Imagine, I choose $a$ as the volume of the cone. Find an expression with $a$ as variabele for the minimum surface.
My Work:
- $$\text{Volume}=\frac{\pi r^2h}{3}=a\to r=\sqrt{\frac{3a}{\pi h}}$$
- $$\text{Surface Cone}=\pi r^2+\pi r\sqrt{r^2+h^2}$$
- Filling in my $r$ in my surface expression: $$\text{Surface Cone}=\pi\left[\sqrt{\frac{3a}{\pi h}}\right]^2+\pi \left[\sqrt{\frac{3a}{\pi h}}\right]\sqrt{\left[\sqrt{\frac{3a}{\pi h}}\right]^2+h^2}=\frac{3a}{h}+\sqrt{3}\sqrt{\frac{a}{h}}\sqrt{\frac{3a}{h}+h^2\pi}$$
- Take the derivative, to find the minumum: $$\frac{\text{d}}{\text{d}h}\text{Surface Cone}=0\Longleftrightarrow\space\space\space h=2\sqrt[3]{a}\sqrt[3]{\frac{3}{\pi}}$$
- Substitute the found $h$ in the $\text{Surface Cone}$ expression:
$$\text{Minimum Surface Cone}=2\sqrt[3]{3^2}\sqrt[3]{\pi}\sqrt[3]{a^2}$$
Am I right?
Let $V=\frac{1}{3}\pi r^2h$ and we have to minimize $$A(r,h)=\pi r^2+\pi r\sqrt{r^2+h^2}$$ we have $$h=\frac{3V}{\pi r^2}$$ and our objective function will be $$A(r,\frac{3V}{\pi r^2})=\pi r^2+\sqrt{r^2+\frac{9V^2}{\pi^2 r^4}}$$ this must be differentiated with respect to $r$.