Find the dimensions of a cylinder with minimum surface area and volume 1 liter.

Step 1 ：Given a cylinder with volume 1 liter, we need to find the dimensions that minimize the surface area. The volume of a cylinder is given by \(\pi r^2 h = V\), where \(r\) is the radius, \(h\) is the height, and \(V\) is the volume. In this case, \(V = 1\) liter.

Step 2 ：Substituting \(V = 1\) into the volume equation, we get \(\pi r^2 h = 1\). Solving for \(h\), we get \(h = \frac{1}{\pi r^2}\).

Step 3 ：The total surface area of a cylinder, including the two circular ends, is given by \(A = 2 \pi r^2 + 2 \pi rh\). Substituting \(h = \frac{1}{\pi r^2}\) into the surface area equation, we get \(A = 2 \pi r^2 + 2 \pi r \cdot \frac{1}{\pi r^2} = 2 \pi r^2 + \frac{2}{r}\).

Step 4 ：To minimize the surface area, we take the derivative of \(A\) with respect to \(r\) and set it equal to zero. The derivative of \(A\) is \(\frac{dA}{dr} = 4 \pi r - \frac{2}{r^2}\). Setting this equal to zero, we get \(4 \pi r - \frac{2}{r^2} = 0\).

Step 5 ：Solving for \(r\), we get \(r = \left(\frac{1}{2 \pi}\right)^{1/3}\). Substituting this value of \(r\) into the equation \(h = \frac{1}{\pi r^2}\), we get \(h = 2r\).

Step 6 ：So, the dimensions of the cylinder that minimize the surface area are radius \(r = \left(\frac{1}{2 \pi}\right)^{1/3}\) and height \(h = 2r\).

Step 7 ：Finally, we check that these dimensions indeed minimize the surface area by taking the second derivative of \(A\) and verifying that it is positive. The second derivative of \(A\) is \(\frac{d^2A}{dr^2} = 4 \pi + \frac{4}{r^3}\), which is positive for all \(r > 0\). Therefore, the surface area is minimized for these dimensions.

Step 8 ：The final answer is \(r = \left(\frac{1}{2 \pi}\right)^{1/3}\) and \(h = 2r\), which can be simplified to \(r = \boxed{\left(\frac{1}{2 \pi}\right)^{1/3}}\) and \(h = \boxed{2 \left(\frac{1}{2 \pi}\right)^{1/3}}\).