Answer
The height to radius ratio = $2$ and $h=\sqrt[3] {\frac{3k}{\pi}}$, where $k$ is the constant volume; see explanations.
Work Step by Step
Step 1. Assume the cylinder of the silo has a radius $R$ and height $h$. The radius of the hemisphere is then $R$.
Step 2. The total volume is fixed; we have $V=\pi R^2h+\frac{2}{3}\pi R^3=k$ where $k$ is a constant, and this gives the relation $h=\frac{3k-2\pi R^3}{3\pi R^2}=\frac{k}{\pi R^2}-\frac{2R}{3}$
Step 3. Assume the unit area cost for the cylinder is $p_1$ and that of the hemisphere is $p_2$; we have $p_2=2p_1$ as a given condition.
Step 4. The total cost can be expressed as $S=p_1A_1+p_2A_2=p_1(2\pi Rh)+p_2(2\pi R^2)$ where $A_1, A_2$ are the surface area of the cylinder and the hemisphere respectively.
Step 5. Replace $h$ with $R$ and rewrite the above equation as $S=2p_1\pi R(\frac{k}{\pi R^2}-\frac{2R}{3})+4p_1\pi R^2)=2p_1\pi R(\frac{k}{\pi R^2}-\frac{2R}{3}+2R)=2p_1\pi R(\frac{k}{\pi R^2}+\frac{4R}{3})=2p_1\pi (\frac{k}{\pi R}+\frac{4R^2}{3})$
Step 6. Take the derivative of the above equation to get $S'=2p_1\pi ( \frac{8R}{3}-\frac{k}{\pi R^2})$. Let $S'=0$, we get $R^3=\frac{3k}{8\pi}$ or $R=\sqrt[3] {\frac{3k}{8\pi}}$ and this gives $h=\frac{3k-2\pi R^3}{3\pi R^2}=\frac{3k-3k/4}{3\pi R^2}=\frac{9k}{12\pi}\times\sqrt[3] {(\frac{8\pi}{3k})^2}=\sqrt[3] {(\frac{9k}{12\pi})^3(\frac{8\pi}{3k})^2}=\sqrt[3] {\frac{3k}{\pi}}$
Step 7. Check $S''=2p_1\pi ( \frac{8}{3}+\frac{2k}{\pi R^3})\gt0$ and the function is concave up indicating a minimum of the cost at this point.
Step 8. We can find the ratio of the height to radius as $\frac{h}{R}=\sqrt[3] {\frac{3k}{\pi}}\times\sqrt[3] {\frac{8\pi}{3k}}=2$
