A closed cylindrical tin must hold $250\pi$ cm³. What radius, in cm, uses the least metal?

Explanation

1. 01

   $\pi r^2 h = 250\pi$, so $h = \frac{250}{r^2}$, and the surface is $S = 2\pi r^2 + \frac{500\pi}{r}$.

2. 02

   $S' = 4\pi r - \frac{500\pi}{r^2} = 0$ gives $r^3 = 125$, $r = 5$.

3. 03

   Then $h = 10$: the best tin has its height equal to its diameter.