The perimeter of a sector is a constant. If i | vedclass

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(D) Let $r$ be the radius and $	heta$ be the sectorial angle in radians. The perimeter $P$ is given by $P = 2r + r	heta = k$ (constant). Thus,$r = \

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$ 2^{c} $

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

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

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