The most economical can

Setting up. Volume: $\pi r^2h=V$, so $h=\dfrac V{\pi r^2}$. Surface: $S=2\pi r^2+2\pi rh$. Substituting,
$$S(r)=2\pi r^2+\frac{2V}r,\qquad r>0.$$

Critical point. $S'(r)=4\pi r-\dfrac{2V}{r^2}=0\iff r^3=\dfrac V{2\pi}$. Then
$$h=\frac V{\pi r^2}=\frac{2\pi r^3}{\pi r^2}=2r.$$

It is a minimum. $S'$ is negative before $r_0$ and positive after (the sign of $4\pi r^3-2V$), and $S\to+\infty$ at both ends $0^+$ and $+\infty$. So $r_0$ is the global minimum.

Numerical answer. For $V=1000$: $r=\left(\frac{500}\pi\right)^{1/3}\approx5.42$ cm and $h\approx10.84$ cm, with $S\approx553.6\text{ cm}^2$.

Remark. The optimal can has a square side view (height equals diameter). Real cans are usually taller than this, because of other constraints such as the cost of sealing the edges.