2023 ONE(e)

The graph of $y=x(x-2m{)}^2$, where $m>0$, is shown.

The total shaded area between the curve and the $x$-axis
from $x=0$ to $x=2m$ is given by $A=\frac{4m^4}{3}$.

A right-angled triangle is now constructed with one
vertex at $(0,0)$ and another on the curve $y=x(x-2m{)}^2$,
as shown below.

Show that the maximum area of such a triangle is $\frac{3}{8}$ of the total shaded area.

*You must use calculus and show any derivatives that you need to find when solving this problem.*
*You do not have to prove that the area you have found is a maximum.*