The maximum value of $Z=12 x+13 y$, subject to constraints $x \geq 0, y \geq 0, x+y \leq 5$ and $3 x+y \leq 9$ is

The feasible region is defined by the constraints $x \geq 0$, $y \geq 0$, $x + y \leq 5$, and $3x + y \leq 9$.

The corner points of the feasible region are found by solving the intersection of the boundary lines:
1. The intersection of $x = 0$ and $y = 0$ is $(0, 0)$.
2. The intersection of $x = 0$ and $x + y = 5$ is $(0, 5)$.
3. The intersection of $y = 0$ and $3x + y = 9$ is $(3, 0)$.
4. The intersection of $x + y = 5$ and $3x + y = 9$ is found by subtracting the equations: $(3x + y) - (x + y) = 9 - 5$, which gives $2x = 4$, so $x = 2$. Substituting $x = 2$ into $x + y = 5$ gives $y = 3$. The point is $(2, 3)$.

We evaluate $Z = 12x + 13y$ at each corner point:
At $(0, 0)$: $Z = 12(0) + 13(0) = 0$.
At $(0, 5)$: $Z = 12(0) + 13(5) = 65$.
At $(3, 0)$: $Z = 12(3) + 13(0) = 36$.
At $(2, 3)$: $Z = 12(2) + 13(3) = 24 + 39 = 63$.

Comparing the values $0, 65, 36, 63$, the maximum value is $65$.

Answer: 65