Optimal radius and height minimizing surface area of a closed cylinder

From the constraint $\pi r^2 h = 355$, solve for $h$ to get $h = \dfrac{355}{\pi r^2}$. Substituting into $S = 2\pi r^2 + 2\pi r h$ gives $S(r) = 2\pi r^2 + \dfrac{710}{r}$, a function of the single variable $r$, defined for $r>0$.

Differentiate: $S'(r) = 4\pi r - \dfrac{710}{r^2}$. Setting $S'(r)=0$ gives $4\pi r^3 = 710$, so $r^3 = \dfrac{710}{4\pi} = \dfrac{355}{2\pi}$, hence $r = \left(\dfrac{355}{2\pi}\right)^{1/3} \approx 3.84\text{ cm}$.

Since $S''(r) = 4\pi + \dfrac{1420}{r^3} > 0$ for all $r>0$, this critical point is a minimum. Substituting back, $h = \dfrac{355}{\pi r^2} \approx \dfrac{355}{\pi (3.84)^2} \approx 7.68\text{ cm}$, which is approximately twice the radius, matching the expected result that $h=2r$ minimizes surface area for fixed volume.