$$
\begin{aligned}
\mathrm{S} & =\left\{x \in \mathrm{R} / x^2+30 \leq 11 x\right\} \\
& =\left\{x \in \mathrm{R} / x^2-11 x+30 \leq 0\right\} \\
& =\{x \in \mathrm{R} /(x-5)(x-6) \leq 0\} \\
& =\{x \in \mathrm{R} / x \in[5,6]\} \\
\mathrm{f}(x) & =3 x^3-18 x^2+27 x-40 \\
\therefore \quad \mathrm{f}^{\prime}(x) & =9 x^2-36 x+27 \\
& =9(x-1)(x-3)>0 \quad \forall x \in[5,6]
\end{aligned}
$$
$\Rightarrow \mathrm{f}(x)$ is increasing in $[5,6]$.
$$
\begin{aligned}
\therefore \quad \text { Maximum value } & =\mathrm{f}(6) \\
& =3(6)^3-18(6)^2+27(6)-40 \\
& =122
\end{aligned}
$$