A conical paper cup is to hold a fixed volume of water. Find the ratio of the height to base radius of the cone which minimizes the amount of paper needed to make the cup. Use the formula πr √r^2+h^2 for the area of the side of a cone, called the lateral area of the cone.

Step 1 ：Given the volume of the cone is fixed, we can use the formula for the volume of a cone, \(V = \frac{1}{3}\pi r^2 h\), to express \(h\) in terms of \(r\) and \(V\). This gives us \(h = \frac{3V}{\pi r^2}\).

Step 2 ：We can then substitute this into the formula for the lateral area of the cone, \(A = \pi r \sqrt{r^2 + h^2}\), to get \(A = \pi r \sqrt{r^2 + \left(\frac{3V}{\pi r^2}\right)^2}\).

Step 3 ：We differentiate this with respect to \(r\) to find the minimum. This gives us \(\frac{dA}{dr} = \pi r \left(-\frac{18V^2}{\pi^2 r^5} + r\right) \div \sqrt{\frac{9V^2}{\pi^2 r^4} + r^2} + \pi \sqrt{\frac{9V^2}{\pi^2 r^4} + r^2}\).

Step 4 ：Solving this equation gives us the solutions for \(r\), which are \(-\frac{2^{5/6} 3^{1/3} (V^2)^{1/6}}{2 \pi^{1/3}}\), \(\frac{2^{5/6} 3^{1/3} (V^2)^{1/6}}{2 \pi^{1/3}}\), and complex solutions which we discard as they are not physically meaningful.

Step 5 ：Substituting the real solutions for \(r\) back into the equation for \(h\) gives us \(h = \frac{6^{1/3} V}{\pi^{1/3} (V^2)^{1/3}}\).

Step 6 ：Finally, we find the ratio \(h/r\) to be \(-\sqrt{2} V / \sqrt{V^2}\).

Step 7 ：Final Answer: The ratio of the height to the base radius of the cone which minimizes the amount of paper needed to make the cup is \(\boxed{-\sqrt{2}}\).