Q.State whether the following statement is True or False: In a LPP, the minimum value of the objective function $Z = ax + by$ is always $0$ if origin is one of the corner point of the feasible region.

Concept: Corner Point Theorem — the optimal value of a linear objective function in a bounded LPP occurs at a corner point of the feasible region. The value at the origin depends on the coefficients $a$ and $b$, not on the fact that the origin is a corner point.

Step 1: If the origin $(0,0)$ is a corner point, then $Z(0,0) = a(0) + b(0) = 0$. …

The key idea here is the Corner Point Theorem (also called the Fundamental Theorem of Linear Programming): if an optimal solution exists, it occurs at a corner point of the feasible region. But the theorem says nothing about which corner point gives the optimum — it only says the optimum is among them. So just because the origin is a corner point does not mean it is the optimal point.

Let’s break this down carefully.

1. What does the statement claim?

   It says: "In a LPP, the minimum value of $Z = ax + by$ is always $0$ if origin is one of the corner point of the feasible region."

   This means: whenever $(0,0)$ is a corner of the feasible region, the smallest possible value of $Z$ must be $0$.

2. What is $Z$ at the origin?

   At $(0,0)$, $Z = a(0) + b(0) = 0$. So the origin gives $Z = 0$. That part is always true.

3. But is $0$ necessarily the minimum?

   Not at all. The minimum is the smallest value of $Z$ over all feasible points. If another corner point gives $Z < 0$, then $0$ is not the minimum — it’s just one value among many.

   For example, consider:

   • Objective: $Z = 2x - 3y$
   • Constraints: $x \ge 0$, $y \ge 0$, $x + y \le 4$, $2x - y \ge -2$

   The feasible region includes the origin $(0,0)$ as a corner. At the origin, $Z = 0$.

   But check another corner: $(0,4)$ gives $Z = 2(0) - 3(4) = -12$, which is much smaller. So the minimum is $-12$, not $0$.

4. When would the statement be true? …

Method: Deciding whether the origin's value is the optimum

The trap in any "the optimum equals the value at point $P$" claim is confusing "$P$ is a corner" with "$P$ is the best corner." The general move is to separate the two.

Steps

Step 1: Compute the objective at the named point.

At the origin, $Z=a(0)+b(0)=0$ — always. So "origin gives $0$" is true, but it says nothing about whether $0$ is the smallest.

Step 2: Recall what the Corner Point Theorem actually promises.

It says the optimum sits at some vertex, not at the one you happen to be looking at. So you must still compare $Z$ across all corners.

Step 3: Probe for a smaller (or larger) corner. …

Common Mistakes

Mistake 1: Assuming that because the origin is a corner giving $Z=0$, the minimum must be $0$.

Why it's wrong: being a corner makes the origin a candidate, not the winner. Correct approach: evaluate $Z=ax+by$ at every corner and compare — another vertex may give a smaller value.

Mistake 2: Forgetting the objective can be negative. …