‌S={x∈R∕x2+30≤11x} ‌={x∈R∕x2−11x+30≤0} ‌={x∈R∕(x−5)(x−6)≤0} ‌={x∈R∕x∈[5,6]} ‌f(x)=3x3−18x2+27x−40 ‌∴‌‌f′(x)=9x2−36x+27 ‌=9(x−1)(x−3)>0‌‌∀x∈[5,6] ‌⇒f(x)‌ is increasing in ‌[5,6]‌. ‌ ‌∴‌‌‌ Maximum value ‌=f(6) ‌=3(6)3−18(6)2+27(6)−40 ‌=122 ‌