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First, we graph the constraints on the $x$-$y$ plane. The constraints are: $x - y \ge 0$ $-x + 2y \ge 2$ $x \ge 3$ $y \le 4$ $y \ge 0$
The feasible region is the intersection of all the constraint regions. We need to find the vertices of this feasible region.
We find the intersection points of the lines that form the boundaries of the feasible region. \r\n1. Intersection of $x = y$ and $-x + 2y = 2$: \r\nSubstituting $x = y$ into the second equation, we get $-y + 2y = 2$, so $y = 2$. Thus, $x = 2$. However, $x \ge 3$, so this intersection is not in the feasible region. \r\n2. Intersection of $x = 3$ and $x = y$: \r\nSince $x = 3$, $y = 3$. So the point is $(3, 3)$. \r\n3. Intersection of $x = 3$ and $y = 4$: \r\nThis point is $(3, 4)$. \r\n4. Intersection of $y = 4$ and $-x + 2y = 2$: \r\nSubstituting $y = 4$ into the second equation, we get $-x + 2(4) = 2$, so $-x + 8 = 2$, which means $x = 6$. So the point is $(6, 4)$. \r\n5. Intersection of $y = 0$ and $x=y$: \r\nThis point is $(0,0)$. However, $x \ge 3$ and $-x+2y \ge 2$, so this point is not in the feasible region. \r\n6. Intersection of $y = 0$ and $-x+2y=2$: \r\nSubstituting $y=0$ into the second equation, we get $-x+2(0)=2$, so $x=-2$. However, $x \ge 3$, so this point is not in the feasible region. \r\n7. Intersection of $x=3$ and $-x+2y=2$: \r\nSubstituting $x=3$ into the second equation, we get $-3+2y=2$, so $2y=5$, which means $y=2.5$. So the point is $(3, 2.5)$.
The corner points of the feasible region are $(3, 3)$, $(3, 4)$, and $(6, 4)$, and $(3, 2.5)$.
We evaluate the objective function $Z = x - 5y$ at each corner point: \r\nAt $(3, 3)$: $Z = 3 - 5(3) = 3 - 15 = -12$ \r\nAt $(3, 4)$: $Z = 3 - 5(4) = 3 - 20 = -17$ \r\nAt $(6, 4)$: $Z = 6 - 5(4) = 6 - 20 = -14$ \r\nAt $(3, 2.5)$: $Z = 3 - 5(2.5) = 3 - 12.5 = -9.5$
The minimum value of $Z$ is $-17$, which occurs at the point $(3, 4)$.
Final Answer: -17 at (3, 4)