If the area of a circular sector of perimeter | vedclass

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(B) Given,the perimeter of the circular sector is $P = 60 \ m$.Let the radius be $r \ m$ and the arc length be $l \ m$.The perimeter of a circular sec

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$15$

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(B) Given,the perimeter of the circular sector is $P = 60 \ m$.Let the radius be $r \ m$ and the arc length be $l \ m$.The perimeter of a circular sector is given by $P = l + 2r$.Substituting the given perimeter,we have $60 = l + 2r$,which implies $l = 60 - 2r$.The area $A$ of a circular sector is given by $A = \frac{1}{2}lr$.Substituting $l = 60 - 2r$ into the area formula,we get:$A = \frac{1}{2}(60 - 2r)r = 30r - r^2$.To maximize the area,we find the derivative of $A$ with respect to $r$ and set it to zero:$\frac{dA}{dr} = \frac{d}{dr}(30r - r^2) = 30 - 2r$.Setting $\frac{dA}{dr} = 0$ gives $30 - 2r = 0$,so $2r = 30$,which means $r = 15 \ m$.To confirm this is a maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2$,which is less than $0$,confirming that $r = 15 \ m$ provides the maximum area.

If the area of a circular sector of perimeter $60 \ m$ is to be maximized,then its radius must be......... $m$.

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