The slant height of a right circular cone is $3 \mathrm{~cm}$. The height of the cone for maximum volume is
The slant height of a right circular cone is $3 \mathrm{~cm}$. The height of the cone for maximum volume is
- $5 \mathrm{~cm}$
- $\sqrt{5} \mathrm{~cm}$
- $3 \mathrm{~cm}$
- $\sqrt{3} \mathrm{~cm}$
Solution
Volume of cone $=\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h}$
We have slant height $\ell=3 \mathrm{~cm}$ and we know that $\ell^2=\mathrm{r}^2+\mathrm{h}^2$
$\begin{aligned}
& \therefore 9=\mathrm{r}^2+\mathrm{h}^2 \Rightarrow \mathrm{r}^2=9-\mathrm{h}^2 \\
& \therefore \mathrm{v}=\frac{1}{3} \pi\left(9-\mathrm{h}^2\right) \mathrm{h}=(3 \pi) \mathrm{h}-\left(\frac{\pi}{3}\right) \mathrm{h}^3 \\
& \frac{\mathrm{dv}}{\mathrm{dh}}=3 \pi-\left(\frac{\pi}{3}\right)\left(3 \mathrm{~h}^2\right)=3 \pi-\pi \mathrm{h}^2
\end{aligned}$
When $\frac{\mathrm{dv}}{\mathrm{dh}}=0$, we get $3 \pi=\pi \mathrm{h}^2 \Rightarrow \mathrm{h}=\sqrt{3}$
$\frac{\mathrm{d}^2 \mathrm{v}}{\mathrm{dh}^2}=0-2 \pi \mathrm{h} \quad \Rightarrow\left(\frac{\mathrm{d}^2 \mathrm{v}}{\mathrm{dh}^2}\right)_{\mathrm{b}-\sqrt{3}}=-2 \sqrt{3} \pi < 0$
$\therefore$ Volume of cone is maximum when $\mathrm{h}=\sqrt{3}$
Asked in: MHT CET 2021 (22 Sep Shift 2)
Practice more Applications of Derivatives questions on Aicharya