Q.Minimise $Z = x + 2y$ subject to $2x + y \ge 3$, $x + 2y \ge 6$, $x, y \ge 0$. Show that the minimum of $Z$ occurs at more than two points.

Concept: Linear Programming Optimality — when the objective line is parallel to a binding constraint edge, the minimum occurs at infinitely many points along that edge.

Reasoning:

1. Plot the constraints:

   $2x + y \ge 3$ passes through $(0,3)$ and $(1.5,0)$.

   $x + 2y \ge 6$ passes through $(0,3)$ and $(6,0)$.

   The feasible region is unbounded above and to the right, with corner points at $(0,3)$ and $(6,0)$.

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

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

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

   Both give the same value. …

We are minimising $Z = x + 2y$ under the constraints:

The key observation: the objective function $Z = x + 2y$ has exactly the same coefficients as the second constraint $x + 2y \ge 6$. This is not a coincidence — it means the objective is parallel to that constraint boundary. In linear programming, when the objective line is parallel to a binding constraint, the optimum is not a single point but an entire edge of the feasible region.

Let's work through it step by step.

1. Draw the feasible region. First, treat the inequalities as equalities to find the boundary lines:

The non-negativity constraints $x \ge 0, y \ge 0$ restrict us to the first quadrant.

Find the intersection of $L_1$ and $L_2$:

Multiply the second equation by 2: $2x + 4y = 12$. Subtract the first: $(2x + 4y) - (2x + y) = 12 - 3 \implies 3y = 9 \implies y = 3$. Then $x + 2(3) = 6 \implies x = 0$. So the intersection is $(0, 3)$.

The lines meet the axes:

• $L_1$: when $x = 0$, $y = 3$; when $y = 0$, $x = 1.5$.
• $L_2$: when $x = 0$, $y = 3$; when $y = 0$, $x = 6$.

Since both constraints are "$\ge$", the feasible region is the region above both lines (and in the first quadrant). The corner points are:

• $A = (0, 3)$ — intersection of $L_1$ and $L_2$.
• $B = (6, 0)$ — where $L_2$ meets the $x$-axis.
• Checking where $L_1$ meets the $x$-axis ($y=0$): $2x=3 \Rightarrow x=1.5$, giving the point $(1.5,0)$. But this point must also satisfy $x+2y\ge 6$: $1.5+0=1.5<6$, so it fails — $(1.5,0)$ is not in the feasible region. The actual feasible region is unbounded, with corner points only at $(0,3)$ and $(6,0)$, extending upward and rightward from there.

2. Evaluate $Z$ at the corner points. At $A = (0, 3)$: $Z = 0 + 2(3) = 6$. At $B = (6, 0)$: $Z = 6 + 0 = 6$. …

Method: Detecting and reporting multiple optimal solutions in a graphical LPP

When the objective's coefficients match a constraint's, expect the optimum along a whole edge, not at a single point. The method is to spot the parallelism, then report the segment.

Steps

Step 1: Compare the objective with each constraint.

If $Z=ax+by$ has the same coefficient ratio as a binding constraint $ax+by\ge c$, the objective line is parallel to that boundary — the multiple-optima signal.

Step 2: Evaluate at the two relevant corners.

$Z=ax+by$ …

Common Mistakes

Mistake 1: Reporting the minimum at a single point.

Why it's wrong: the objective $Z=x+2y$ is parallel to the constraint $x+2y\ge6$, so $Z=6$ all along the edge from $(0,3)$ to $(6,0)$ — infinitely many optimal points. Correct approach: recognise the parallelism and state the whole segment.

Mistake 2: Including $(1.5,0)$ as a corner. …