Largest cone that can be inscribed in a sphere

Question

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.

My Attempt

r : radius of the base

h : height of the cone

$$ x=\sqrt{R^2-r^2}\implies h=R+x=R+\sqrt{R^2-r^2}\\ V(r)=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi r^2(R+\sqrt{R^2-r^2})\\ V'(r)=\frac{\pi}{3}\big( 2r.(R+\sqrt{R^2-r^2})+r^2\frac{-r}{\sqrt{R^2-r^2}} \big)\\=\frac{\pi}{3}r\big( 2R+2\sqrt{R^2-r^2})-\frac{r^2}{\sqrt{R^2-r^2}} \big)\\=\frac{\pi}{3}r\big( \frac{2R\sqrt{R^2-r^2}+2R^2-2r^2-r^2}{\sqrt{R^2-r^2}} \big)\\=\frac{-\pi}{3}r\big( \frac{2r^2-2R\sqrt{R^2-r^2}-2R^2}{\sqrt{R^2-r^2}} \big) $$

How do I proceed further and find the points where $V'(r)=0$ and prove the above statement ?

Note: I would like to stick with $r$ as the variable.

Answer 1

$2R\sqrt {R^2 - r^2} +2R^2 - 3r^2 = 0\\ 4R^2 (R^2 - r^2) =(2R^2 - 3r^2)^2\\ 4R^4 - 4R^2r^2 = 4R^4 - 12R^2r^2 + 9r^2\\ 8R^2 = 9r^2\\ r = \sqrt {\frac 89} R$

Answer 2

I would write $$V=\frac{1}{3}\pi r^2h$$ then we get $$r^2+x^2=R^2$$ and $$x+R=h$$ and so $$V(x)=\frac{1}{3}\pi(R^2-x^2)(x+R)$$ Or you can write: $$h=x+R$$ and $$x=\sqrt{R^2-r^2}$$ then we have $$V=\frac{1}{3}\pi r^2(\sqrt{R^2-r^2}+R)$$ $$V'(r)=1/3\,{\frac {\pi\,r \left( 2\,{R}^{2}-3\,{r}^{2}+2\,\sqrt { \left( R-r \right) \left( R+r \right) }R \right) }{\sqrt { \left( R-r \right) \left( R+r \right) }}} $$ Solving the equation $$V'(r)=0$$ we get $$r_{opt}=\frac{2\sqrt{2}R}{3}$$