Mark the up, flat and down parts on the way from $A$ to $B$ as $x$, $y$, $z$ respectively. Then:
$$x + y + z = 15, \frac{x}{4} + \frac{y}{5} + \frac{z}{6} = 3,\ 1 \le x, y, z \le 13.$$
From the first equation: $y = 15 - x - z$, substitute it into the second equation:
$$\frac{x}{4} + \frac{15-z-x}{5} + \frac{z}{6} = 3 \Leftrightarrow \frac{x}{4} - \frac{x}{5} = \frac{z}{5} - \frac{z}{6} \Leftrightarrow \frac{x}{20} = \frac{z}{30} \Leftrightarrow z = \frac{3}{2}x.$$
$$\text{Then } y = 15 - x - z = 15 - x - \frac{3}{2}x = 15 - \frac{5}{2}x.$$
So the required time is:
$$t = \frac{x}{6} + \frac{y}{5} + \frac{z}{4} = \frac{x}{6} + 3 - \frac{x}{2} + \frac{3}{8}x = 3 + \frac{x}{24}.$$

The maximum (the minimum) $t$ can be in case of $x$ is maximum (minimum).
Put down the limitation for $x$, which follow from the condition of the problem:
$$1 \le x \le 13,\ 1 \le z = \frac{3}{2}x \le 13 \Leftrightarrow \frac{2}{3} \le x \le \frac{26}{3},\ 1 \le y = 15 - \frac{5}{2}x \le 13 \Leftrightarrow \frac{4}{5} \le x \le \frac{28}{5}.$$
Since all conditions have to be fulfilled simultaneously, we have such limitation for $x$:
$$1 \le x \le \frac{28}{5}.$$
If $x=1$, then $z = \frac{3}{2}$ and $y = \frac{25}{2}$. If $x = \frac{28}{5}$, then $z = \frac{42}{5}$ and $y = 1$.

$$t = \frac{x}{6} + \frac{y}{5} + \frac{z}{4} = \frac{x}{6} + 3 - \frac{x}{2} + \frac{3}{8}x = 3 + \frac{x}{24}.$$

The maximum (the minimum) $t$ can be in case of $x$ is maximum (minimum).
Put down the limitation for $x$, which follow from the condition of the problem:

$$t_{\max} = 3 + \frac{1}{24} \cdot \frac{28}{5} = 3 + \frac{7}{30} = \frac{97}{30},\ t_{\min} = 3 + \frac{1}{24} \cdot 1 = \frac{73}{24}.$$