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

Feasible region determination
The objective function $z = 3x + 5y$ is subject to constraints: $x + 3y \leq 60$, $x + y \geq 10$, $x = y$, and $x \geq 0$, $y \geq 0$.

Substituting $x = y$ into the inequality constraints yields $4y \leq 60$ and $2y \geq 10$, simplifying to $5 \leq y \leq 15$. With $x = y$, the non-negativity conditions are satisfied. The feasible region is the line segment from $(5,5)$ to $(15,15)$.

Evaluation of objective function
The function values at the endpoints are:
$z(5,5) = 3(5) + 5(5) = 40$
$z(15,15) = 3(15) + 5(15) = 120$

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