Let the first part = y, then other part = $(20-y)$.
$f(y) = (20 - y) y^3$ or $f(y) = 20y^3 - y^4$
For maximum or minimum, put $f'(y) = 0$
$\Rightarrow f'\left(y\right)= 60y^{2} - 4y^{3} = 0 $
$\Rightarrow 4y^{2} \left(15 - y\right)= 0$
$ \Rightarrow y=0 or y=15$
Now, $ f''\left(y\right) =120 y -12y^{2}$
$ \Rightarrow f''\left(15\right)= 120\left(15\right)- 12\left(15\right)^{2} <0 $
$\therefore \:\: f\left( y \right) $ is maximum at y=15
Hence, 20 can be divided in 15 and 5