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 is $l = r	heta$.Perimeter $P = 2r + r	heta

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 is $l = r	heta$.Perimeter $P = 2r + r	heta = r(2 + 	heta)$.Thus,$r = \frac{P}{2 + 	heta}$.The area of the sector is $A = \frac{1}{2} r^2 	heta$.Substituting $r$,we get $A = \frac{1}{2} \left( \frac{P}{2 + 	heta} ight)^2 	heta = \frac{P^2}{2} \cdot \frac{	heta}{(2 + 	heta)^2}$.To maximize $A$,we differentiate with respect to $	heta$ and set it to $0$:$\frac{dA}{d	heta} = \frac{P^2}{2} \left[ \frac{(2 + 	heta)^2 - 	heta \cdot 2(2 + 	heta)}{(2 + 	heta)^4} ight] = 0$.$(2 + 	heta)^2 - 2	heta(2 + 	heta) = 0$.Since $2 + 	heta 
eq 0$,we have $2 + 	heta - 2	heta = 0$,which gives $	heta = 2$.Therefore,the area is maximum when the sectorial angle 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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