Problem Statement
Solve the following Linear Programming Problem graphically:
$$\text{Minimize and Maximize } Z = 5x + 10y$$
subject to the constraints:
$$\begin{aligned} x + 2y &\le 120 \\ x + y &\ge 60 \\ x - 2y &\ge 0 \\ x, y &\ge 0 \end{aligned}$$
Verified Solution & Marking Scheme
Convert inequalities to equations and find intercepts
1. $x + 2y = 120$: Passes through $(120, 0)$ and $(0, 60)$.
2. $x + y = 60$: Passes through $(60, 0)$ and $(0, 60)$.
3. $x - 2y = 0 \implies x = 2y$: Passes through $(0, 0)$ and $(60, 30)$.
4. $x \ge 0, y \ge 0$: First quadrant.
Determine corner points of feasible region
Plotting and shading according to inequalities gives a bounded polygonal region with corner points:
- Point $A$: Intersection of $x + y = 60$ and $x - 2y = 0 \implies 2y + y = 60 \implies y = 20, x = 40 \implies A(40, 20)$.
- Point $B$: Intersection of $x + 2y = 120$ and $x - 2y = 0 \implies 2y + 2y = 120 \implies y = 30, x = 60 \implies B(60, 30)$.
- Point $C$: Intersection of $x + 2y = 120$ with $x$-axis ($y = 0$) $\implies C(120, 0)$.
- Point $D$: Intersection of $x + y = 60$ with $x$-axis ($y = 0$) $\implies D(60, 0)$.
Evaluate objective function Z = 5x + 10y at all corner points
- At $A(40, 20)$: $Z = 5(40) + 10(20) = 200 + 200 = 400$.
- At $B(60, 30)$: $Z = 5(60) + 10(30) = 300 + 300 = 600$.
- At $C(120, 0)$: $Z = 5(120) + 10(0) = 600$.
- At $D(60, 0)$: $Z = 5(60) + 10(0) = 300$.
Notice that the maximum value $600$ occurs at two corner points: $B(60, 30)$ and $C(120, 0)$, and hence at all points on the line segment joining $B$ and $C$.