Optimal solutions in LPP

An LPP is not finished once you have drawn the feasible region and marked the corner points — the real question is what those corner values mean. An optimal solution is a feasible point where the objective function $Z$ reaches its best possible value, largest for a maximisation problem, smallest for a minimisation problem. But "best possible" hides four different endings: a single winning corner, a tie between two adjacent corners that hands you infinitely many optimal points, an unbounded region where $Z$ never stops growing, or constraints so contradictory that no feasible point exists at all. This article stays entirely inside the CBSE Class 12 two-variable graphical method. Building on the Corner Point Theorem, we look closely at how to read the final $Z$ -table correctly: spotting ties, recognising when an unbounded feasible region still has no optimum, and separating "unbounded LPP" from "infeasible LPP" — the two endings students confuse most often in the exam hall.

The core insight

An optimal solution is the feasible point (or points) where a linear objective function $Z$ attains its extreme value at a corner of the feasible region, and correctly reading the corner-point table tells you whether that optimum is unique, infinite, unbounded, or absent.

The core idea

What "optimal" means. Among all feasible solutions of an LPP — points satisfying every constraint — the one(s) that give $Z$ its required maximum or minimum value are called the optimal solution(s), and that extreme value of $Z$ is the optimal value. Nothing about this depends on how many variables the problem has in general, but the CBSE syllabus restricts attention to two variables, solved graphically.

Why corners are enough. For a bounded feasible region, the Corner Point Theorem guarantees that a linear $Z$ achieves both its maximum and its minimum at a corner (vertex) of the region — never strictly inside it, and never on the interior of an edge unless that whole edge ties. This is why the entire method reduces to one arithmetic step:

Z = ax + by, \qquad Z_{\text{corner } i} = a x_i + b y_i \text{ for each vertex } (x_i, y_i).

where: $a, b$ are the fixed coefficients of the objective function, and $(x_i, y_i)$ runs over every corner point of the feasible region found by solving pairs of boundary equations simultaneously. Once this table of $Z$ -values is built, reading off the answer is a matter of care, not new computation.

The four endings. Scan the completed $Z$ -table for these patterns: A unique optimal solution occurs when the required extreme value of $Z$ appears at exactly one corner — the ordinary, most common case.

Multiple (infinitely many) optimal solutions occur when the same optimal value of $Z$ is attained at two adjacent corners. Since $Z$ is linear, it then takes that identical value at every point of the line segment joining them — because the objective line is exactly parallel to that boundary edge. An unbounded solution (no finite optimum) occurs when the feasible region itself is unbounded and $Z$ keeps increasing (for a maximisation problem) or decreasing (for a minimisation problem) without limit as you move along the unbounded direction — checked using the open half-plane test. An infeasible problem occurs when the constraints contradict one another so completely that no point satisfies all of them together; the feasible region is empty, and the LPP has no solution at all.

Figure

A feasible triangle with two circled corner points tied at the same Z value and the connecting edge highlighted to show infinitely many optimal solutions

Derivation AWhy corner ties give infinitely many optimal solutions

The goal is to justify, in one clean argument longer than a slogan, why finding an optimal solution is always a matter of reading a finite table of corner values rather than searching the whole feasible region point by point, and why a tie between two adjacent corners forces infinitely many optima rather than just two.

Step 1. Take any point $P$ strictly inside the feasible region or strictly inside an edge (not a corner). Because $Z = ax + by$ is linear, its level curves $ax + by = k$ are parallel straight lines. Sliding this line through $P$ in the direction of increasing $k$ (for maximisation) keeps it inside a bounded feasible region for a little further before it exits — so $P$ cannot already hold the maximum; some corner further along beats it. The same argument run in the decreasing direction rules out $P$ as the minimum. Hence no interior point, and no point strictly inside an edge, can be optimal unless the entire edge is tied — the search space collapses from infinitely many feasible points to the finite list of corners.

Step 2. Build the corner-point table by solving the boundary lines pairwise for their intersections, keeping only points that also satisfy every other constraint, then substitute each corner into $Z = ax + by$ . Because there are finitely many corners, this is always a finite calculation, however complicated the constraints looked when drawn.

Step 3. Now suppose two adjacent corners $A$ and $B$ (joined by one edge of the region) give the same value $Z(A) = Z(B) = k$ . Every point on segment $AB$ can be written as $(1-t)A + tB$ for $0 \le t \le 1$ , and because $Z$ is linear, $Z\big((1-t)A + tB\big) = (1-t)Z(A) + tZ(B) = (1-t)k + tk = k$ . So the entire segment shares the value $k$ — this is exactly the geometric statement that the objective line $ax+by=k$ is parallel to, and lies along, the edge $AB$ .

