This is a question inspired from What is the relation between height and radius of base of a right circular cone when its surface area is given and volume is to be maximum?
In the linked post, OP has solved for critical points for the problem and asks for verification, and he/she doesn't determine whether the critical points are local maximum.
Is there a way to solve this constrained optimization problem using elementary inequality, like a.m.-g.m.-inequality or cauchy-schwarz-inequality?
I've got the motivation since verification of local maximality is usually a second-order condition. (i.e. involves second-derivatives) The calculation can get messy in multivariable-calculus. An elementary inequality approach often gives more elegant and concise solution.
My attempt:
Keep the notations in the linked post.
- $r$: radius of base
- $h$: height of cone
- $V = \frac13\pi r^2h$: volume of cone
- $A = \pi r (r + \sqrt{r^2 + h^2})$: surface area of cone
Apply a.m.-g.m.-inequality on $rh$:
$$rh \le \frac{r^2+h^2}{2}.$$
$$V = \frac13\pi r^2h \le \frac13 \pi r \,\frac{r^2+h^2}{2} = \frac16 \pi r \left(\sqrt{r^2+h^2}\right)^2 \le \frac16 \pi r \left(r + \sqrt{r^2+h^2}\right)^2$$
Then I don't know how to continue.
Best Answer
We have $$A=\pi r(r+\sqrt{r^2+h^2})$$ (the surface area). solving this formula for $h$ we get $$h=\sqrt{\frac{A^2}{\pi^2 r^2}-\frac{2A}{\pi}}$$ plugging this in the formula for $V$ $$V=\frac{\pi}{3}\sqrt{\frac{A^2r^2}{\pi^2}-\frac{2A}{\pi}r^4}$$ Now we consider the radicand only which is equal to $$-\frac{2A}{\pi}\left(r^2-\frac{A}{4\pi}\right)^2+\frac{A^3}{8\pi^3}$$ So we have $$V=\frac{\pi}{3}\sqrt{\frac{A^3}{8\pi^3}-\frac{2A}{\pi}\left(r^2-\frac{A}{4\pi}\right)^2}$$ So we get $$V\le \frac{\pi}{3}\sqrt{\frac{A^3}{8\pi^3}}$$ and the equal sign holds if $$r^2=\frac{A}{4\pi}$$