Solution. From $y \leq x$ it follows that

$$
x+3 y+2 x \geq x+3 y+2 y, \text { i.e. } 3(x+y) \geq x+5 y \geq 105
$$

Therefore, $a \leq 35$. Since for $x=y=17.5$ we have $x+5 y=105$, it follows that $a=35$. From $y \geq 0$ it follows that $4 y \geq 0$, so $x+y+4 y \geq x+y$, i.e., $105=x+5 y \geq x+y$. Therefore, $b \geq 105$. Since for $x=105$, $y=0$ we have $x+5 y=105$, it follows that $b=105$.

Finally, the required numbers are $a=35$ and $b=105$, and thus $b-a=70$.
5. Determine the real numbers $a$ and $b, a \leq b$ for which the following conditions are satisfied:

a) for any real numbers $x$ and $y$ such that $0 \leq y \leq x$ and $x+5 y=105$ it holds that $a \leq x+y \leq b$,

b) the difference $b-a$ is minimal.