The maximum value of the function f(x)=3x^3-1 | vedclass

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(A) First,we determine the set $S$ by solving the inequality $x^2+30 \leq 11x$. $x^2-11x+30 \leq 0$ $(x-5)(x-6) \leq 0$ Thus,$S = [5, 6]$. Next,we fin

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$122$

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(A) First,we determine the set $S$ by solving the inequality $x^2+30 \leq 11x$. $x^2-11x+30 \leq 0$ $(x-5)(x-6) \leq 0$ Thus,$S = [5, 6]$. Next,we find the derivative of $f(x) = 3x^3-18x^2+27x-40$: $f'(x) = 9x^2-36x+27 = 9(x^2-4x+3) = 9(x-1)(x-3)$. For $x \in [5, 6]$,both $(x-1)$ and $(x-3)$ are positive,so $f'(x) > 0$. Since $f'(x) > 0$ on the interval $[5, 6]$,the function $f(x)$ is strictly increasing on this interval. Therefore,the maximum value occurs at the right endpoint $x=6$. $f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40$ $f(6) = 3(216) - 18(36) + 162 - 40$ $f(6) = 648 - 648 + 162 - 40 = 122$.

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

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