A cylindrical can is to hold $20 \mathrm{~m}^{3}$. The material for the top and bottom costs $\$ 10 / \mathrm{m}^{2}$ and the material for the side costs $\$ 8 / \mathrm{m}^{2}$. Find the radius and height of the most economical can. (3 pts)

Step 1 ：Define the symbols for the radius \(r\).

Step 2 ：Define the cost function \(C = 20\pi r^2 + \frac{320}{r}\).

Step 3 ：Compute the derivative of the cost function \(C'(r) = 40\pi r - \frac{320}{r^2}\).

Step 4 ：Solve the equation \(C'(r) = 0\) to find the value of \(r\). The solutions are \(r = \frac{2}{\pi^{1/3}}\), \(r = -\frac{1}{\pi^{1/3}} - \frac{\sqrt{3}i}{\pi^{1/3}}\), and \(r = -\frac{1}{\pi^{1/3}} + \frac{\sqrt{3}i}{\pi^{1/3}}\).

Step 5 ：Since the radius of a cylinder cannot be a complex number, we discard the complex solutions and keep only the real solution \(r = \frac{2}{\pi^{1/3}}\).

Step 6 ：Compute the height for the solution of \(r\), \(h = \frac{20}{\pi r^2} = \frac{5}{\pi^{1/3}}\).

Step 7 ：Final Answer: The radius and height of the most economical can are approximately \(\boxed{\frac{2}{\pi^{1/3}}}\) meters and \(\boxed{\frac{5}{\pi^{1/3}}}\) meters, respectively.