The difference between the maximum and minimum values of the objective function $Z = 3x + 5y$, subject to the constraints $x + 3y \leq 60$, $x + y \geq 10$, $x - y \leq 0$, $x, y \geq 0$ is

The feasible region is bounded by the lines $x + 3y = 60$, $x + y = 10$, $x = y$, and the y-axis ($x = 0$).

The corner points of the feasible region are obtained by solving the equations of the lines pairwise:
1. Intersection of $x = y$ and $x + y = 10$ gives $(5, 5)$.
2. Intersection of $x = y$ and $x + 3y = 60$ gives $(15, 15)$.
3. Intersection of $x = 0$ and $x + y = 10$ gives $(0, 10)$.
4. Intersection of $x = 0$ and $x + 3y = 60$ gives $(0, 20)$.

Evaluating the objective function $Z = 3x + 5y$ at these corner points:
At $(5, 5)$, $Z = 3(5) + 5(5) = 40$
At $(15, 15)$, $Z = 3(15) + 5(15) = 120$
At $(0, 10)$, $Z = 3(0) + 5(10) = 50$
At $(0, 20)$, $Z = 3(0) + 5(20) = 100$

The maximum value of $Z$ is $120$ and the minimum value is $40$.

The difference between the maximum and minimum values is $120 - 40 = 80$.

Answer: $80$