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Convert the inequalities into equations to find the boundary lines: $$L_1: x + 2y = 12$$ $$L_2: 2x + y = 12$$ $$L_3: 4x + 5y = 20$$ The non-negativity constraints $x \ge 0$ and $y \ge 0$ restrict the solution to the first quadrant.
For $L_1$, intercepts are $(12, 0)$ and $(0, 6)$. For $L_2$, intercepts are $(6, 0)$ and $(0, 12)$. For $L_3$, intercepts are $(5, 0)$ and $(0, 4)$. Plot these lines on a Cartesian plane.
The feasible region is the polygon bounded by the intersection of the half-planes defined by the constraints. The vertices of the feasible region are found by solving the intersection points of the lines: Intersection of $L_1$ and $L_2$: $x+2y=12$ and $2x+y=12$ gives $(4, 4)$. Intersection of $L_1$ and $L_3$: $x+2y=12$ and $4x+5y=20$ gives $(-20/3, 28/3)$ (outside first quadrant). The vertices are $(5, 0), (6, 0), (4, 4), (0, 6), (0, 4)$.
Calculate $Z = 600x + 400y$ at each vertex: At $(5, 0): Z = 600(5) + 400(0) = 3000$ At $(6, 0): Z = 600(6) + 400(0) = 3600$ At $(4, 4): Z = 600(4) + 400(4) = 2400 + 1600 = 4000$ At $(0, 6): Z = 600(0) + 400(6) = 2400$ At $(0, 4): Z = 600(0) + 400(4) = 1600$
Final Answer: Maximum Z = 4000 at point (4, 4)
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