I hate conditions

Let us first check the turning points of $f(x) = 2x^3-15x^2+36x-48$ by finding $f'(x)=0$ and then $f''(x)$ .

$\begin{aligned} f(x) & = 2x^3-15x^2+36x-48 \\ f'(x) & = 6x^2 - 30x + 36 & \small \color{#3D99F6}{\text{Equating to }f(x) = 0} \\ 6x^2 - 30x + 36 & = 0 \\ x^2 - 5x + 6 & = 0 \\ (x-2)(x-3) & = 0 \\ f''(x) & = 12x - 30 \end{aligned}$

The turning points are $x=2$ and $x=3$ , since $f''(2) = 24-30 < 0$ and $f''(3) = 36-30 > 0$ , $f(2)$ and $f(3)$ are the local maximum and local minimum respectively. Therefore, $f(x)$ is increasing for $x \in (-\infty, 2)$ , then decreasing for $x \in (2, 3)$ and increasing again for $x \in (3, \infty)$ .

The condition is:

$\begin{aligned} x^2+20 & \le 9x \\ x^2 - 9x + 20 & \le 0 \\ (x-4)(x-5) & \le 0 \\ \implies A & = \{ x| x \in [4,5]\} \end{aligned}$

Since $x \in [4/5]$ is in the range of $x \in (3,\infty)$ , which is increasing, $f(x)$ is maximum when $x$ is maximum, therefore, $K = f(5) = 2(5^3) - 15(5^2) + 36(5) - 48 = \boxed{7}$

The convex constraint set is given by $A=[4,5]$ and it can be easily checked (by conputing its second derivative) that the function $f$ is convex in $A$ . Hence, the maximum value of $f$ in $K$ occurs at its boundary. We note $f(4)=-16$ and $f(5)=7$ . Hence, $K=7$ .