Q.The feasible region of a linear programming problem is the unbounded region in the first quadrant ($x \ge 0$, $y \ge 0$) satisfying $x + y \ge 3$ and $x + 2y \ge 4$. Its corner points are $(0, 3)$, $(2, 1)$ and $(4, 0)$. Evaluate $Z = 4x + y$ at each corner point and find the minimum value of $Z$, if it exists.

Concept

For an unbounded feasible region the corner-point value is only a candidate minimum. It is the true minimum only if the open half-plane $Z < (\text{that value})$ has no point in common with the region.

Corner points

The region is $\{\,x\ge 0,\ y\ge 0,\ x+y\ge 3,\ x+2y\ge 4\,\}$. Its vertices are:

• $(0,3)$ — where $x+y=3$ meets the $y$-axis;
• $(2,1)$ — intersection of $x+y=3$ and $x+2y=4$ (subtracting gives $y=1,\ x=2$);
• $(4,0)$ — where $x+2y=4$ meets the $x$-axis.

Evaluate $Z = 4x + y$ …

Method: Optimum-Existence Test on an Unbounded Region (Minimisation)

Use this to find a minimum of $Z=ax+by$ when the feasible region is unbounded — a corner value is only a candidate until you prove nothing beats it.

Steps

Step 1: Evaluate $Z$ at the corners to get a candidate minimum $m$.

Tabulate $Z=ax+by$ at each vertex and let $m$ be the smallest.

Step 2: Test the open half-plane $ax+by<m$.

Ask whether any feasible point makes $ax+by<m$. Two equivalent checks:

• graphically, does the region $ax+by<m$ overlap the feasible region?
• algebraically, combine the constraints — e.g. from $x+y\ge c$ and $x\ge0$ derive a lower bound on $ax+by$. …

Common Mistakes

Mistake 1: Declaring the smallest corner value the minimum without checking existence.

Why it's wrong: on an unbounded region a corner value is only a candidate; you must confirm no feasible point beats it. Here the candidate is $Z=3$ at $(0,3)$. Correct approach: test the open half-plane $4x+y<3$ — using $x\ge0$ and $x+y\ge3$, $4x+y=(x+y)+3x\ge3$, so nothing undercuts $3$ and the minimum genuinely exists.

Mistake 2: Assuming "unbounded" means no minimum exists. …