CBSE Class 12-science Answered

Dear Student,

Here is the solution:

Let ABCD be the rectangle and BC = x. Let radius of the semicircle be r. Perimeter of the window = 2r + 2x + πr = 20,

Or, x = 1/2 (20 – 2r – πr) ---------------------- (i)

Area of the figure, A = 2rxX + 1/2 πr² = 2rx1/2 (20 – 2r – πr) + 1/2 πr²

= 20r – 2r² – 1/2 πr².

Then dA/dr = 20 – 4r – πr and d2A/dr² = – 4 – π.

Now, dA/dr = 0 => 20 – 4r – πr = 0 => r = 20/(4 + π).

When r = 20/(4 + π), d2A/dr² = – (4 + π) < 0.

Therefore, A is maximum when r = 20/(4 + π) and then x = 1/2 [20 – (2 + π).

20/(4 + π) = 20/(4 + π).

Hence, maximum light will be admitted when the radius of the semi-circle is 20/(4 + π) and the side BC = 20/(4 + π).

Regards

Team Topperlearning