# Maximum volume of a cone given surface area?

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?

• How did you go from an equation with $r$ and $h$ to a solution for $r$ without $h$? Apr 9, 2020 at 22:56
• Sorry! I used the Volume equation to solve for r, which after subbing in h, becomes all in terms of r. Apr 9, 2020 at 23:15 $$V = \frac{1}{3} \pi r^2 h$$

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

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:

$$r = \frac{\sqrt{A}}{\sqrt{3} \sqrt{\pi }}$$

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