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

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(D) Let the radius of the sector be $r$ and the sectorial angle be $	heta$ (in radians). The arc length $l = r	heta$. The perimeter $P = l + 2r = r\

Vedclass · Accepted Answer

$ 2^c $

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(D) Let the radius of the sector be $r$ and the sectorial angle be $	heta$ (in radians). The arc length $l = r	heta$. The perimeter $P = l + 2r = r	heta + 2r = r(	heta + 2)$. Thus,$r = \frac{P}{	heta + 2}$. The area of the sector $A = \frac{1}{2}r^2	heta$. Substituting $r$,we get $A = \frac{1}{2} \left( \frac{P}{	heta + 2} ight)^2 	heta = \frac{P^2}{2} \cdot \frac{	heta}{(	heta + 2)^2}$. To maximize $A$,we differentiate with respect to $	heta$ and set to $0$: $\frac{dA}{d	heta} = \frac{P^2}{2} \left[ \frac{(	heta + 2)^2 - 	heta \cdot 2(	heta + 2)}{(	heta + 2)^4} ight] = 0$. This implies $(	heta + 2) - 2	heta = 0$,so $2 - 	heta = 0$,which gives $	heta = 2 	ext{ radians}$. Therefore,the sectorial angle for maximum area is $2^c$.

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

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