Question

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: 25 (2) 30 (3) 12.5 (4) 10
A. `12.5`
B. 10
C. 25
D. 30

Answer 1

Correct Answer - C
Total length `=2r+r theta=20`
`rArr theta=(20-2r)/(r)`
Now, are of flower-bed,

`A=(1)/(2) r^(2) theta`
`rArr A(1)/(2) r^(2)((20-2r)/(r))`
`rArr A=10r-r^(2)`
`:. (dA)/(dr)=10-2r`
For maximum or minima, put, `(dA)/(dr)=0`
`rArr 10-2r=0 rArr r=5`
`:. A_("max") =(1)/(2)(5)^(2)[ (20-2(5))/(5)]`
`=(1)/(2)xx25xx2=25` sq. m`