Question

A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is 30 m, find the dimensions so that the greatest possible amount of light may be admitted.

Answer 1

Let x be the length, y be the breadth of the rectangle and r be the radius of the semicircle. Then perimeter of the window = x + 2y + πr, where x = 2r

This is given to be 30 m

The greatest possible amount of light may be admitted if the area of the window is maximized. 

Let A be the area of the window.

Hence, the required dimensions of the window are as follows:

Length of rectangle = \(\left(\frac{60}{\pi + 4}\right)\) meters

breadth of rectangle = \(\left(\frac{30}{\pi + 4}\right)\) meters

radius of the semicircle = \(\left(\frac{30}{\pi + 4}\right)\) meters