A food company in São Paulo packs one litre of olive oil, which is 1000 cm³, in closed cylindrical…

Step 1. One variable. $h = \dfrac{1000}{\pi r^2}$, so $S = 2\pi r^2 + 2\pi r\,\dfrac{1000}{\pi r^2} = 2\pi r^2 + \dfrac{2000}{r}$, for $r > 0$.

Step 2. The derivative. $S' = 4\pi r - \dfrac{2000}{r^2} = 0$ gives $r^3 = \dfrac{500}{\pi} = 159.15$, so $r = 5.42\ \text{cm}$.

Step 3. The check. $S'' = 4\pi + \dfrac{4000}{r^3} > 0$, so it is a minimum. It is the only critical point, so it is the least value.

Step 4. The height. $h = \dfrac{1000}{\pi \times 5.4193^2} = 10.84\ \text{cm}$, which is $2r$.

Step 5. The least area. $S = 2\pi \times 5.4193^2 + \dfrac{2000}{5.4193} = 184.54 + 369.04 = 553.58\ \text{cm}^2$.

Step 6. The rival tin. For $r = 5$, $S = 2\pi \times 25 + \dfrac{2000}{5} = 157.08 + 400 = 557.08\ \text{cm}^2$. It uses $557.08 - 553.58 = 3.5\ \text{cm}^2$ more.

Why a minimum. For a very small $r$, the term $\dfrac{2000}{r}$ is huge. For a very large $r$, the term $2\pi r^2$ is huge. So the least area lies between, at the critical point.