Question

Show that the right circular cone of the least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

Answer 1

Step 1. Determine the value of the first derivative.

The volume($V$) and surface($S$) of the right circular cone are as follows:

$\begin{array}{rcl}V& =& \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}\\ & \Rightarrow & h=\frac{3V}{{\mathrm{\pi r}}^2}\cdots \left(1\right)\\ \mathrm{S}& =& \mathrm{\pi r}\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}\cdots \left(2\right)\end{array}$

Substitute the value of $h$ into $S$.

$\begin{array}{rcl}\mathrm{S}& =& \mathrm{\pi r}\sqrt{{\mathrm{h}}^2+{\mathrm{r}}^2}\\ \mathrm{S}& =& \mathrm{\pi r}\sqrt{{\left(\frac{3V}{{\mathrm{\pi r}}^2}\right)}^2+{\mathrm{r}}^2}\\ S^2& =& {\pi }^2{r}^2\left[\frac{9V^2}{{\pi }^2{r}^4}+{\mathrm{r}}^2\right]\\ S^2& =& \frac{9V^2}{r^2}+{\pi }^2{r}^4\end{array}$

Determine the first derivative with respect to $r$.

$\begin{array}{rcl}& & \left(S^2\right)\end{array}\text{'}=-\frac{18V^2}{r^3}+4{\mathrm{\pi }}^2{\mathrm{r}}^3$

Hence, the first derivative is $\begin{array}{rcl}& & \left(S^2\right)\end{array}\text{'}=-\frac{18V^2}{r^3}+4{\mathrm{\pi }}^2{\mathrm{r}}^3$.

Step 2. Set the first derivative equal to zero to find $r$.

$\begin{array}{rcl}-\frac{18V^2}{r^3}+4{\mathrm{\pi }}^2{\mathrm{r}}^3& =& 0\\ & \Rightarrow & -18{\mathrm{V}}^2+4{\mathrm{\pi }}^2{\mathrm{r}}^6=0\\ & \Rightarrow & 4{\mathrm{\pi }}^2{\mathrm{r}}^6=18{\mathrm{V}}^2\\ & \Rightarrow & 4{\mathrm{\pi }}^2{\mathrm{r}}^6=18\times \frac{1}{9}{\mathrm{\pi }}^2{\mathrm{r}}^4{\mathrm{h}}^2\left[\therefore \mathrm{V}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}\right]\\ & \Rightarrow & 4{\mathrm{r}}^2=2{\mathrm{h}}^2\\ & \Rightarrow & \mathrm{r}=\frac{\mathrm{h}}{\sqrt{2}}\end{array}$

Hence, the value is $r=\frac{h}{\sqrt{2}}$.

Step 3. Prove that the least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

First, determine the second derivative and then substitute the value of $r$ to prove the given statement.

$\begin{array}{rcl}\begin{array}{c}\left(S^2\right)\end{array}\text{'}\text{'}& =& \frac{54V^2}{r^4}+12{\mathrm{\pi }}^2{\mathrm{r}}^2\\ \begin{array}{l}\Rightarrow \left({\mathrm{S}}^2\right)\end{array}\text{'}\text{'}& =& \frac{54{\mathrm{V}}^2}{{\left({\displaystyle \frac{\mathrm{h}}{\sqrt{2}}}\right)}^4}+12{\mathrm{\pi }}^2{\left(\frac{\mathrm{h}}{\sqrt{2}}\right)}^2\\ & \Rightarrow & \begin{array}{l}\left({\mathrm{S}}^2\right)\end{array}\text{'}\text{'}>0\end{array}$

S is minimum for $r=\frac{h}{\sqrt{2}}\Rightarrow h=\sqrt{2}r$.

Hence, it is proved that the right circular cone of the least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.