Problem Statement
Solve the following Linear Programming Problem graphically:
$\text{Minimize and Maximize } Z = 5x + 10y$
subject to constraints:
$x + 2y \le 120, \quad x + y \ge 60, \quad x - 2y \ge 0, \quad x \ge 0, \, y \ge 0$
Verified Solution & Marking Scheme
Determine boundary lines and test points
1. $x + 2y = 120$: Passes through $(120, 0)$ and $(0, 60)$. Test $(0,0): 0 \le 120$ (True, towards origin).
2. $x + y = 60$: Passes through $(60, 0)$ and $(0, 60)$. Test $(0,0): 0 \ge 60$ (False, away from origin).
3. $x - 2y = 0$: Passes through $(0, 0)$ and $(40, 20)$. Test $(10, 0): 10 - 0 \ge 0$ (True, towards positive $x$-axis).
4. $x \ge 0, y \ge 0$ restricts to first quadrant.
Find corner points of bounded feasible region
The feasible region is bounded with vertices:
- $A(60, 0)$: Intersection of $x + y = 60$ and $y = 0$.
- $B(120, 0)$: Intersection of $x + 2y = 120$ and $y = 0$.
- $C(60, 30)$: Intersection of $x + 2y = 120$ and $x - 2y = 0$ (since $x = 2y \implies 4y = 120 \implies y = 30, x = 60$).
- $D(40, 20)$: Intersection of $x + y = 60$ and $x - 2y = 0$ (since $x = 2y \implies 3y = 60 \implies y = 20, x = 40$).
Evaluate objective function Z = 5x + 10y at all vertices
- At $A(60, 0)$: $Z = 5(60) + 10(0) = 300$ (Minimum)
- At $B(120, 0)$: $Z = 5(120) + 10(0) = 600$ (Maximum)
- At $C(60, 30)$: $Z = 5(60) + 10(30) = 300 + 300 = 600$ (Maximum)
- At $D(40, 20)$: $Z = 5(40) + 10(20) = 200 + 200 = 400$
Conclusion regarding multiple optima
Minimum value is $300$ at point $(60, 0)$.
Maximum value is $600$ at all points on the line segment joining $B(120, 0)$ and $C(60, 30)$.