Question

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

• A. 122
• B. -122
• C. -222
• D. 222

Hint:

Calculus Tip: When finding absolute extrema on a closed interval $[a,b]$, if the function lacks critical points inside the interval, the maximum and minimum will strictly lie on the boundary points $x=a$ and $x=b$.

Answer 1

The Correct Option is A

Concept: Calculus - Application of Derivatives (Maxima and Minima on a Closed Interval).

Step 1: Solve the inequality to find the domain set $S$. The given set condition is $x^{2}+30 \le 11x$. Rearrange into a standard quadratic inequality: $x^{2} - 11x + 30 \le 0$. Factor the quadratic: $(x-5)(x-6) \le 0$. Using the wavy curve method, the solution is the closed interval $[5, 6]$. Thus, we only evaluate the function for $x \in [5, 6]$.

Step 2: Find the first derivative of the function. Differentiate $f(x)=3x^{3}-18x^{2}+27x-40$ with respect to $x$. $f^{\prime}(x) = 9x^{2} - 36x + 27$. Factor out the common term 9: $f^{\prime}(x) = 9(x^{2} - 4x + 3)$. Further factor the quadratic expression: $f^{\prime}(x) = 9(x-1)(x-3)$.

Step 3: Analyze the sign of the derivative on the specified interval. We need to check the behavior of $f(x)$ strictly within our domain $[5, 6]$. For any value of $x$ between 5 and 6, $(x-1)$ is positive and $(x-3)$ is positive. Therefore, $f^{\prime}(x) = 9(\text{positive})(\text{positive})>0$ for all $x \in [5, 6]$.

Step 4: Determine the monotonicity to find the maximum point. Because the first derivative $f^{\prime}(x)$ is strictly greater than 0 on the entire interval $[5, 6]$, the function $f(x)$ is strictly increasing over this domain. For a strictly increasing function, the maximum value will always occur at the rightmost endpoint of the interval.

Step 5: Calculate the maximum value at the endpoint. Substitute the upper bound $x = 6$ into the original function: $f(6) = 3(6)^{3} - 18(6)^{2} + 27(6) - 40$. $f(6) = 3(216) - 18(36) + 162 - 40$. $f(6) = 648 - 648 + 162 - 40 = 122$. $$ \therefore \text{The maximum value of the function on the given set is } 122. $$