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Q.

Consider a rectangular sheet of perimeter 6 meters. A circular sector of radius equal to smaller side of sheet and centre at one corner is removed such that the area of remaining portion is maximum, then area of removed portion is

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a

$\frac{π}{(4 + π)^{2}}$

b

$\frac{9}{(4 + π)^{2}}$

c

$\frac{36 π}{(4 + π)^{2}}$

d

$\frac{9 π}{(4 + π)^{2}}$

answer is D.

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Detailed Solution

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Area of remaining portion $A = ab - \frac{1}{4} {πa}^{2} = a (3 - a) - \frac{1}{4} {πa}^{2}$

$\Rightarrow \frac{dA}{da} = 3 - 2 a - \frac{πa}{2} = 0 \Rightarrow a = \frac{6}{4 + π} and \Rightarrow \frac{d^{2} A}{{da}^{2}} = - 2 - \frac{π}{2} < 0 \Rightarrow a = \frac{6}{4 + π}$ at $a = \frac{6}{4 + π}, A$ is maximum $\Rightarrow$ Removed portion $= \frac{1}{4} π \cdot a^{2}$

$\Rightarrow \frac{1}{4} π \cdot \frac{36}{(4 + π)^{2}} = \frac{9 π}{(4 + π)^{2}}$

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