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

Finding the range of $Z = 3x + 5y$ subject to constraints

The feasible region is bounded by the constraints $x + 3y \leq 60$, $x + y \geq 10$, $y \geq x$, with $x \geq 0$, $y \geq 0$.

The corner points occur at intersections:
$(5,5)$ from $x+y=10$ and $y=x$
$(15,15)$ from $x+3y=60$ and $y=x$
$(0,10)$ from $x+y=10$ and $x=0$
$(0,20)$ from $x+3y=60$ and $x=0$

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

The minimum value of $Z$ is $40$ at $(5,5)$
The maximum value of $Z$ is $120$ at $(15,15)$

The difference is $120 - 40 = 80$