Maximum Value of Cubic Polynomial on Closed Interval

To find the maximum value of the function $f(x) = 2x^3 - 15x^2 + 36x - 48$ on the set $A = \{x \mid x^2 + 20 \le 9x\}$, we proceed through the following steps:

1. Determine the set $A$:
   The inequality is given by:
   $x^2 - 9x + 20 \le 0$
   Factoring the quadratic expression:
   $(x - 4)(x - 5) \le 0$
   Thus, the set $A$ is the closed interval:
   $A = [4, 5]$

2. Analyze the derivative of $f(x)$:
   Compute the first derivative of $f(x)$:
   $f'(x) = \frac{d}{dx}\left(2x^3 - 15x^2 + 36x - 48\right) = 6x^2 - 30x + 36$
   Factoring the derivative:
   $f'(x) = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$

   The critical points of $f(x)$ are at $x = 2$ and $x = 3$, neither of which lies in the interval $[4, 5]$.

3. Check the monotonicity of $f(x)$ on $[4, 5]$:
   For any $x \in [4, 5]$:
   $x - 2 > 0 \quad \text{and} \quad x - 3 > 0$
   Therefore:
   $f'(x) = 6(x - 2)(x - 3) > 0 \quad \text{for all } x \in [4, 5]$
   This implies that $f(x)$ is strictly increasing on the interval $[4, 5]$.

4. Calculate the maximum value:
   Since $f(x)$ is strictly increasing on $[4, 5]$, the maximum value occurs at the right endpoint, $x = 5$:
   $f(5) = 2(5)^3 - 15(5)^2 + 36(5) - 48$
   $f(5) = 2(125) - 15(25) + 180 - 48$
   $f(5) = 250 - 375 + 180 - 48 = 7$

Thus, the maximum value of the function on the given set is $7$.