Example 8. There is a rectangular sheet of size $80 \times 50$. Now, we need to cut off a square of the same size from each corner and then make it into an open box. What should be the side length $y$ of the square to be cut off so that the volume of this open box is maximized?
Let the side length of the cut-out square be $x$, then the volume of the box made is
$$
V=x(80-2 x)(50-2 x) .
$$

To find the maximum value of $V$, a common approach is to try to make the product a constant, which involves converting $x$ into $4x$.
$$
\begin{aligned}
V & =\frac{1}{4} \cdot 4 x(80-2 x)(50-2 x) \\
& \leqslant \frac{1}{4} \cdot\left(\frac{4 x+80-2 x+50-2 x}{3}\right)^{3} \\
& =\frac{130^{3}}{108} .
\end{aligned}
$$

However, the condition for equality to hold should be $4 x=80 - 2 x=50-2 x$, which is impossible. Thus, this solution is incorrect. To achieve equality and find the extremum, the correct approach is
$$
\begin{array}{l}
V=\frac{1}{12} \cdot 6 x \cdot(80-2 x) \cdot 2(50-2 x) \\
\leqslant \frac{1}{12} \cdot\left[\frac{6 x+(80-2 x)+(100-4 x)}{3}\right]^{3} \\
=18000 .
\end{array}
$$

Equality holds when and only when $6 x=80-2 x=100-4 x$, i.e., $x=10$, at which $V$ reaches its maximum value.
This solution is concise, but why is it transformed this way?

This can be explained by introducing parameters into the inequality to obtain the answer.
$$
\begin{array}{l}
V=\frac{1}{\alpha \beta} \cdot \alpha x \cdot \beta(50-2 x) \cdot(80-2 x) \\
\leqslant \frac{1}{\alpha \beta}\left(\frac{\alpha x+50 \beta-2 \beta x+80-2 x}{3}\right)^{3} \\
=\frac{1}{\alpha \beta}\left[\frac{(50 \beta+80)+(\alpha-2 \beta-2) x}{3}\right]^{3} .
\end{array}
$$

To make (1) independent of $x$, it is necessary that
$$
\alpha-2 \beta-2=0 .
$$

For equality to hold, it must also be true that
$$
\alpha x=50 \beta-2 \beta x=80-2 x .
$$

From (2) and (3), we can obtain $\alpha=6, \beta=2$.