Question 2. The sum of several positive numbers is 1983. When is their product the largest, and what is the maximum value?

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From the previous problem, if the sum of $n$ positive numbers is 1983, then their product reaches its maximum value when these $n$ numbers are equal, and the maximum value $M_{2}$ is $\left(\frac{1983}{n}\right)^{n}$. However, unlike in problem 1, in this problem, $n$ is not a predetermined number; we need to determine $n$ such that $\left(\frac{1983}{n}\right)^{n}$ is maximized.
Taking the logarithm (we use the natural logarithm),
$$
\log \left(\frac{1983}{n}\right)^{n}=n \log 1983-n \log n.
$$

The problem we face is to find the value of $n$ for which $n \log 1983-n \log n$ is maximized.
Consider the function
$$
y=x \log 1983-x \log x.
$$

Since its derivative is
$$
y'=\log 1983-1-\log x=\log \frac{1983}{e x},
$$

when $x<\frac{1983}{e}$, $y'>0$; when $x=\frac{1983}{e}$, $y'=0$; when $x>\frac{1983}{e}$, $y'<0$;
when $x<\frac{1983}{e}$, the function $y$ is increasing; when $x=\frac{1983}{e}$, the function reaches its maximum; when $x>\frac{1983}{e}$, the function $y$ is decreasing; at $x=\frac{1983}{e}$, the function reaches its maximum value $\frac{1983}{e}$.

However, $\frac{1983}{e}$ is not an integer (as is well known, the base of the natural logarithm $e=2.71828 \cdots$), so the maximum value of the function $y=x \log 1983-x \log x$ is not the maximum value of $n \log 1983-n \log n$!

How do we find the maximum value of $n \log 1983-n \log n$? We need to use the properties of the function $y=x \log 1983-x \log x$.

When $n=\left[\frac{1983}{e}\right]$, $n \log 1983-n \log n$ is maximized. Here, $[x]$ denotes the integer part of $x$, for example, $\left[\frac{1983}{e}\right]=729$. Similarly, when $n>\frac{1983}{e}$, $n \log 1983-n \log n$ is maximized at $n=\left[\frac{1983}{e}\right]+1$. After calculation, we find
$$
\begin{array}{l}
{\left[\frac{1983}{e}\right] \log 1983-\left[\frac{1983}{e}\right] \log \left[\frac{1983}{e}\right]} \\
<\left(\left[\frac{1983}{e}\right]+1\right) \log 1983-\left(\left[\frac{1983}{e}\right]+1\right) \\
\log \left(\left[\frac{1983}{e}\right]+1\right). \text{ Therefore, } n \log 1983-n \log n \text{ is maximized at } \\
n=\left[\frac{1983}{e}\right]+1=730, \text{ and the maximum value is } \\
\left(\left[\frac{1983}{e}\right]+1\right) \log 1983-\left(\left[\frac{1983}{e}\right]+1\right) \\
\log \left(\left[\frac{1983}{e}\right]+1\right)=730 \log 1983-730 \log 730.
\end{array}
$$
Thus, the answer to this problem is: the maximum value of the product is $\left(\frac{1983}{730}\right)^{730}$, and it is achieved when the number of addends is 730, and each addend is equal to $\frac{1983}{730}$.

In general, if the sum of several positive numbers is a fixed value $m$, then their product's maximum value is $\left(\frac{m}{\left[\frac{m}{e}\right]}\right)^{\left[\frac{m}{e}\right]}$ or $\left(\frac{m}{\left[\frac{m}{e}\right]+1}\right)^{\left[\frac{m}{e}\right]+1}$, and the maximum value is achieved when these numbers are all equal to $\frac{m}{\left[\frac{m}{e}\right]}$ or all equal to $\frac{m}{\left[\frac{m}{e}\right]+1}$.

It is worth noting that the maximum value of the function $y=f(x)$ and $f(n)$ (where $n$ is a natural number) is not necessarily the same. Readers can consider another example: the quadratic function $y=2 x^{2}-9 x+1$ and $2 n^{2}-9 n+1$ have their minimum values at $x=\frac{9}{4}$ and $n=2$, respectively.