Q.In a LPP if the objective function $Z = ax + by$ has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same _________ value.

The key idea is that in Linear Programming, if the objective function attains its maximum at two distinct corner points, the entire line segment connecting them is also optimal — this is the multiple optimal solutions case. …

The key idea here is convexity of the feasible region and linearity of the objective function. In Linear Programming, the feasible region is always a convex polygon (or polyhedron). The objective function $Z = ax + by$ is a linear function — its level sets are straight lines.

When two distinct corner points give the same maximum value, say $Z = M$, both points lie on the same level line $ax + by = M$. Since the feasible region is convex, the entire line segment between them lies inside the region. And because $Z$ is linear, its value at any point on that segment is a weighted average of its values at the endpoints — which is still $M$.

Let’s walk through this step by step.

1. Understand the geometry of a linear objective.

   The function $Z = ax + by$ is linear. For any constant $c$, the set of points where $Z = c$ is a straight line. As $c$ changes, these lines shift parallel to each other. Optimising $Z$ means finding the farthest such line (in the direction of increase) that still touches the feasible region.

2. What happens at a corner point optimum?

   Normally, the maximum occurs at a single corner point — the last point the shifting level line touches before leaving the region. That’s the typical case.

3. What does it mean when two corner points give the same maximum?

   If two distinct corner points $P$ and $Q$ both give $Z = M$, then both lie on the same level line $ax + by = M$. This line is a supporting line to the feasible region — it touches the region along an entire edge (or face) that contains both $P$ and $Q$.

4. Why does every point on the segment also give $Z = M$?

   Take any point $R$ on the line segment $PQ$. Since the feasible region is convex, $R$ is inside the region. And because $Z$ is linear:

But $R$ is a convex combination of $P$ and $Q$: $R = \lambda P + (1-\lambda)Q$ for some $0 \le \lambda \le 1$. …

Method: Using linearity along a segment

For a statement about the value of $Z$ at every point of a segment joining two equally-optimal corners, argue from the linearity of the objective.

The reasoning pattern

Step 1: Note the objective is linear.

$Z = ax + by$ changes at a constant rate along any straight line, so along a segment its value moves linearly between the two endpoint values.

Step 2: Equal endpoints force a constant.

Any point $R$ on the segment from $P$ to $Q$ is a weighted average $R = \lambda P + (1-\lambda)Q$. Then

$Z(R) = \lambda Z(P) + (1-\lambda)Z(Q).$ …

Common Mistakes

Mistake 1: Filling the blank with "optimal" or "minimum" instead of "maximum."

Why it's wrong: the sentence itself states the two corners share the same maximum value, and by linearity every point on the joining segment inherits that same value -- so the blank must repeat "maximum." Writing "minimum" c …