The maximum valume (in cu.m) of the right circluar cone having slant height 3 m is

Options

A. `(4)/(3) pi`
B. `2sqrt(3)pi`
C. `3sqrt(3)pi`
D. `6pi`

Answer

B

Explanation

Let h= height of the cone,
r= radius of circular base ltbr `=sqrt((3)^(2)-h^(2))" " [ :. L^(2)=h^(2)+r^(2)]`
`=sqrt(9-h^(2))" "...(i)`

Now, volume (V) of cone `=(1)/(3) pir r^(2)h`
`rArr V(h)=(1)/(3)pi(9-h^(2))h " "` [ From Eq. (ii)]
`=(1)/(3)pi[ 9h-h^(2)] " "...(ii)`
For maximum volumne `V'(h) =0 and V'' (h) le 0`
Here ` V'(h)=0 rArr (9-3h^(2))=0`
`rArr h=sqrt(3)" " [ :. h cancellt]`
and `V''(h)=(1)/(3) pi (-6h) le 0"for" h= sqrt(3)`
Thus, volume is maximum when `h= sqrt(3)`
Now, maximum volume
`V(sqrt(3)=(1)/(3)pi (9sqrt(3)-3sqrt(3))` [ From Eq. (iii)]
`=2sqrt(3)pi`