The perimeter of a sector is p. What is its r | vedclass

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(D) Let $r$ be the radius and $	heta$ be the central angle in radians.The perimeter of the sector is given by $P = 2r + r	heta = p$.From this,we hav

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$p/4$

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(D) Let $r$ be the radius and $	heta$ be the central angle in radians.The perimeter of the sector is given by $P = 2r + r	heta = p$.From this,we have $	heta = \frac{p - 2r}{r} = \frac{p}{r} - 2$.The area $A$ of the sector is given by $A = \frac{1}{2} r^2 	heta$.Substituting the value of $	heta$,we get $A = \frac{1}{2} r^2 \left( \frac{p - 2r}{r} ight) = \frac{1}{2} r(p - 2r) = \frac{1}{2} pr - r^2$.To find the maximum area,we differentiate $A$ with respect to $r$:$\frac{dA}{dr} = \frac{1}{2} p - 2r$.Setting $\frac{dA}{dr} = 0$,we get $\frac{1}{2} p = 2r$,which implies $r = \frac{p}{4}$.To verify the maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2 < 0$,which confirms that the area is maximum at $r = \frac{p}{4}$.

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

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