Find the largest cylinder inscribed inside a sphere. Is this calculation correct so far?

Question

A right circular cylinder is inscribed in a sphere of radius $r$. Find the largest possible volume of such a cylinder.

I have that the radius of the cylinder is $r$, the radius of the sphere is $R$, and the height of the inscribed cylinder is $h$. So $$(2r)^2+h^2=(2R)^2$$ $$4r^2 + h^2 = 4R^2$$ $$h=2\sqrt{R^2-r^2}$$ So volume of the cylinder is: $$V=2\pi r^2\sqrt{R^2-r^2}$$ $$V'=4\pi r\sqrt{R^2-r^2}+2\pi r^2\frac d{dr}\sqrt{R^2-r^2}$$ And I think: $$\frac d{dr}\sqrt{R^2-r^2}=\frac r{\sqrt{R^2-r^2}}$$ so $$V'=4\pi r\sqrt{R^2-r^2}+\frac{2\pi r^3}{\sqrt{R^2-r^2}}$$ $$=4\pi r(R^2-r^2)+2\pi r^3$$ $$4\pi rR^2-4\pi r^3+2\pi r^3=4\pi rR^2-2\pi r^3$$ So finding critical values: $$4\pi r R^2-2\pi r^3=0$$ $$4\pi r R^2=2\pi r^3$$ $$4\pi R^2=2\pi r^2$$ $$\frac{4\pi R^2}{2\pi}=r^2$$ $$r=\sqrt2R$$ Someone else who is better at math is getting $r=\sqrt{\frac23}R$. Where did I go wrong?

Answer 1

You made a sign error in computing the derivative of $\sqrt{R^2-r^2}$. $$\frac d{dr}\sqrt{R^2-r^2}=\frac{-r}{\sqrt{R^2-r^2}}$$ Thus $$V'=4\pi r\sqrt{R^2-r^2}-\frac{2\pi r^3}{\sqrt{R^2-r^2}}=0$$ $$4\pi r(R^2-r^2)-2\pi r^3=0$$ $$2(R^2-r^2)=r^2$$ $$2R^2=3r^2$$ $$r=\sqrt{\frac23}R$$

Answer 2

You might enjoy the fact that you actually do not need derivatives. By the AM-GM inequality

$$V^2=16\pi^2\cdot \frac{r^2}{2}\cdot \frac{r^2}{2}\cdot(R^2-r^2)\leq 16\pi^2\left(\frac{R^2}{3}\right)^3 $$ i.e. $V\leq \frac{4\pi R^3}{3\sqrt{3}}$, with equality attained at $\frac{r^2}{2}=R^2-r^2$, i.e. at $r=R\sqrt{\frac{2}{3}}$.