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(D) Let $r$ be the radius of the circular sector and $l$ be the length of the arc. The perimeter of the sector is given by $P = 2r + l = 20 \ m$.There

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$5$

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(D) Let $r$ be the radius of the circular sector and $l$ be the length of the arc. The perimeter of the sector is given by $P = 2r + l = 20 \ m$.Therefore,the arc length is $l = 20 - 2r$.The area of a circular sector is given by $A = \frac{1}{2} r l$.Substituting the value of $l$,we get $A(r) = \frac{1}{2} r (20 - 2r) = 10r - r^2$.To find the maximum area,we differentiate $A(r)$ with respect to $r$ and set it to zero:$\frac{dA}{dr} = 10 - 2r$.Setting $\frac{dA}{dr} = 0$,we get $10 - 2r = 0$,which implies $r = 5 \ m$.To verify,the second derivative is $\frac{d^2A}{dr^2} = -2$,which is less than $0$,confirming that the area is maximum at $r = 5 \ m$.

$20 \ m$ of wire is available to fence a flowerbed in the form of a circular sector. If the flowerbed is to have maximum surface area,then the radius of the circle is (in $m$)

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