A Window is in the form of rectangle surmounted by semi-circular opening.If the perimeter of the window is 20m,find the dimensions of the window so that the maximum possible light is admitted through the whole opening.

Asked by ANIKETKHADILKAR | 9th Feb, 2011, 05:15: PM

Expert Answer:

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

Answered by  | 9th Feb, 2011, 06:22: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.