Three, (25 points) "If $a, b, c$ are positive real numbers, then $\sqrt[3]{a b c} \leqslant \frac{a+b+c}{3}$, where the equality holds if and only if $a=b=c$." Using the above conclusion, find the maximum value of the function $y=2 x(4-x)(3-2 x)\left(0<x<\frac{3}{2}\right)$.
$$
\begin{array}{l}
\text { Given the known inequality } a b c \leqslant\left(\frac{a+b+c}{3}\right)^{3}. \\
\text { To find the maximum value of } y, \text { it is necessary that the sum of the three factors of } y \text { is a constant and the three factors are equal when } x \text { exists. } \\
\text { Clearly, the three factors of } y \text { do not meet the above conditions, so we need to introduce parameters. } \\
\text { Let } a \text { and } b \text { be positive numbers. Then } \\
y=\frac{2}{a b}[x(4 a-a x)(3 b-2 b x)] \\
\leqslant \frac{2}{a b}\left[\frac{x+(4 a-a x)+(3 b-2 b x)}{3}\right]^{3} \\
=\frac{2}{a b}\left[\frac{4 a+3 b-(a+2 b-1) x}{3}\right]^{3} . \\
\text { When }\left\{\begin{array}{l}
a+2 b-1=0, \\
x=4 a-a x=3 b-2 b x
\end{array}\right. \text {, we solve to get } \\
(x, a, b)=\left(\frac{2}{3}, \frac{1}{5}, \frac{2}{5}\right) \text { or }(3,3,-1) \text { (discard). } \\
\end{array}
$$

Therefore, the maximum value of $y$ is $\frac{2}{a b}\left(\frac{4 a+3 b}{3}\right)^{3}=\frac{200}{27}$.