Four, (Full marks 20 points) A grain warehouse is to be built, with its internal volume $V$ being constant, and $h$, $R$, and $H$ being variables. Determine the values of $\frac{h}{R}$ and $\frac{H}{R}$ to minimize the surface area, i.e., to use the least amount of material.

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Four, $\pi R^{2} H+\frac{\pi}{3} R^{2} h=V$,
$$
S=\pi R \cdot \sqrt{R^{2}+h^{2}}+2 \pi R H+\pi R^{2} .
$$

As shown in Figure 3, let $h=R \tan x\left(0 < x < \frac{\pi}{2}\right)$, then
$$
\begin{aligned}
S & =\pi R \sqrt{R^{2}+R^{2} \tan ^{2} x}+2 \pi R H+\pi R^{2} \\
& =\pi R^{2} \sec x+2 \pi R H+\pi R^{2} \\
& =\pi R^{2} \sec x+2 \pi R\left(\frac{V}{\pi R^{2}}-\frac{1}{3} R \tan x\right)+\pi R^{2} \\
& =\pi R^{2} \sec x+\frac{2 V}{R}-\frac{2}{3} \pi R^{2} \tan x+\pi R^{2} \\
& =\left(\sec x-\frac{2}{3} \tan x+1\right) \pi R^{2}+\frac{2 V}{R} .
\end{aligned}
$$
Since $0 < x < \frac{\pi}{2}$, we have $\sec x > 1 > \frac{1}{3}$, and $\tan x > 0$,
$$
\therefore \sec x-\frac{2}{3} \tan x \\
>\frac{1}{3} \sec x \geqslant \frac{1}{3} . \\
\therefore S>\frac{4}{3} \pi R^{2}+\frac{2}{R} V .
$$
Since when $0 < x < \frac{\pi}{2}$,
$$
\therefore \sec x-\frac{2}{3} \tan x
$$
Multiplying both sides of equation (3) by $\cos x$, we get
$$
\left(S-\frac{2 V}{R}-\pi R^{2}\right) \cos x+\frac{2}{3} \pi R^{2} \sin x=\pi R^{2} \text {. }
$$
Since if $a \cos x+b \sin x=c$ has a solution, then $a^{2}+b^{2} \geqslant c^{2}$,
$$
\therefore\left(S-\frac{2 V}{R}-\pi R^{2}\right)^{2}+\left(\frac{2}{3} \pi R^{2}\right)^{2} \geqslant\left(\pi R^{2}\right)^{2} \text {. }
$$

From (6) $S \geqslant \frac{3+\sqrt{5}}{3} \pi R^{2}+\frac{2 V}{R}$.
When $S=\frac{3+\sqrt{5}}{3} \pi R^{2}+\frac{2 V}{R}$, equation (5) becomes
$$
\frac{\sqrt{5}}{3} \cos x+\frac{2}{3} \sin x=1,
$$

which is $\sin \left(x+\arctan \frac{\sqrt{5}}{2}\right)=1$.
It is easy to see that $x=\frac{\pi}{2}-\arctan \frac{\sqrt{5}}{2}, \tan x=\frac{2 \sqrt{5}}{5}$.
The right side of equation (7) $=\frac{3+\sqrt{5}}{3} \pi R^{2}+\frac{V}{R}+\frac{V}{R}$
$$
\geqslant 3 \sqrt[3]{\frac{3+\sqrt{5}}{3} \pi V^{2}},
$$

When $\frac{3+\sqrt{5}}{3} \pi R^{2}=\frac{V}{R}$, i.e., $R=\sqrt[3]{\frac{3 V}{(3+\sqrt{5}) \pi}}$ the equality holds.
In summary, when $\frac{h}{R}=\tan x=\frac{2 \sqrt{5}}{5}$, and $R=\sqrt[3]{\frac{3 V}{(3+\sqrt{5}) \pi}}$,

then $S_{\text {min }}=3 \sqrt[3]{\frac{3+\sqrt{5}}{3} \pi V^{2}}=\sqrt[3]{9(3+\sqrt{5}) \pi V^{2}}$.
At this time, $\frac{H}{R}=\frac{V}{\pi R^{3}}-\frac{1}{3} \tan x=\frac{15+3 \sqrt{5}}{15}=\frac{5+\sqrt{5}}{5}$.