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The perimeter of a sector is p. The area...

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

A

`sqrtp`

B

`(1)/(sqrtp)`

C

`(p)/(2)`

D

`(p)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
D

Perimeter of a sector `=p`. AOB be the sector with radius r. If angle of the sector be `theta` radians, then area of sector `(A)=(1)/(2)r^(2)theta`
Lenth of arc `(d)=rtheta or theta=(s)/(r )`.
Therefore perimeter of the sector
`=(p)=r+s+r=2r+s`
substituting `theta=(s)/(r )` in (i), `A=((1)/(2)r^(2))((s)/(r))=(1)/(2)rs`
`impliess=(2A)/(r )`. Now substituting the value of s in (ii), we get `p=2r+((2A)/(r))` or `2A=pr-2r^(2)`. Differentiating with respect to r. we get `2(dA)/(dr)=p-4r`. We know that for the maximum
value of area `(dA)/(dr)=0` or `p-4r=0` or `r=(p)/(4)`.
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