The perimeter of a sector is a constant. If its area is to be maximum, the sectorial angle is

Options

A. `(pi)/6 `rad
B. `(pi)/4` rad
C. `4` rad
D. 2 rad

Answer

D

Explanation

Let length of sector is `l` and radius of sector is `r`.
`:.l=(2pi r theta)/(360^(@))`
Perimeter of sector `P=(2pi r theta)/(360^(@))+ 2r`
`implies r=P/(((2pi theta)/(360^(@))+2))`
`:.` Area of sector `A=(pi r^(2) theta)/(360^(@))=(pi)/(360^(@))[ (P^(2))/(((2pi theta)/(360^(@))+2)^(2))] theta`
`implies A=(piP^(2))/(360^(@)) [ (theta)/(((2pi theta)/(360^(@))+2)^(2))]`
`implies (dA)/( d theta)=(piP^(2))/(360^(@))[(((2pi theta)/(360^(@))+2)^(2)-theta .2 ((2pi theta)/(360^(@))+2)(2pi)/(360^(@)))/(((2 pi theta)/(360^(@))+2)^(4))]`
For maxima and minima put `(dA)/(d theta) =0`
`((2 pi theta)/(360^(@))+2)[(2 pi theta)/(360^(@))+2-(4 pi theta)/(360^(@))]=0`
`implies ((pi theta)/(360^(@))+1)(1-(pi theta)/(360^(@)))=0implies(pi theta)/(360^(@))=1`
`implies (pi theta)/(180^(@))=2implies theta =2` rad
Thus, area of sector will be maximum, if sectorial angle is of 2 rad.