MathsISC Class 12 · Linear Programming

Unbounded and Infeasible Feasible Regions in Linear Programming

Updated 15 October 2026 · Fact-checked

On an unbounded feasible region an optimum may not exist. The smallest corner value m is the minimum only if the open half-plane $ax + by < m$ misses the region; for a maximum M, test $ax + by > M$. If no point satisfies all the constraints, the LPP is infeasible and has no solution.

The red half-plane Z < 8 misses the region, so 8 is the minimum; the region runs on forever, so Z has no maximum.

Illustration drawn for QuestionBuddy.

Understand Unbounded and Infeasible Regions in LPP

Bounded or unbounded. A feasible region is bounded if it can be enclosed in a circle; otherwise it is unbounded. Regions defined only by " $\geq$ " constraints (diet and cost problems) are usually unbounded: they extend away from the origin. A region can be unbounded even with a " $\leq$ " constraint if some direction is left open, as with $x - y \leq 2$ , $y \geq 1$ .

Corner point theorem, unbounded case. If the region is unbounded, a maximum or minimum may not exist. But if it does exist, it occurs at a corner point. So the corner method still works, with one extra check:
1. Evaluate Z at the corners. Let $m$ be the smallest value and $M$ the largest.
2. Minimum: draw the open half-plane $a x + b y < m$ . If it has no point in common with the feasible region, $m$ is the minimum; otherwise Z has no minimum.
3. Maximum: draw $a x + b y > M$ . If it has no point in common with the region, $M$ is the maximum; otherwise there is no maximum.
Draw the line $a x + b y = m$ (or $M$ ) dashed, because the points on it are not part of the open half-plane.

Why the test works. Every point with a smaller Z than $m$ lies in the half-plane $a x + b y < m$ . If that half-plane misses the region completely, no feasible point beats $m$ .

A quick algebraic version. Along an unbounded edge, write a general point and see what happens to Z as you move outward. If Z keeps increasing, there is no maximum; if it keeps decreasing, no minimum. For example, with positive coefficients a, b and only " $\geq$ " constraints, Z can be made as large as you like but has a minimum at a corner.

Multiple optima on an unbounded region. If the objective line is parallel to an unbounded edge, Z can be optimal at every point of that ray, giving infinitely many optimal solutions.

Infeasible problems. If no point satisfies all the constraints together, the feasible region is empty and the LPP is infeasible: there is no feasible solution and hence no optimal solution. Graphically, the shaded half-planes have no common part. Algebraically, combining the constraints gives a contradiction such as $0 \geq 3$ . In practice an infeasible model means the requirements contradict the resources: something must be relaxed.

Summary of outcomes.

Region	Possible outcome
Bounded, non-empty	Max and min both exist, at corners
Unbounded	Max or min (or both, or neither) may fail to exist; test with open half-planes
Empty	Infeasible: no solution

The 2027 ISC syllabus names "feasible (bounded/unbounded) and infeasible regions" and "optimum feasible solution (may/may not exist)", so you should be able to recognise and justify each case.

Unbounded and Infeasible Regions in LPP: formulas to remember

Minimum test (unbounded): ${a x + b y < m} \cap R = \emptyset \Rightarrow Z_{m i n} = m$; m = smallest corner value.
Maximum test (unbounded): ${a x + b y > M} \cap R = \emptyset \Rightarrow Z_{m a x} = M$; M = largest corner value.
Edge test: $Z (p + t d) = Z (p) + t (a d_{1} + b d_{2})$; Sign of $a d_{1} + b d_{2}$ along an unbounded edge decides.
Infeasible: $R = \emptyset$; Constraints contradict; no feasible solution.

How to solve Unbounded and Infeasible Regions in LPP questions

Treat an unbounded region exactly like a bounded one, then add one justification line.

