55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

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: $$\frac{x}{4}+\frac{15-z-x}{5}+\frac{z}{6}=3\iff \frac{x}{4}-\frac{x}{5}=\frac{z}{5}-\frac{z}{6}\iff \frac{x}{20}=\frac{z}{30}\iff z=\frac{3}{2}x.$$ $$\text{Then }y=15-x-z=15-x-\frac{3}{2}x=15-\frac{5}{2}x.$$ 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: $$1\le x\le 13,1\le z=\frac{3}{2}x\le 13\iff \frac{2}{3}\le x\le \frac{26}{3},1\le y=15-\frac{5}{2}x\le 13\iff \frac{4}{5}\le x\le \frac{28}{5}.$$ 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:

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