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(D) Let $r$ be the radius and $	heta$ be the sectorial angle in radians. The perimeter $P$ of the sector is given by $P = 2r + r	heta = r(2 + 	heta

Vedclass · Accepted Answer

$2^c$

Vedclass · Answer

(D) Let $r$ be the radius and $	heta$ be the sectorial angle in radians. The perimeter $P$ of the sector is given by $P = 2r + r	heta = r(2 + 	heta)$.Since the perimeter is constant,let $P = k$,where $k$ is a constant.Thus,$r(2 + 	heta) = k$,which implies $r = \frac{k}{2 + 	heta}$.The area $A$ of the sector is given by $A = \frac{1}{2} r^2 	heta$.Substituting the value of $r$,we get $A = \frac{1}{2} \left(\frac{k}{2 + 	heta}ight)^2 	heta = \frac{k^2}{2} \cdot \frac{	heta}{(2 + 	heta)^2}$.To find the maximum area,we differentiate $A$ with respect to $	heta$:$\frac{dA}{d	heta} = \frac{k^2}{2} \left[ \frac{(2 + 	heta)^2(1) - 	heta(2)(2 + 	heta)}{(2 + 	heta)^4} ight] = \frac{k^2}{2} \cdot \frac{(2 + 	heta) - 2	heta}{(2 + 	heta)^3} = \frac{k^2}{2} \cdot \frac{2 - 	heta}{(2 + 	heta)^3}$.Setting $\frac{dA}{d	heta} = 0$,we get $2 - 	heta = 0$,which implies $	heta = 2$.For $	heta = 2$,the second derivative $\frac{d^2A}{d	heta^2}$ is negative,confirming that the area is maximum at $	heta = 2^c$.

The perimeter of a sector is constant. If its area is to be maximum,the sectorial angle should be

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