The maximum value of the function f(x)=3x³−18x²+27x−40 on the set  S = {xx²+30 is   

S={xx²+30−11x

={x

Now f(x)=3x³−18x²+27x−40

⇒f ’(x)=9(x−1)(x−3),

Which is positive in[5, 6]

⇒f(x) increasing in[5, 6]

Hence maximum value = f(6) = 122