11. (20 points) For any real numbers $a_{1}, a_{2}, \cdots, a_{5}$ $\in[0,1]$, find the maximum value of $\prod_{1 \leq i<j \leq 5}\left|a_{i}-a_{j}\right|$.
11. Let \( f\left(a_{1}, a_{2}, \cdots, a_{5}\right)=\prod_{1 \leqslant i < j \leqslant 5} \left(a_{i}-a_{j}\right) \) where \( 1 > a_{1} > a_{2} > \cdots > a_{5} > 0 \).

To maximize \( f\left(a_{1}, a_{2}, \cdots, a_{5}\right) \), it must be that \( a_{1}=1 \) and \( a_{5}=0 \).
Thus, \( f\left(1, a_{2}, a_{3}, a_{4}, 0\right) \)
\[
=\left(a_{2}-a_{3}\right)\left(a_{2}-a_{4}\right)\left(a_{3}-a_{4}\right) \prod_{k=2}^{4} a_{k}\left(1-a_{k}\right).
\]

Let \( a_{2}-a_{4}=x \). Then \( x \in (0,1) \).
By the AM-GM inequality, we have
\[
\begin{aligned}
f & \left(1, a_{2}, a_{3}, a_{4}, 0\right) \\
= & \left(1-a_{2}\right) a_{4}\left(1-a_{3}\right) a_{3}\left(1-a_{4}\right) a_{2} . \\
& \left(a_{2}-a_{3}\right)\left(a_{3}-a_{4}\right)\left(a_{2}-a_{4}\right) \\
\leqslant & \left(\frac{1-a_{2}+a_{4}}{2}\right)^{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1-a_{4}+a_{2}}{2}\right)^{2} . \\
& \left(\frac{a_{2}-a_{4}}{2}\right)^{2}\left(a_{2}-a_{4}\right) \\
= & \left(\frac{1-x}{2}\right)^{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1+x}{2}\right)^{2}\left(\frac{x}{2}\right)^{2} x \\
= & \frac{1}{2^{8}}\left(x^{7}-2 x^{5}+x^{3}\right) \triangleq g(x),
\end{aligned}
\]

Equality holds when \( a_{2}+a_{4}=1 \) and \( a_{3}=\frac{1}{2} \).
\[
\begin{array}{l}
\text { Let } g^{\prime}(x)=\frac{1}{2^{8}}\left(7 x^{6}-10 x^{4}+3 x^{2}\right) \\
=\frac{1}{2^{8}} x^{2}\left(7 x^{2}-3\right)\left(x^{2}-1\right)=0 .
\end{array}
\]

Then in the interval \( (0,1) \), there is a unique solution \( x_{0}=\frac{\sqrt{21}}{7} \).
When \( 0 < x < x_{0} \), \( g^{\prime}(x) > 0 \) and \( g(x) \) is strictly increasing;

When \( x_{0} < x < 1 \), \( g^{\prime}(x) < 0 \) and \( g(x) \) is strictly decreasing.
Thus, the maximum value of \( g(x) \) is
\[
\begin{array}{l}
g\left(x_{0}\right)=\frac{3 \sqrt{21}}{2^{4} \times 7^{4}}=\frac{3 \sqrt{21}}{38416} . \\
\text { When } a_{2}-a_{4}=\frac{\sqrt{21}}{7}, \text { it follows that when } a_{1}=1, a_{2}= \\
\frac{7+\sqrt{21}}{14}, a_{3}=\frac{1}{2}, a_{4}=\frac{7-\sqrt{21}}{14}, a_{5}=0, \\
f\left(a_{1}, a_{2}, \cdots, a_{5}\right) \text { achieves its maximum value } \frac{3 \sqrt{21}}{38416} .
\end{array}
\]

From \( a_{2}-a_{4}=\frac{\sqrt{21}}{7} \), it follows that when \( a_{1}=1, a_{2}= \)
\( f\left(a_{1}, a_{2}, \cdots, a_{5}\right) \) achieves its maximum value \( \frac{3 \sqrt{21}}{38416} \).