Ratio of Height to the Radius of its base A conical tent is of given capacity.For the least amount of canvas required for it,find the ratio of its height to the radius of its base.
(capacity usually means its volume.Only units are different)
 A: Let $h$ be an altitude of the tent, $r$ be a radius of the base and $V$ be a volume of the tent. 
 Also, let $h=rx$.
Thus, $V=\frac{\pi r^2h}{3}$ or
$$V=\frac{\pi r^3x}{3},$$
which gives 
$$r=\sqrt[3]{\frac{3V}{\pi{x}}}.$$
Thus, for an area of the canvas by AM-GM we obtain:
$$f(x)=\pi{r}\sqrt{h^2+r^2}=\sqrt[3]{9\pi V^2}\cdot\frac{\sqrt{x^2+1}}{\sqrt[3]{x^2}}=$$
$$=\sqrt[3]{9\pi V^2}\cdot\frac{\sqrt{\frac{x^2}{2}+\frac{x^2}{2}+1}}{\sqrt[3]{x^2}}\geq\sqrt[3]{9\pi V^2}\cdot\frac{\sqrt{3\sqrt[3]{\left(\frac{x^2}{2}\right)^2\cdot1}}}{\sqrt[3]{x^2}}=3\sqrt[3]{\frac{\sqrt3\pi V^2}{2}}.$$
The equality occurs for $\frac{x^2}{2}=1$, which gives $x=\sqrt2$.
Done!
A: Without loss of generality, one can consider the volume of the tent to be $V=\frac{\pi}{3}$.
Thus we have a minimization problem:
$$S(r, h)=\pi\cdot r\cdot \sqrt{h^2+r^2} \to min$$
subject to 
$$V=\frac{\pi}{3}=\frac{\pi r^2h}{3} \Rightarrow r=\frac{1}{\sqrt{h}}.$$
By substitution:
$$S(h)=\pi \cdot \frac{1}{\sqrt{h}}\sqrt{h^2+\frac{1}{h}}=\pi\cdot \sqrt{h+\frac{1}{h^2}}.$$
$$S'(h)=0 \Rightarrow h=\sqrt[3]{2}.$$
$$S''(\sqrt[3]{2})>0.$$
$$r=\frac{1}{\sqrt{h}}=\frac{1}{\sqrt[6]{2}}.$$
Hence:
$$\frac{h}{r}=\sqrt{2}.$$
