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For maximum light to pass, area should be maximum.

AIM: Maximize area

Let the length of rectangle be L and breadth be B.

which makes each side of triangle be B.

P(Perimeter)= 2*L + 3*B = 16

L= (16-3B)/2

Area= B*L + (31/2B*B)/4 = 8B-(3B2/2)+(31/2B2/4)

dA/dB= 8-3B+(31/2B/2)

At maxima or minima dA/dB=0,

B= 16/(6-31/2)

We will see that d2A/dB 2 @ B= 16/(6-31/2) is smaller than zero which means

that it is a point of maxima.

Now find the value of L and put both of them in the Area equation.

I'm too tired to do so.

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Q: A special window has the shape of a rectangle surmounted by an equilateral triangle If the perimeter of the window is sixteen feet what dimensions will emit the most light?
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