Minimizing surface area of a cylindrical tank with hemispherical top

The surface area is the sum of the flat bottom disk, the cylindrical side, and the hemispherical cap: $S = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h.$ The volume constraint is $V = \pi r^2 h + \frac{2}{3}\pi r^3 = 250\pi,$ so solving for $h$ gives $h = \frac{250}{r^2} - \frac{2r}{3}.$

Substituting into $S$: $S(r) = 3\pi r^2 + 2\pi r\left(\frac{250}{r^2} - \frac{2r}{3}\right) = 3\pi r^2 + \frac{500\pi}{r} - \frac{4\pi r^2}{3} = \frac{5\pi r^2}{3} + \frac{500\pi}{r}.$ Taking the derivative and setting it to zero: $S'(r) = \frac{10\pi r}{3} - \frac{500\pi}{r^2} = 0 \implies r^3 = 150 \implies r = 150^{1/3}.$ Checking $S''(r) = \frac{10\pi}{3} + \frac{1000\pi}{r^3} > 0$ confirms a minimum, and $h = 250/r^2 - 2r/3$ evaluates to a positive value, so the answer is physically valid.

The second derivative was found to be ten pi over three plus one thousand pi over r cubed, which is always positive for positive r. This positive value confirms the critical point is a minimum, not a maximum. So the answer is positive.