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First, we need to graph the constraints to find the feasible region. The constraints are: $x+2y\le120$ $x+y\ge60$ $x-2y\ge0$ $x, y\ge0$
Convert the inequalities to equations to find the boundary lines: $x+2y=120$ $x+y=60$ $x-2y=0$
Find the intersection points of these lines: Intersection of $x+2y=120$ and $x+y=60$: Subtract the second equation from the first: $y = 60$. Then $x = 60 - y = 60 - 60 = 0$. So the intersection point is $(0, 60)$. Intersection of $x+2y=120$ and $x-2y=0$: Add the two equations: $2x = 120$, so $x = 60$. Then $60 - 2y = 0$, so $2y = 60$ and $y = 30$. So the intersection point is $(60, 30)$. Intersection of $x+y=60$ and $x-2y=0$: From the second equation, $x = 2y$. Substitute into the first equation: $2y + y = 60$, so $3y = 60$ and $y = 20$. Then $x = 2y = 2(20) = 40$. So the intersection point is $(40, 20)$.
The feasible region is bounded by the points $(40, 20)$, $(60, 30)$, and $(0, 60)$.
Evaluate the objective function $Z = 5x + 10y$ at each corner point: At $(40, 20)$: $Z = 5(40) + 10(20) = 200 + 200 = 400$ At $(60, 30)$: $Z = 5(60) + 10(30) = 300 + 300 = 600$ At $(0, 60)$: $Z = 5(0) + 10(60) = 0 + 600 = 600$
The maximum value of $Z$ is 600, which occurs at the points $(60, 30)$ and $(0, 60)$. Since these are two points, all the points on the line segment joining $(60, 30)$ and $(0, 60)$ will give the maximum value.
Final Answer: The maximum value of Z is 600, which occurs at all points on the line segment joining (60, 30) and (0, 60).