Twenty metres of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in `s qdotm)` of the flower-bed is: (1) 25 (2) 30 (3) 12.5 (4) 10

Options

A. 12.5
B. 10
C. 25
D. 30

Answer

C

Explanation

Let r be the radius and `theta` ( in radians ) be the sector angle of the circular sector. Further let A be the area of the sector . Then
`A=1/2r^2 theta`

If is given that perimeter of the sector of the of OAB is 20 meters
`therefore r+r theta = 20 rArr =(20-2r)/(r)`
From (i) and (ii) we obtain
` A = 1/2 r^2 ((20 -2r )/(r))ne 10r -r^2`
`rArr (dA)/(dr) =10 -2r and (d^2A)/(dr^2)=-2`
For maximum and minimum values of A we must have
`(dA)/(dr)=0 rArr 10-2r -0 rArr = 5`
Putting r=5 in (ii) we obtain `theta=2`
`therefore A = 1/2xx5^2 xx2 =25 m^2`