Q.Find the value of the following: Maximise $Z = -x + 2y$, subject to the constraints: $x \ge 3$, $x + y \ge 5$, $x + 2y \ge 6$, $y \ge 0$.

The feasible region is unbounded, so we find the corner points, evaluate $Z$, and then check whether it can grow without bound.

Step 1 - Corner points. Intersecting the boundary lines and keeping only points that satisfy every constraint ($x \ge 3$, $x + y \ge 5$, $x + 2y \ge 6$, $y \ge 0$):

• $x = 3$ and $x + y = 5 \Rightarrow (3, 2)$
• $x + y = 5$ and $x + 2y = 6 \Rightarrow (4, 1)$
• $x + 2y = 6$ and $y = 0 \Rightarrow (6, 0)$

Step 2 - Evaluate $Z = -x + 2y$.

The largest corner value is $1$, at $(3, 2)$. …

The constraints are $x \ge 3$, $x + y \ge 5$, $x + 2y \ge 6$, $y \ge 0$. Each is a half-plane; their intersection is the feasible region.

1. Corner points of the feasible region. Solving the boundary lines in pairs and keeping only points that satisfy every constraint:

• $x = 3$ and $x + y = 5 \Rightarrow (3, 2)$
• $x + y = 5$ and $x + 2y = 6 \Rightarrow (4, 1)$
• $x + 2y = 6$ and $y = 0 \Rightarrow (6, 0)$

So the corner points are $(3, 2)$, $(4, 1)$ and $(6, 0)$.

2. Evaluate $Z = -x + 2y$ at each corner.

The largest value at a corner is $1$, at $(3, 2)$. …

Method: Maximising over an unbounded region — when no maximum exists

The full graphical method plus the mandatory unbounded check; here the check is what reveals that the "best corner" is not the maximum.

Steps

Step 1: Build the feasible region and its corners.

Shade each half-plane, keep only intersections satisfying all constraints. With several "$\ge$" constraints the region is typically unbounded.

Step 2: Evaluate $Z$ and get the best corner value $M$.

Step 3: Apply the open-half-plane test — never skip it. …

Common Mistakes

Mistake 1: Quoting the best corner value as the maximum without the unbounded check.

Why it's wrong: the region is unbounded and $Z=-x+2y$ keeps growing along $x=3$ as $y\to\infty$, so the corner value $1$ is not the maximum. Correct approach: test the open half-plane $-x+2y>1$; it meets the region, so no maximum exists.

Mistake 2: Listing an infeasible point such as $(3,1.5)$ as a corner. …