Find the maximum value of the function \[f(x)=3{{x}^{3}}-18{{x}^{2}}+27x-40\] on the set \[S=\left\{ x\text{ }\in \text{ }R:\text{ }{{x}^{2}}+30\text{ }\le \text{ }11x \right\}\] is:
A) 122
B) -222
C) -122
D) 222

Hint: Perform factorization of both the quadratic and cubic equation and then use the concept of increasing function after finding the first derivative of the cubic equation.

Complete step-by-step answer:
Given: \[S=\left\{ x\text{ }\in \text{ }R:\text{ }{{x}^{2}}+30\text{ }\le \text{ }11x \right\}\]
Transpose 11x to left hand side, we get
\[S=\left\{ x\text{ }\in \text{ }R:\text{ }{{x}^{2}}+30-11x\text{ }\le \text{ 0} \right\}\]
Now factorise the quadratic equation using Splitting of middle term method or Determinant method.
Middle term can be splitted as -6x and -5x since the sum of -6x and -5x is -11x and the product is 11x2.
\[\Rightarrow {{x}^{2}}-6x-5x+30\]
\[\Rightarrow \left( x-5 \right)\left( x-6 \right)\text{ }\le \text{ }0\]
On equating, we get
\[S=\left\{ x\text{ }\in \text{ }R:\text{ 5 }\le \text{ }x\text{ }\le \text{ 6} \right\}\]
Now, \[f(x)=3{{x}^{3}}-18{{x}^{2}}+27x-40\]
\[\Rightarrow f'(x)=9{{x}^{2}}-36x+27\]
On taking 9 common, we get
\[\Rightarrow f'(x)=9({{x}^{2}}-4x+3)\]
Now, split the middle term -4x
\[\Rightarrow f'(x)=9(x-1)(x-3)\]
We get to know that f(x) is increasing in [5,6] because f’(x) is positive in [5,6] because we get f’(x) > 0 on putting both 5 and 6 as x and by first derivative test, f’(x) > 0 implies f(x) is increasing within the interval in which f’(x) > 0.
Now, find the value of f(x) at 6 to find the maximum value because the value of f(x) is maximum at the peak point of the interval, and the function is increasing at.
\[f(6)=3{{(6)}^{3}}-18{{(6)}^{2}}+27(6)-40\]
\[=648-648+162-40\]
\[=122\]

Hence, the correct option is A (122).

Note: First derivative test states that f(x) is increasing when f’(x)>0 and f(x) is decreasing when f’(x)<0.