6. Given that $a$ is a real number, and for any $k \in[-1,1]$, when $x \in(0,6]$, $6 \ln x+x^{2}-8 x+a \leqslant k x$ always holds, then the maximum value of $a$ is $\qquad$ .
Solve
The original system is equivalent to $\left\{\begin{array}{l}6 \ln x+x^{2}-7 x+a \leqslant 0, \\ 6 \ln x+x^{2}-9 x+a \leqslant 0\end{array} \Rightarrow 6 \ln x+x^{2}-7 x+a \leqslant 0\right.$. Let $f(x)=6 \ln x+x^{2}-7 x, f^{\prime}(x)=\frac{6}{x}+2 x-7=\frac{2 x^{2}-7 x+6}{x}=\frac{(2 x-3)(x-2)}{x}$,
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$x$ & $\left(0, \frac{3}{2}\right)$ & $\frac{3}{2}$ & $\left(\frac{3}{2}, 2\right)$ & 2 & $(2,6)$ & 6 \\
\hline$f^{\prime}(x)$ & + & 0 & - & 0 & + & \\
\hline$f(x)$ & $\nearrow$ & $6 \ln \frac{3}{2}-\frac{33}{4}$ & $\searrow$ & $6 \ln 2-10$ & $\nearrow$ & $6 \ln 6-6$ \\
\hline
\end{tabular}

From the table, we know that $-a \geqslant 6 \ln 6-6 \Rightarrow a \leqslant 6-6 \ln 6$. Therefore, the maximum value of $a$ is $6-6 \ln 6$.