Q.Consider the following Linear Programming Problem: Minimise $Z = x + 2y$ Subject to $2x + y \geq 3$, $x + 2y \geq 6$, $x, y \geq 0$. Show graphically that the minimum of $Z$ occurs at more than two points.

Concept: Graphical Method – Unboundedness and Multiple Optimal Solutions

The feasible region is unbounded (open away from the origin), but the objective function has a lower bound. The minimum occurs along an entire edge of the region, not at a single corner.

Steps:

1. Plot the constraints:

   $2x + y \geq 3$ (line through $(0,3)$ and $(1.5,0)$)

   $x + 2y \geq 6$ (line through $(0,3)$ and $(6,0)$)

   $x, y \geq 0$ (first quadrant).

2. The feasible region is the intersection of the half-planes above both lines, in the first quadrant. The corner points are $A(0,3)$ and $B(6,0)$.

3. Evaluate $Z = x + 2y$ at the corners:

   At $A(0,3)$: $Z = 0 + 6 = 6$

   At $B(6,0)$: $Z = 6 + 0 = 6$

   Both give the same value. …

Draw the constraint boundaries.

• $2x + y = 3$ meets the axes at $(1.5, 0)$ and $(0, 3)$.
• $x + 2y = 6$ meets the axes at $(6, 0)$ and $(0, 3)$.

Both inequalities are "$\ge$", so the feasible region lies above/right of each line, with $x, y \ge 0$. The region is unbounded, and its two corner points are $(0,3)$ and $(6,0)$.

Evaluate $Z = x + 2y$ at the corners.

Both corners give the same value, $Z = 6$.

Why the minimum repeats along a whole edge. …

Method: Detecting Infinitely Many Optima (Objective Parallel to an Edge)

Use this when asked to show that an optimum is attained at more than one point.

Steps

Step 1: Build the feasible region and list its corners.

Plot the constraint lines, shade with the origin test, and find the vertices where boundaries meet.

Step 2: Evaluate $Z=ax+by$ at the corners and look for a tie.

If two adjacent corners give the same optimal value, the objective is not optimised at a single point.

Step 3: Confirm the objective is parallel to that edge. …

Common Mistakes

Mistake 1: Reporting a single optimal corner.

Why it's wrong: both $(0,3)$ and $(6,0)$ give $Z=6$, and the objective $x+2y$ is parallel to the edge $x+2y=6$, so the whole segment is optimal — naming just one point misses the point of the question. Correct approach: after finding the tie, state that every point on the joining edge is a minimiser.

Mistake 2: Shading the wrong side of the $\ge$ constraints. …