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First, we graph the inequalities to find the feasible region. The constraints are: $x+y\le800$ $2x+y\le1000$ $x\le400$ $x\ge0$ $y\ge0$
The corner points of the feasible region are the intersections of the constraint lines. \r\nIntersection of $x+y=800$ and $2x+y=1000$: Subtracting the first equation from the second, we get $x=200$. Substituting $x=200$ into the first equation, we get $200+y=800$, so $y=600$. Thus, the intersection point is $(200, 600)$. \r\nIntersection of $x+y=800$ and $x=400$: Substituting $x=400$ into the first equation, we get $400+y=800$, so $y=400$. Thus, the intersection point is $(400, 400)$. \r\nIntersection of $2x+y=1000$ and $x=400$: Substituting $x=400$ into the second equation, we get $2(400)+y=1000$, so $800+y=1000$, which gives $y=200$. Thus, the intersection point is $(400, 200)$. \r\nOther corner points are $(0,0)$, $(0,800)$, and $(400,0)$. \r\nSo, the corner points are $(0,0)$, $(0,800)$, $(400,0)$, $(400, 200)$, $(200, 600)$, and $(400,400)$.
Now, we evaluate the objective function $z=4x+3y$ at each corner point: \r\nAt $(0,0)$: $z=4(0)+3(0)=0$ \r\nAt $(0,800)$: $z=4(0)+3(800)=2400$ \r\nAt $(400,0)$: $z=4(400)+3(0)=1600$ \r\nAt $(400,200)$: $z=4(400)+3(200)=1600+600=2200$ \r\nAt $(200,600)$: $z=4(200)+3(600)=800+1800=2600$ \r\nAt $(400,400)$: The point (400,400) does not satisfy the constraint $2x+y \le 1000$ since $2(400)+400 = 1200 > 1000$. So, (400,400) is not a feasible point. \r\nHowever, the intersection of $x+y=800$ and $x=400$ is $(400,400)$, which we already considered and found to be infeasible. \r\nLet's re-examine the feasible region. The corner points are (0,0), (0,800), (400,0), (400,200), and (200,600). \r\n
Comparing the values of $z$ at the feasible corner points, we find that the maximum value is $z=2600$ at the point $(200, 600)$.
Final Answer: The maximum value of $z$ is 2600 at $(x, y) = (200, 600)$.
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