4.12 The sides of rectangle $R$ have lengths $a$ and $b$, which are positive integers. By using a set of lines parallel to any side, $R$ is divided into congruent smaller rectangles with integer side lengths. The number of ways to do this (including the case where $R$ is not divided) is denoted by $D(a, b)$. Try to find the perimeter $p$ of $R$ such that
$$
\frac{D(a, b)}{p}
$$

is maximized.
(Austrian-Polish Mathematical Competition, 1988)
[Solution] The number of divisions parallel to one side equals the number of divisors of the other side.
Let the number of positive divisors of $a$ be $d(a)$, then
$$
\begin{aligned}
f(a, b) & =\frac{D(a, b)}{p} \\
& =\frac{d(a)+d(b)-1}{2(a+b)} .
\end{aligned}
$$

We have
$$
f(2,2)=\frac{2+2-1}{2(2+2)}=\frac{3}{8} .
$$

Below we prove: for any $a, b \in \mathbb{N}$
$$
f(a, b) \leqslant \frac{3}{8} .
$$

Equality holds only when $a=b=2$. Since
$$
\begin{aligned}
& \frac{3}{8}-f(a, b) \\
= & \frac{3}{8}-\frac{d(a)+d(b)-1}{2(a+b)} \\
= & \frac{2+3a-4d(a)+2+3b-4d(b)}{8(a+b)} .
\end{aligned}
$$

It is sufficient to prove that for any $n \in \mathbb{N}$,
$$
2+3n \geqslant 4d(n),
$$

and equality holds only when $n=2$.
$$
n=1 \text { when } 2+3 \cdot 1 \geqslant 4=4d(1) \text {. }
$$
(1) is obviously true.
When $n=2$, we have the equality
$$
2+3 \cdot 2=4d(2)=8 .
$$

Below we prove that when $n \geqslant 3$, $2+3n>4d(n)$.
Let the standard factorization of $n$ be
$$
n=\prod_{i=1}^{k} \prod_{i_{i}}^{\alpha_{i}}
$$

where $p_{i}$ are primes, $\alpha_{i}$ are natural numbers, $i=1,2, \cdots, k$.
Then the number of all divisors of $n$ is
$$
d(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right) .
$$

Below we use mathematical induction (on $\alpha$) to prove: for any $p \geqslant 2, \alpha \in \mathbb{N}$, we always have
$$
p^{\alpha} \geqslant 1+\alpha \text {. }
$$
When $\alpha=1$, $p^{\alpha}=p^{1} \geqslant 2=1+1$, (2) holds;
Assume when $\alpha=t$, we have $\quad p^{t} \geqslant 1+t$.
Then when $\alpha=t+1$,
$$
p^{t+1}=p^{t} \cdot p \geqslant p(1+t) \geqslant 1+(t+1) .
$$

Therefore,
$$
p^{\alpha} \geqslant 1+\alpha .
$$

If $n$ has a prime factor $p \geqslant 3$, then for any $\alpha \in \mathbb{N}$, it is easy to prove by mathematical induction that
$$
3p^{\alpha}>4(1+\alpha) \text {. }
$$

Thus,
$$
\begin{array}{c}
3 \prod_{i=1}^{k} \prod_{i}^{\alpha_{i}}>4 \prod_{i=1}^{k} \prod_{i}^{\alpha_{i}}\left(1+\alpha_{i}\right), \\
2+3n>4d(n) .
\end{array}
$$

If $n=2^{\alpha}$, then for $\alpha \in \mathbb{N}$, and $\alpha>1$, it is easy to prove by mathematical induction that

Thus,
$$
\begin{array}{l}
3 \cdot 2^{\alpha}>2(1+2\alpha) \\
2+3n>4(1+\alpha)
\end{array}
$$

From the above,
$$
\begin{aligned}
\frac{3}{8}-f(a, b) & \geqslant 0, \\
f(a, b) & \leqslant \frac{3}{8} .
\end{aligned}
$$

Therefore, $f(a, b)$ has a maximum value of $\frac{3}{8}$ only when $a=b=2$.