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(D) Let the radius of the sector be $r$ and the angle subtended by the radii at the centre be $	heta$ (in radians).The length of the arc is $l = r	h

Vedclass · Accepted Answer

$2$

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(D) Let the radius of the sector be $r$ and the angle subtended by the radii at the centre be $	heta$ (in radians).The length of the arc is $l = r	heta$.The perimeter of the sector is $P = 2r + l = 2r + r	heta = r(2 + 	heta)$.From this,we can express the radius as $r = \frac{P}{2 + 	heta}$.The area of the sector is $A = \frac{1}{2}r^2	heta$.Substituting the expression for $r$ in terms of $P$ and $	heta$:$A = \frac{1}{2} \left(\frac{P}{2 + 	heta}ight)^2 	heta = \frac{P^2}{2} \cdot \frac{	heta}{(2 + 	heta)^2}$.To find the maximum area,we differentiate $A$ with respect to $	heta$ and set it to zero:$\frac{dA}{d	heta} = \frac{P^2}{2} \cdot \frac{(2 + 	heta)^2(1) - 	heta(2)(2 + 	heta)}{(2 + 	heta)^4} = 0$.This simplifies to:$(2 + 	heta)^2 - 2	heta(2 + 	heta) = 0$.Factoring out $(2 + 	heta)$:$(2 + 	heta)(2 + 	heta - 2	heta) = 0$.$(2 + 	heta)(2 - 	heta) = 0$.Since $	heta > 0$,we have $	heta = 2$ radians.Thus,the angle at the centre for maximum area is $2$ radians.

Among all sectors of a fixed perimeter,choose the one with maximum area. Then,the angle at the centre of this sector (i.e.,the angle between the bounding radii) is

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