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 \(\boxed{-\sqrt{2}}\).