Answer
$h=r=\frac{10}{\sqrt[3] \pi}$
Work Step by Step
Step 1. Assume the cylinder has a base radius of $r$ and a height of $h$; its volume can then be expressed as $V=\pi r^2h=1000cm^3$, which gives $h=\frac{1000}{\pi r^2}$
Step 2. The surface area of this open top cylinder can be found as $S=\pi r^2+2\pi rh=\pi r^2+\frac{2000}{r}$
Step 3. For a lightest can, we need to minimize the surface area. Take the derivative of the above equation to get $S'=2\pi r-\frac{2000}{ r^2}$
Step 4. Let $S'=0$ to get $r^3=\frac{1000}{\pi}, r=\frac{10}{\sqrt[3] \pi}\approx6.83cm$, $h=\frac{1000}{\pi r^2}=\frac{1000\sqrt[3] {\pi^2}}{100\pi}=\frac{10}{\sqrt[3] \pi}=r$ and $S=\pi r^2+2\pi r^2=3\pi r^2\approx439.4cm^2$ .
Step 5. The above $S$ value is a minimum because $S''=2\pi+\frac{4000}{r^3}\gt0$ and the function is concave up around $ r=\frac{10}{\sqrt[3] \pi}$.
Step 6. The result from Example 2, where both the top and bottom surfaces were considered, gives $h=2r$ for a minimum surface area, compared to $h=r$ in this case. Thus the can in Example 2 was tall and thin compared to the can in this case at a minimum of weight.
