Show that the altitude of the right circular cone of maximum volume
Question:

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4 r}{3}$.

Solution:

A sphere of fixed radius (r) is given.

Let R and h be the radius and the height of the cone respectively.

The volume (V) of the cone is given by,

$V=\frac{1}{3} \pi R^{2} h$

Now, from the right triangle BCD, we have:

$\mathrm{BC}=\sqrt{r^{2}-R^{2}}$

$\therefore h=r+\sqrt{r^{2}-R^{2}}$

$\begin{aligned} \therefore V &=\frac{1}{3} \pi R^{2}\left(r+\sqrt{r^{2}-R^{2}}\right)=\frac{1}{3} \pi R^{2} r+\frac{1}{3} \pi R^{2} \sqrt{r^{2}-R^{2}} \\ \therefore \frac{d V}{d R} &=\frac{2}{3} \pi R r+\frac{2}{3} \pi R \sqrt{r^{2}-R^{2}}+\frac{\pi R^{2}}{3} \cdot \frac{(-2 R)}{2 \sqrt{r^{2}-R^{2}}} \\ &=\frac{2}{3} \pi R r+\frac{2}{3} \pi R \sqrt{r^{2}-R^{2}}-\frac{\pi R^{3}}{3 \sqrt{r^{2}-R^{2}}} \\ &=\frac{2}{3} \pi R r+\frac{2 \pi R\left(r^{2}-R^{2}\right)-\pi R^{3}}{3 \sqrt{r^{2}-R^{2}}} \\ &=\frac{2}{3} \pi R r+\frac{2 \pi R r^{2}-3 \pi R^{3}}{3 \sqrt{r^{2}-R^{2}}} \end{aligned}$

Now, $\frac{d V}{d R^{2}}=0$

$\Rightarrow \frac{2 \pi r R}{3}=\frac{3 \pi R^{3}-2 \pi R r^{2}}{3 \sqrt{r^{2}-R^{2}}}$

$\Rightarrow 2 r \sqrt{r^{2}-R^{2}}=3 R^{2}-2 r^{2}$

$\Rightarrow 4 r^{2}\left(r^{2}-R^{2}\right)=\left(3 R^{2}-2 r^{2}\right)^{2}$

$\Rightarrow 4 r^{4}-4 r^{2} R^{2}=9 R^{4}+4 r^{4}-12 R^{2} r^{2}$

$\Rightarrow 9 R^{4}-8 r^{2} R^{2}=0$

$\Rightarrow 9 R^{2}=8 r^{2}$

$\Rightarrow R^{2}=\frac{8 r^{2}}{9}$

Now, $\frac{d^{2} V}{d R^{2}}=\frac{2 \pi r}{3}+\frac{3 \sqrt{r^{2}-R^{2}}\left(2 \pi r^{2}-9 \pi R^{2}\right)-\left(2 \pi R r^{2}-3 \pi R^{3}\right)(-6 R) \frac{1}{2 \sqrt{r^{2}-R^{2}}}}{9\left(r^{2}-R^{2}\right)}$

$=\frac{2 \pi r}{3}+\frac{3 \sqrt{r^{2}-R^{2}}\left(2 \pi r^{2}-9 \pi R^{2}\right)+\left(2 \pi R r^{2}-3 \pi R^{3}\right)(3 R) \frac{1}{2 \sqrt{r^{2}-R^{2}}}}{9\left(r^{2}-R^{2}\right)}$

Now, when $R^{2}=\frac{8 r^{2}}{9}$, it can be shown that $\frac{d^{2} V}{d R^{2}}<0 .$

$\therefore$ The volume is the maximum when $R^{2}=\frac{8 r^{2}}{9}$.

When $R^{2}=\frac{8 r^{2}}{9}$, height of the cone $=r+\sqrt{r^{2}-\frac{8 r^{2}}{9}}=r+\sqrt{\frac{r^{2}}{9}}=r+\frac{r}{3}=\frac{4 r}{3}$

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4 r}{3}$.

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.