[Example 2.4.1] There is a rectangular iron sheet with dimensions $80 \times 50$. We need to cut out a square of the same size from each of the four corners and then fold it to form an open-top box. What should be the side length of the square cut out from each corner to maximize the volume of the open-top box?
We can set the length of the cut edges to be $x$, and the volume of the lidless box to be $V$, then
$$
V=x(80-2 x)(50-2 x) \text {. }
$$

How can we find the maximum value of $V$? To find the maximum value of the product, using the arithmetic mean-geometric mean inequality, we should make the "sum" a constant. Generally, we consider
$$
(80-2 x)+(50-2 x)=130-4 x .
$$

To eliminate $x$, we need to transform the $x$ in the expression of $V$ into $\frac{1}{4} \cdot 4 x$. The calculation process is
$$
\begin{aligned}
V & =\frac{1}{4} \cdot 4 x \cdot(80-2 x) \cdot(50-2 x) \\
& \leqslant \frac{1}{4} \cdot\left(\frac{4 x+80-2 x+50-2 x}{3}\right)^{3} \\
& =\frac{130^{3}}{108} .
\end{aligned}
$$
$x$ is eliminated, but the condition for equality to hold is:
$$
4 x=80-2 x=50-2 x
$$

This equation clearly has no solution.
Therefore, the above solution is incorrect.
To obtain the condition for equality to hold and to find the maximum value of the volume $V$, the correct solution is
$$
\begin{aligned}
V & =\frac{1}{12} \cdot 6 x \cdot(80-2 x) \cdot 2(50-2 x) \\
& \leqslant \frac{1}{12} \cdot\left[\frac{6 x+(80-2 x)+2(50-2 x)}{3}\right]^{3} \\
& =\frac{1}{12} \cdot \frac{180^{3}}{3^{3}} \\
& =18000 .
\end{aligned}
$$

In the above inequality, the condition for equality to hold is
$$
6 x=80-2 x=2(50-2 x)
$$

That is
$$
x=10 \text {. }
$$

Therefore, when and only when $x=10$, the volume $V$ has a maximum value, and the maximum value is 18000.
This solution is very simple, but why multiply $x$ by 6 and $50-2 x$ by 2? That is, why transform
$$
V=x(80-2 x)(50-2 x)
$$

into
$$
V=\frac{1}{12} \cdot 6 x \cdot(80-2 x) \cdot 2(50-2 x)
$$

to make the sum in the inequality a constant and the equality hold?
The coefficients in the above expression are the result of a general problem-solving strategy.
We first express the coefficients of $x, 80-2 x, 50-2 x$ in the expression of $V$ using a generalized parameter, at this time
$$
\begin{aligned}
V & =\frac{1}{\alpha \beta \gamma} \cdot \alpha x \cdot \beta(50-2 x) \cdot \gamma(80-2 x) \\
& \leqslant \frac{1}{\alpha \beta \gamma} \cdot\left(\frac{\alpha x+50 \beta-2 \beta x+80 \gamma-2 \gamma x}{3}\right)^{3} \\
& =\frac{1}{\alpha \beta \gamma} \cdot\left[\frac{(50 \beta+80 \gamma)+(\alpha-2 \beta-2 \gamma) x}{3}\right]^{3} .
\end{aligned}
$$

To make the above expression independent of $x$, we need
$$
\alpha-2 \beta-2 \gamma=0 .
$$

To make the equality hold, we also need
$$
\alpha x=50 \beta-2 \beta x=80 \gamma-2 \gamma x .
$$

For computational convenience, we can set $\gamma=1$, then the above two equations can be transformed into
$$
\left\{\begin{array}{l}
\alpha=2 \beta+2, \\
\alpha x=50 \beta-2 \beta x=80-2 x,
\end{array}\right.
$$

Solving these, we get
$$
\beta=-\frac{4}{5} \quad \text { or } \beta=2 \text {. }
$$

We take $\beta=2$, then $\alpha=6$.
This is the reason why we transform $V$ into
$$
V=\frac{1}{12} \cdot 6 x \cdot(80-2 x) \cdot 2(50-2 x)
$$

In solving this problem, we generalize the coefficients of $x, 80-2 x, 50-2 x$, treating $x, 80-2 x, 50-2 x$ as expressions with parameters $2 x, \beta(80-2 x)$, $\gamma(50-2 x)$. These parameters $\alpha, \beta, \gamma$ are dynamic and are fixed at a certain moment to meet the needs of the solution. The generalized parameters then become specific values. Without this generalization process, the maximum value of $V$ would be difficult to find.