20 m of wire is available to fence a flower | vedclass

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(C) Let $r$ be the radius and $\ell$ be the arc length of the circular sector as shown in the figure. The perimeter of the sector is given by $P = 2r

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let $r$ be the radius and $\ell$ be the arc length of the circular sector as shown in the figure. The perimeter of the sector is given by $P = 2r + \ell = 20 \ m$.Thus,$\ell = 20 - 2r$.The area $A$ of the circular sector is given by $A = \frac{1}{2} \ell r$.Substituting the value of $\ell$,we get $A = \frac{1}{2}(20 - 2r)r = 10r - r^2$.To find the maximum area,we differentiate $A$ with respect to $r$:$\frac{dA}{dr} = 10 - 2r$.Setting $\frac{dA}{dr} = 0$,we get $10 - 2r = 0$,which implies $r = 5 \ m$.To verify the maximum,we find the second derivative: $\frac{d^2A}{dr^2} = -2$.Since $\frac{d^2A}{dr^2} < 0$,the area is maximum at $r = 5 \ m$.

$20 \ m$ of wire is available to fence a flower bed in the form of a circular sector. If the flower bed should have the greatest possible surface area,then the radius of the circle is (in $m$)

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