Prove that a conical tent of given

Solution:

Let the surface area of conical tent be $S=\pi \mathrm{r} \sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$

Let the volume of the conical tent $\mathrm{V}=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}$

$\Rightarrow \mathrm{h}=\frac{3 \mathrm{~V}}{\pi \mathrm{r}^{2}}$

$\therefore S=\pi r \sqrt{r^{2}+\left(\frac{3 V}{\pi r^{2}}\right)^{2}}$

$\Rightarrow \mathrm{S}=\frac{1}{\mathrm{r}} \sqrt{\pi^{2} \mathrm{r}^{6}+9 \mathrm{~V}^{6}}$

Now differentiating with respect to $r$ we get,

$\frac{d S}{d r}=\frac{d}{d r}\left[\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{6}}\right]$

$=\frac{1}{r} \frac{6 \pi^{2} r^{5}}{2\left(\sqrt{\pi^{2} r^{6}+9 V^{6}}\right)}-\frac{\sqrt{\pi^{2} r^{6}+9 V^{6}}}{r^{2}}$

For minima putting $\frac{d S}{d r}=0$ we get,

$\frac{3 \pi^{2} r^{4}}{\sqrt{\pi^{2} r^{6}+g V^{6}}}=\frac{\sqrt{\pi^{2} r^{6}+g V^{6}}}{r^{2}}$

$\Rightarrow 3 \pi^{2} r^{6}=\pi^{2} r^{6}+9 V^{6}$

$\Rightarrow 2 \pi^{2} r^{6}=9 V^{6}$

Substitutting the value of $V$ we get,

$2 \pi^{2} r^{6}=9\left(\frac{1}{3} \pi r^{2} h\right)^{2}$

$\Rightarrow 2 \pi^{2} r^{6}=\pi^{2} r^{4} h^{2}$

$\Rightarrow 2 r^{2}=h^{2}$

$\therefore h=\sqrt{2} r$