Q.Solve the following linear programming problem graphically (Graph sheet is not required): Maximize Z = 4x + 3y subject to x + y ≤ 50, x + 2y ≤ 80, 2x + y ≥ 20 and x ≥ 0, y ≥ 0.

Find every corner point of the feasible region formed by the constraints, then evaluate $Z$ at each — the maximum occurs at a corner point (Corner Point Theorem).

Constraints: $x+y\le50$, $x+2y\le80$, $2x+y\ge20$, $x\ge0$, $y\ge0$.

Finding the corner points of the feasible region:

• Intersection of $x+y=50$ and $x+2y=80$: subtracting gives $y=30$, so $x=20$ — point $(20,30)$.
• On the x-axis ($y=0$): the boundary $2x+y=20$ gives $x=10$ (point $(10,0)$); $x+y=50$ gives $x=50$ (point $(50,0)$, and this also satisfies $x+2y=50\le80$ and $2x+y=100\ge20$, so it is feasible). Feasible segment on $y=0$ runs from $(10,0)$ to $(50,0)$.
• On the y-axis ($x=0$): $2x+y\ge20$ gives $y\ge20$ (point $(0,20)$); $x+2y\le80$ gives $y\le40$ (point $(0,40)$, and this also satisfies $x+y=40\le50$). Feasible segment on $x=0$ runs from $(0,20)$ to $(0,40)$. …