The maximum value of the function $f(x) = 2x^{3} - 15x^{2} + 36x - 48$ on the set $A = \{x \mid x^{2} + 20 \leq 9x\}$ is

The maximum value of the function f(x) = 2x³ - 15x² + 36x - 48 on the set A = {x mid x² + 20 ≤ 9x} is

$f(x) = 2x^3 - 15x^2 + 36x - 48$

$A = \{x : x^2 + 20 \le 9x\}$

$\text{To find: } \max_{x\in A} f(x)$

$x^2 - 9x + 20 \le 0 \;\Rightarrow\; (x-4)(x-5)\le 0$

$\Rightarrow x\in[4,5]$

$A = [4,5] \quad (\text{closed interval})$

$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$

$\text{critical points: } x=2,\ x=3$

$x=2,3 \notin [4,5] \;\Rightarrow\; \text{no critical point inside } [4,5]$

$x\in[4,5]:\ x>3 \;\Rightarrow\; f'(x)=6(x-2)(x-3) > 0$

$\Rightarrow f \text{ strictly increasing on } [4,5]$

$f(4) = 128 - 240 + 144 - 48 = -16$

$f(5) = 250 - 375 + 180 - 48 = 7$

$\boxed{\max_{x\in A} f(x) = f(5) = 7}$