Find radius that minimizes the surface of the solid

I have solved the following problem:

"A solid, with a volume of $8cm^3$, consists of a cylinder and two equilateral cones, external to the cylinder and each with a base in common with the cylinder itself. Find the base radius so that the surface of the solid is minimal."

but the result doesn't agree with the result of the book, which is $\sqrt[3]{\frac{3+\sqrt{3}}{\pi}}$. I don't see where my mistake is, so I would appreciate some feedback on my solution, thanks.

My solution:

Since the cones are equilateral their lateral surface is $\pi r\cdot 2r=2\pi r^2$; the lateral surface of the cyilinder is $2\pi rh$, where $h$ is the height of the cylinder so the total lateral surface is $S=2\pi r^2+2\pi r h$.

From $V_{tot}=8$ we can find $h$ as a function of $r$ by noting that $V_{cone}=\frac{1}{3}\pi r^2\cdot \sqrt{3}r$, $V_{cyl}=\pi r^2h$ so $V_{tot}=2V_{cone}+V_{cyl}=2\left( \frac{1}{3}\pi r^2\cdot \sqrt{3}r\right)+\pi r^2h=8$ which implies that $h=\frac{8}{\pi r^2}-\frac{2}{\sqrt{3}}r$.

So, $S=2\pi r^2+2\pi r\left(\frac{8}{\pi r^2}-\frac{2}{\sqrt{3}}r \right)=2\pi r^2(1-\frac{2}{\sqrt{3}})+\frac{16}{r}$ hence $S'(r)=4\pi r(1-\frac{2}{\sqrt{3}})-\frac{16}{r^2}=0\Leftrightarrow\fbox{$r=\sqrt[3]{\frac{4\sqrt{3}}{\pi (\sqrt{3}-2)}}$}$.

The surface area of one cone will be $2\pi r^2$ (apothem will be $2r$) so two cones will give you $4\pi r^2$. Thus, $S'(r)=8\pi r(1-\frac{1}{\sqrt{3}})-\frac{16}{r^2}=0\implies r=\sqrt[3]{\frac{2\sqrt{3}}{\pi (\sqrt{3}-1)}}=\sqrt[3]{\frac{3+\sqrt{3}}{\pi }}$