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

Here, $x^2-11x+30\le 0$

$\begin{array}{r}\Rightarrow x^2-5x-6x+30\le 0 \Rightarrow (x-5)(x-6)\le 0\\ \Rightarrow 5\le x\le 6\therefore S=\{x\in R,5\le x\le 6\}\end{array}$

Now, $f\left(x\right)=3x^3-18x^2+27x-40$

$\therefore f^{\mathrm{\prime }}\left(x\right)=9x^2-36x+27=9(x-1)(x-3)$,

which is positive in [5, 6]. So $So,f\left(x\right) is increa\sin g in [5,6]$

Hence, maximum value of $f\left(x\right)=f\left(6\right)=122$