Maximum volume of a cone given surface area?

Question

Given a surface area of $2m^2$, what is the maximum volume of an open-top cone?

h=height of cone r=radius of base L=slant height=$√(h^2+r^2)$

$2=πr√(h^2+r^2)$ -> $h=√(4-π^2r^4)/πr$

Plugging the height formula into the Volume Formula: $(πr^2h)/3$

Solving for $r$, I get $0.606m$, giving a max. volume of $0.33m^3$.

Could someone verify this or tell me where I went wrong?

Answer 1

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

$$A = \pi (h^2 + r^2) \frac{2 \pi r}{2 \pi \sqrt{h^2 + r^2}} = \pi r \sqrt{h^2 + r^2}$$

Unrolled and laid flat, the cone looks like:

For a given area $A$, we have the height obeys:

$$\sqrt{ \left( \frac{A}{\pi r} \right)^2 - r^2} = h$$

So

$$V = \frac{1}{3} \pi r^2 \sqrt{ \left( \frac{A}{\pi r} \right)^2 - r^2}$$

$$\frac{dV}{dr} = \frac{A^2-3 \pi ^2 r^4}{3 r \sqrt{\frac{A^2-\pi ^2 r^4}{r^2}}}$$

Set this to $0$ to find:

$$r = \frac{\sqrt{A}}{\sqrt[4]{3} \sqrt{\pi }}$$

.... and you can easily solve for $h$.