The minimum value of $Z=3 x+5 y$, given subject to the constraints $x+y \geq 2, x+3 y \geq 3, x, y \geq 0$ is

The objective function is $Z = 3x + 5y$.

The constraints are $x + y \geq 2$, $x + 3y \geq 3$, $x \geq 0$, and $y \geq 0$.

To find the feasible region, determine the intersection points of the boundary lines:
1. $x + y = 2$ passes through $(2, 0)$ and $(0, 2)$.
2. $x + 3y = 3$ passes through $(3, 0)$ and $(0, 1)$.

The intersection of $x + y = 2$ and $x + 3y = 3$ is found by subtracting the equations:
$(x + 3y) - (x + y) = 3 - 2 \Rightarrow 2y = 1 \Rightarrow y = 0.5$.
Substituting $y = 0.5$ into $x + y = 2$, we get $x = 1.5$.
The intersection point is $(1.5, 0.5)$.

The vertices of the unbounded feasible region are $(3, 0)$, $(1.5, 0.5)$, and $(0, 2)$.

Evaluate $Z = 3x + 5y$ at these vertices:
At $(3, 0)$: $Z = 3(3) + 5(0) = 9$.
At $(1.5, 0.5)$: $Z = 3(1.5) + 5(0.5) = 4.5 + 2.5 = 7$.
At $(0, 2)$: $Z = 3(0) + 5(2) = 10$.

Comparing the values, the minimum value of $Z$ is 7.

Answer: 7