The perimeter of a sector is $P$ . The area of the sector is maximum when its radius is

1 / \sqrt{P}

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P / 2

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P / 4

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\sqrt{P}

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Solution

The correct option is A $P / 4$
Now perimeter of the sector will be , $P = 2 r + r θ \to θ = \frac{P - 2 r}{r}$
$A r e a = \frac{π r^{2}}{2 π} θ = \frac{r^{2} θ}{2}$ Putting value,
$A r e a = \frac{r (P - 2 r)}{2}$
For maximum area, differentiating and equating it to $0$ , we have
$\frac{P - 4 r}{2} = 0 \to r = \frac{P}{4}$
Hence, option C is correct.

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The perimeter of a sector is P. The area of the sector is maximum when its radius is

The correct option is A P/4Now perimeter of the sector will be , P=2r+rθ→θ=P−2rrArea=πr22πθ=r2θ2 Putting value,Area=r(P−2r)2 For maximum area, differentiating and equating it to 0, we have P−4r2=0→r=P4 Hence, option C is correct.

The perimeter of a sector is $P$ . The area of the sector is maximum when its radius is