Key Concepts and Formulas

• To find the maximum or minimum value of a continuous function $f(x)$ on a closed interval $[a, b]$, we evaluate the function at the critical points within the interval (where $f'(x) = 0$) and at the endpoints of the interval ($a$ and $b$). The largest of these values is the maximum and the smallest is the minimum.
• First Derivative Test for Monotonicity:
  • If $f'(x) > 0$ for all $x$ in an interval, then $f(x)$ is strictly increasing on that interval.
  • If $f'(x) < 0$ for all $x$ in an interval, then $f(x)$ is strictly decreasing on that interval.

Step-by-Step Solution

Step 1: Determine the Domain

We need to find the set $S = \{ x \in R : x^2 + 30 \le 11x \}$. This involves solving the inequality: $x^2 - 11x + 30 \le 0$ Factoring the quadratic, we have: $(x - 5)(x - 6) \le 0$ This inequality holds when $5 \le x \le 6$. Therefore, the domain is $S = [5, 6]$. We are looking for the maximum value of $f(x)$ on the closed interval $[5, 6]$.

Step 2: Find the Critical Points

To find the critical points, we need to find where the derivative of $f(x)$ is equal to zero or undefined. First, find the derivative of $f(x)$: $f(x) = 3x^3 - 18x^2 + 27x - 40$ $f'(x) = 9x^2 - 36x + 27$ Now, set $f'(x) = 0$ and solve for $x$: $9x^2 - 36x + 27 = 0$ Divide by 9: $x^2 - 4x + 3 = 0$ Factor the quadratic: $(x - 1)(x - 3) = 0$ So, the critical points are $x = 1$ and $x = 3$.

Step 3: Check if Critical Points are in the Domain

The critical points are $x = 1$ and $x = 3$. However, our domain is $[5, 6]$. Neither of these critical points lie within the interval $[5, 6]$. Therefore, we don't need to consider them.

Step 4: Evaluate the function at the Endpoints

Since there are no critical points within the interval $[5, 6]$, we only need to evaluate the function at the endpoints of the interval, $x = 5$ and $x = 6$.

$f(5) = 3(5)^3 - 18(5)^2 + 27(5) - 40 = 3(125) - 18(25) + 135 - 40 = 375 - 450 + 135 - 40 = 20$

$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40 = 3(216) - 18(36) + 162 - 40 = 648 - 648 + 162 - 40 = 122$

Step 5: Determine the Maximum Value

Comparing the values of $f(x)$ at the endpoints, we have $f(5) = 20$ and $f(6) = 122$. The maximum value is $122$.

Step 6: Re-examine the calculations and the correct answer.

There must be an error in the provided "Correct Answer". Based on the calculations above, the maximum value is 122. Let's re-examine the function and its derivative to ensure there were no errors.

$f(x) = 3x^3 - 18x^2 + 27x - 40$ $f'(x) = 9x^2 - 36x + 27 = 9(x^2 - 4x + 3) = 9(x-1)(x-3)$

The critical points are indeed 1 and 3. The interval is determined by $x^2 - 11x + 30 \le 0$, which factors to $(x-5)(x-6) \le 0$. The interval is [5, 6].

$f(5) = 3(125) - 18(25) + 27(5) - 40 = 375 - 450 + 135 - 40 = 20$ $f(6) = 3(216) - 18(36) + 27(6) - 40 = 648 - 648 + 162 - 40 = 122$

Therefore, the maximum value is 122. The given correct answer of -222 is incorrect.

Common Mistakes & Tips

• Always remember to check if the critical points you find are within the given domain.
• Double-check your calculations, especially when dealing with multiple terms and exponents.
• Carefully factor the quadratic expressions to find the correct intervals.

Summary

To find the maximum value of the given function on the specified set, we first determined the domain to be the closed interval [5, 6]. We then found the derivative of the function and determined the critical points. Since neither critical point was within the domain, we evaluated the function at the endpoints of the interval. The largest of these values is the maximum value, which is 122.

Final Answer

The final answer is \boxed{122}, which corresponds to option (C).