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First, we graph the inequalities $x-y\ge0$ and $x-2y\ge-2$, along with $x\ge0$ and $y\ge0$.
Convert the inequalities to equations to find the boundary lines: $x - y = 0 \implies y = x$ $x - 2y = -2 \implies x = 2y - 2$
Find the intersection points of the lines: Intersection of $y = x$ and $x = 2y - 2$: $x = 2x - 2 \implies x = 2$. Thus, $y = 2$. The intersection point is $(2, 2)$. Intersection of $x = 0$ and $y = 0$: $(0, 0)$ Intersection of $x = 0$ and $x - 2y = -2$: $0 - 2y = -2 \implies y = 1$. The intersection point is $(0, 1)$. Intersection of $y = 0$ and $x - y = 0$: $x - 0 = 0 \implies x = 0$. The intersection point is $(0, 0)$. Intersection of $y = 0$ and $x - 2y = -2$: $x - 2(0) = -2 \implies x = -2$. However, $x \ge 0$, so this intersection is not feasible.
The feasible region is determined by the inequalities $x-y\ge0$, $x-2y\ge-2$, $x\ge0$, and $y\ge0$. The vertices of the feasible region are $(0,0)$, $(2,2)$ and $(0,1)$. However, the region is unbounded.
Evaluate $Z = x + 2y$ at the vertices: At $(0, 0)$: $Z = 0 + 2(0) = 0$ At $(2, 2)$: $Z = 2 + 2(2) = 6$ At $(0, 1)$: $Z = 0 + 2(1) = 2$
Since the region is unbounded, we need to check if the value of $Z$ can increase indefinitely. Consider a point $(x, y)$ in the feasible region such that $x = 2y - 2$ and $x > 2$. Then $y > 2$. Let $y = k$, where $k > 2$. Then $x = 2k - 2$. $Z = x + 2y = (2k - 2) + 2k = 4k - 2$. As $k$ increases, $Z$ also increases. Therefore, the maximum value of $Z$ is unbounded.
Final Answer: The maximum value of Z is unbounded.
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