Sixteen meter of wire is available to fence off a flower bed in the form of a sector of a circle. If the flower bed has the maximum surface area, then the radius is

Question

Infinity Learn · Accepted Answer

Let ‘r’ be the radius of the sector and angle subtended at the centre be $θ$

Then, S=Surface area of sector

$= \frac{θ}{360} \times π r^{2}$

$We know θ = \frac{length of arc}{radius}$

$\begin{array}{l} = \frac{16 - 2 r}{r} \\ S = \frac{θ}{2 π} π \cdot r^{2} = \frac{16 - 2 r}{2 r} r^{2} \\ S = (8 - r) \cdot r = 8 r - r^{2} \\ \Rightarrow \frac{d S}{d r} = 8 - 2 r \end{array}$

Now for area to be maximum

$\begin{array}{l} \Rightarrow \frac{d S}{d r} = 0 \\ \Rightarrow r = 4 \end{array}$

Maxima and minima

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Sixteen meter of wire is available to fence off a flower bed in the form of a sector of a circle. If the flower bed has the maximum surface area, then the radius is

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Maxima and minima

Remember concepts with our Masterclasses.

Sixteen meter of wire is available to fence off a flower bed in the form of a sector of a circle. If the flower bed has the maximum surface area, then the radius is

Let ‘r’ be the radius of the sector and angle subtended at the centre be θThen, S=Surface area of sector=θ360×πr2 We know θ= length of arc radius =16−2rrS=θ2ππ⋅r2=16−2r2rr2S=(8−r)⋅r=8r−r2⇒dSdr=8−2rNow for area to be maximum⇒dSdr=0⇒r=4

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