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# 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.

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 Expert 9th February 2011, 6:22 PM
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