Q96 / 180 · KCET 2024 · Mathematics
The maximum volume of the right circular cone with slant height 6 units is
- A$4 \sqrt{3} \pi$ cu units
- Correct answerB$16 \sqrt{3} \pi$ cu units
- C$3 \sqrt{3} \pi$ cu units
- D$6 \sqrt{3} \pi$ cu units
Solution
$\because$ Slant height of the cone, $L=6$ units.
Let the radius be $r$ and height be $h$.
and volume $(V)=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \pi\left(L^2-h^2\right) h \quad\left[\because L^2=r^2+h^2\right]$
$=\frac{1}{3} \pi\left(36-h^2\right) h$
So, $\frac{d V}{d h}=\frac{1}{3} \pi\left(36-3 h^2\right)=0 \Rightarrow h=2 \sqrt{3}$ units
Now, $\frac{d^2 V}{d h^2}=-2 \pi h \lt 0$ at $h=2 \sqrt{3}$ units
Hence, the maximum volume of the cone
$V=\frac{1}{3} \pi\left(36-h^2\right) h$
$=\frac{1}{3} \pi\left(36-(2 \sqrt{3})^2\right) 2 \sqrt{3}$
$=16 \sqrt{3} \pi$ cu units
Hence, the correct answer is option: 'B'.