Step 4. Read the conclusion: if $k$ is the required extreme (max or min) among all corners, then every point of segment $AB$ is optimal, giving infinitely many optimal solutions, while any single non-tied corner with the extreme value is the unique optimum. This is the entire logical content behind the rule "scan the table for ties before declaring a unique winner."

Worked example

Consider the LPP: maximise $Z = 4x + 4y$ subject to $x + y \le 4$ , $x \ge 0$ , $y \ge 0$ .

Step 1. The feasible region is the triangle with vertices $(0,0)$ , $(4,0)$ , $(0,4)$ — bounded, so the Corner Point Theorem guarantees both a maximum and a minimum exist.

Step 2. Evaluate $Z$ at each corner: $Z(0,0) = 0$ ; $Z(4,0) = 4(4)+4(0) = 16$ ; $Z(0,4) = 4(0)+4(4) = 16$ .

Step 3. Two adjacent corners, $(4,0)$ and $(0,4)$ , are tied at $Z=16$ , and they are joined directly by the boundary edge $x+y=4$ . This happens precisely because the objective line $4x+4y=k$ , which has slope $-1$ , is parallel to the constraint line $x+y=4$ , which also has slope $-1$ — exactly the geometric condition from Step 4 above.

Step 4. By the segment argument, every point on $x+y=4$ with $x,y \ge 0$ gives $Z=16$ . The LPP has infinitely many optimal solutions, all with maximum value $Z = 16$ ; the minimum value is $Z=0$ , attained uniquely at $(0,0)$ .

Competitive stretch

The same LPP with $Z = 3x + 4y$ instead has slope $-3/4$ , not parallel to $x+y=4$ (slope $-1$ ), so the tie disappears: $Z(0,0)=0$ , $Z(4,0)=12$ , $Z(0,4)=16$ , giving a unique maximum of $16$ at $(0,4)$ — a small coefficient change is enough to collapse infinitely many optima back to one.

Common traps

Declaring a unique optimum from a partial glance at the table — always compare every corner's $Z$ -value before concluding, since a tie can sit anywhere in the list.
Assuming an unbounded feasible region automatically has no optimum — it can still have a well-defined minimum (or maximum) in the opposite direction; only the open half-plane check settles which.
Reporting only one of two tied corners as "the" answer instead of naming the whole joining segment as infinitely many solutions.
Confusing an unbounded LPP (feasible region exists, but $Z$ grows without limit) with an infeasible LPP (feasible region is empty) — these are opposite failures and need opposite justifications.

Ask the tutor: if your corner-point table shows a tie, or you're unsure whether an unbounded region has a genuine optimum, ask the AI tutor to walk through the open half-plane check with your specific constraints.

If two corners tie, do I need to write out every point on the segment as an answer?

No — you state the result in words, not as an infinite list. Write that the maximum value is the tied $Z$ -value, and that it is attained at every point on the segment joining the two tied corners (name both corner coordinates and the boundary equation of that edge, such as $x+y=4$ ). That single sentence fully describes the infinitely many optimal solutions; examiners want the structure of the answer, not an impossible enumeration.

How do I tell an unbounded LPP apart from one that simply has no optimum in that direction?

Both can look similar on a rough sketch, so use the open half-plane test properly: pick the half-plane defined by $ax+by > k$ (for the current best value $k$ found at a corner) and check whether it shares any point with the feasible region. If it does, the objective keeps improving forever and the LPP is genuinely unbounded in that direction. If it does not, $k$ really is the optimum even though the region itself is unbounded — exactly the situation in the earlier example where a maximum failed to exist but a minimum still did.

Is "infeasible" the same as "unbounded with no solution"?

No, they are opposite failures and must not be mixed up. An LPP is infeasible when the constraints contradict each other so no point satisfies all of them — the feasible region itself is empty, and this has nothing to do with the objective function. An LPP is unbounded (with respect to optimisation) when the feasible region is non-empty and even unbounded in extent, but the objective function keeps improving without limit as you move through it. Always check feasibility first (does any point satisfy every constraint together), and only then ask whether the objective function is bounded on that region.

Optimal solutions in LPP

The core idea

Worked example

Competitive stretch

Common traps

Before this