The perimeter of a sector is p. The area of t | vedclass

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(D) Let $r$ be the radius and $s$ be the arc length of the sector. The perimeter $p$ is given by $p = 2r + s$,so $s = p - 2r$.The area $A$ of the sect

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$\frac{p}{4}$

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(D) Let $r$ be the radius and $s$ be the arc length of the sector. The perimeter $p$ is given by $p = 2r + s$,so $s = p - 2r$.The area $A$ of the sector is given by $A = \frac{1}{2}rs$.Substituting $s = p - 2r$ into the area formula,we get $A = \frac{1}{2}r(p - 2r) = \frac{1}{2}pr - r^2$.To find the maximum area,we differentiate $A$ with respect to $r$ and set the derivative to zero:$\frac{dA}{dr} = \frac{1}{2}p - 2r = 0$.Solving for $r$,we get $2r = \frac{p}{2}$,which implies $r = \frac{p}{4}$.Since $\frac{d^2A}{dr^2} = -2 < 0$,the area is maximum at $r = \frac{p}{4}$.

The perimeter of a sector is $p$. The area of the sector is maximum when its radius is

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