The maximum value of the function \(f(x)=3 x^3-18 x^2+27 x-40\) on the set \(\mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\}\) is

\(\begin{aligned}
& \mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\} \\
& x^2+30 \leq 11 x \\
& x^2-11 x+30 \leq 0 \\
& (x-5)(x-6) \leq 0 \\
& x \in[5,6]
\end{aligned}\)
Now, \(\mathrm{f}(x)=3 x^3-18 x^2+27 x-40\)
\(\begin{aligned}
\mathrm{f}^{\prime}(x) & =9 x^2-36 x+27 \\
\mathrm{f}^{\prime}(x) & =9\left(x^2-4 x+3\right) \\
& =9\left[\left(x^2-4 x+4\right)-1\right] \\
& =9(x-2)^2-9
\end{aligned}\)
\(\therefore \quad \mathrm{f}^{\prime}(x)>0 \forall x \in[5,6]\)
\(\therefore \mathrm{f}(x)\) is strictly increasing in the interval \([5,6]\)
\(\therefore \quad\) Maximum value of \(\mathrm{f}(x)\) when \(x \in[5,6]\) is
\(f(6)=122\)