<s>[INST] Largest cone that can be inscribed in a sphere 
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.
 [/INST] 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 $$r_{opt}=\frac{2\sqrt{2}R}{3}$$
</s>