Let's analyze the problem step by step.

Given:

1. Volume of the cylinder: $V = \pi r^2 h = 32 \, \text{cm}^3$.
2. Surface area to minimize: $A = 32r^2 + 2\pi r h$, where $r$ is the radius and $h$ is the height.

Steps to solve:

1. From the volume equation $\pi r^2 h = 32$, solve for $h$: $h = \frac{32}{\pi r^2}.$

2. Substitute $h$ into the surface area equation $A$: $A = 32r^2 + 2\pi r \left(\frac{32}{\pi r^2}\right).$ Simplify: $A = 32r^2 + \frac{64}{r}.$

3. Differentiate $A$ with respect to $r$ to find the critical points: $\frac{dA}{dr} = 64r - \frac{64}{r^2}.$

4. Set $\frac{dA}{dr} = 0$ to minimize $A$: $64r - \frac{64}{r^2} = 0.$ Multiply through by $r^2$ to eliminate the denominator: $64r^3 = 64 \quad \implies \quad r^3 = 1 \quad \implies \quad r = 1.$

5. Find the corresponding $h$ using $h = \frac{32}{\pi r^2}$: $h = \frac{32}{\pi (1)^2} = \frac{32}{\pi}.$

6. Calculate the ratio $\frac{h}{r}$: $\frac{h}{r} = \frac{\frac{32}{\pi}}{1} = \frac{32}{\pi}.$

Final Answer:

The ratio of $h$ to $r$ for the most economical can is: $\boxed{\frac{32}{\pi}}$

Let me know if you want additional clarification or further steps! Here are some related questions to expand this topic:

1. How would the solution change if the volume of the cylinder were different?
2. Can you explain why minimizing surface area is equivalent to minimizing material waste in this context?
3. What role does differentiation play in solving optimization problems like this?
4. How does the ratio $h/r = \frac{32}{\pi}$ compare to the dimensions of a sphere with the same volume?
5. Could you explore how manufacturing constraints (e.g., material thickness) might affect the optimization problem?

Tip: Always double-check the units in optimization problems, as they often hint at what the variables represent!