1. Find the set $A$

We need to solve $x^2 + 20 \le 9x$ which gives $x^2 - 9x + 20 \le 0$ Factorizing, $(x-4)(x-5) \le 0$ So, $A = [4,5]$

2. Maximize $f(x)$ on $[4,5]$

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

Differentiate: $f'(x)=6x^2-30x+36$ $f'(x)=6(x^2-5x+6)=6(x-2)(x-3)$

For $x \in [4,5]$, both $(x-2)$ and $(x-3)$ are positive, so $f'(x)>0 \quad \text{for all } x\in[4,5]$ Hence $f(x)$ is increasing on $[4,5]$.

Therefore, the maximum occurs at the right endpoint $x=5$.

3. Compute the maximum value

$f(5)=2(5)^3-15(5)^2+36(5)-48$ $=2\cdot 125-15\cdot 25+180-48$ $=250-375+180-48$ $=-125+180-48$ $=55-48=7$

Also, for confirmation, $f(4)=2(64)-15(16)+144-48=128-240+144-48=-16$ So indeed the maximum on $[4,5]$ is $7$.

4. Final answer

The maximum value is $\boxed{7}$

5. Comparison with stored answer

Stored correct answer: $7$

Our derived answer matches the stored answer.