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

Options

A. `-122`
B. `-222`
C. `222`
D. `122`

Answer

To find the maximum value of the function \( f(x) = 3x^3 - 18x^2 + 27x - 40 \) on the set \( S = \{ x \in \mathbb{R} : x^2 + 30 \leq 11x \} \), we will follow these steps:

### Step 1: Determine the set \( S \)

We start with the inequality:
\[ x^2 + 30 \leq 11x \]
Rearranging gives:
\[ x^2 - 11x + 30 \leq 0 \]
Next, we factor the quadratic:
\[ (x - 5)(x - 6) \leq 0 \]
To find the intervals where this inequality holds, we identify the roots \( x = 5 \) and \( x = 6 \).

### Step 2: Analyze the sign of the quadratic

We can test intervals around the roots:
- For \( x < 5 \): Choose \( x = 4 \) → \( (4 - 5)(4 - 6) = (-1)(-2) = 2 \) (positive)
- For \( 5 < x < 6 \): Choose \( x = 5.5 \) → \( (5.5 - 5)(5.5 - 6) = (0.5)(-0.5) = -0.25 \) (negative)
- For \( x > 6 \): Choose \( x = 7 \) → \( (7 - 5)(7 - 6) = (2)(1) = 2 \) (positive)

Thus, the quadratic is negative or zero in the interval:
\[ S = [5, 6] \]

### Step 3: Find the critical points of \( f(x) \)

Next, we differentiate \( f(x) \):
\[ f'(x) = 9x^2 - 36x + 27 \]
Setting the derivative to zero to find critical points:
\[ 9x^2 - 36x + 27 = 0 \]
Dividing through by 9:
\[ x^2 - 4x + 3 = 0 \]
Factoring gives:
\[ (x - 1)(x - 3) = 0 \]
Thus, the critical points are \( x = 1 \) and \( x = 3 \).

### Step 4: Evaluate the function at the endpoints of \( S \)

Since \( S = [5, 6] \), we evaluate \( f(x) \) at the endpoints:
1. At \( x = 5 \):
\[ f(5) = 3(5^3) - 18(5^2) + 27(5) - 40 \]
\[ = 3(125) - 18(25) + 135 - 40 \]
\[ = 375 - 450 + 135 - 40 = 20 \]

2. At \( x = 6 \):
\[ f(6) = 3(6^3) - 18(6^2) + 27(6) - 40 \]
\[ = 3(216) - 18(36) + 162 - 40 \]
\[ = 648 - 648 + 162 - 40 = 122 \]

### Step 5: Determine the maximum value

Comparing the values:
- \( f(5) = 20 \)
- \( f(6) = 122 \)

The maximum value of \( f(x) \) on the set \( S \) is:
\[ \boxed{122} \]