameter of the semi-circle appears to be the base of the rectangle, then;
- the radius
and, the perimeter of the window can now be expressed as:


Given that the perimeter = 600 cm
∴


Since h > 0, then:

By rearrangement and using the inverse rule:



Thus, the largest length x = 233.39 cm
However, the area of the window is given as:

![\mathbf{A = x \Big [ 300 - \Big ( \dfrac{1}{2}+\dfrac{1}{4} \Big) x \Big ] +\dfrac{1}{2}\pi \Big(\dfrac{x}{2} \Big )^2}](https://tex.z-dn.net/?f=%5Cmathbf%7BA%20%3D%20x%20%5CBig%20%5B%20%20300%20-%20%5CBig%20%28%20%5Cdfrac%7B1%7D%7B2%7D%2B%5Cdfrac%7B1%7D%7B4%7D%20%5CBig%29%20x%20%5CBig%20%5D%20%20%2B%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5CBig%28%5Cdfrac%7Bx%7D%7B2%7D%20%5CBig%20%29%5E2%7D)

Now, at maximum, when the area A = 0. Taking the differentiation, we have:


Making x the subject of the formula, we have:

x = 168.03 cm
Taking the second derivative:
![\mathbf{\dfrac{d}{dx} \Big [300 -2x \Big( \dfrac{1}{2} + \dfrac{\pi}{8}\Big) \Big]}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bd%7D%7Bdx%7D%20%5CBig%20%5B300%20-2x%20%5CBig%28%20%5Cdfrac%7B1%7D%7B2%7D%20%2B%20%5Cdfrac%7B%5Cpi%7D%7B8%7D%5CBig%29%20%5CBig%5D%7D)

Therefore, we can conclude that the maximum area that exists for such a window is 168.03 cm
Learn more about derivative here:
brainly.com/question/9964510?referrer=searchResults