1Draw the constraints and shade; decide whether the region is bounded, unbounded or empty.
2If empty, say so: the LPP is infeasible. Support it with the graph or an algebraic contradiction.
3Otherwise list the corners and evaluate Z at each.
4If bounded, choose the largest and smallest values.
5If unbounded, take the best corner value and draw the open half-plane beyond it with a dashed line.
6If the half-plane misses the region, the corner value is optimal; if not, state that there is no maximum (or minimum).

Unbounded and Infeasible Regions in LPP in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

An infeasible plan

Suppose a shop has space for at most 10 sofas, but its suppliers insist on orders of at least 6 of model A and at least 5 of model B.

Can an order meet all three conditions?

Try it first, then show the solutionSolution

Constraints: $x + y \leq 10$ , $x \geq 6$ , $y \geq 5$ .
The last two give $x + y \geq 11$ , contradicting $x + y \leq 10$ .

Answer: No: the region is empty, so the LPP is infeasible. The shop must relax a minimum order or find more space.

Memory hook: Infeasible = the rules contradict each other; no plan, not even a bad one.

A missing constraint shows up as "no maximum"

Imagine a factory model whose only conditions are minimum production levels: at least 10 units of x, at least 5 of y and at least 20 in total, with profit Z = 50x + 40y.

Does Z have a maximum? What does the answer say about the model?

Try it first, then show the solutionSolution

The region $x \geq 10$ , $y \geq 5$ , $x + y \geq 20$ is unbounded.
Along $y = 5$ , $x = t$ with $t \to \infty$ , $Z = 50 t + 200$ grows without limit.

Answer: No maximum. A real factory has limits on labour, machines or demand; their absence from the model makes the profit "unlimited".

Memory hook: An unbounded maximum usually means a constraint was left out.

Tricks to remember Unbounded and Infeasible Regions in LPP

Short memory aids. Say each one aloud once, then use it on a question the same day.

Unbounded: corner, then check
Corner values first; then the dashed open half-plane check.
Use it when: Every unbounded LPP.
Less for min, more for max
Minimum test uses $<$ the corner value; maximum test uses $>$ .
Use it when: Writing the half-plane.
Cost goes down, profit goes up
With positive coefficients and "≥" constraints, a minimum exists but a maximum does not.
Use it when: A quick prediction.
Add to contradict
Combine constraints to reach something like $0 \geq 3$ : infeasible.
Use it when: Proving infeasibility.

Unbounded and Infeasible Regions in LPP: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Recognising an unbounded region · 1 mark · MCQBoard level

The feasible region defined by $x + 2 y \geq 6$ , $x \geq 0$ , $y \geq 0$ is:
(a) bounded
(b) unbounded
(c) empty
(d) a single point

Try it first, then show the solutionSolution

All points beyond the line $x + 2 y = 6$ in the first quadrant qualify, with no upper limit.

Answer: (b) unbounded.

Example 2: Minimum on an unbounded region · 3 marks · Short answerBoard level

Minimise $Z = 2 x + 3 y$ subject to $x + y \geq 5$ , $x + 3 y \geq 9$ , $x, y \geq 0$ . Does Z have a maximum?

Try it first, then show the solutionSolution

Unbounded region with corners (0, 5), (3, 2) (from $2 y = 4$ ) and (9, 0). (1 mark)
Z = 15, 12, 18; smallest 12 at (3, 2). The open half-plane $2 x + 3 y < 12$ has no point in the region (dashed line through (6, 0) and (0, 4) touches it only at (3, 2)), so $Z_{m i n} = 12$ . (1 mark)
Along $y = 0$ , $Z = 2 x \to \infty$ , so there is no maximum. (1 mark)

Answer: $Z_{m i n} = 12$ at (3, 2); no maximum.

Example 3: A maximum that does exist · 3 marks · Short answerHarder board level

Maximise $Z = 2 x - 4 y$ subject to $x - y \leq 2$ , $y \geq 1$ , $x \geq 0$ .

Try it first, then show the solutionSolution

Corners: (0, 1) and (3, 1); the region is unbounded upwards. Z = −4 and 2. (1 mark)
Test $2 x - 4 y > 2$ , i.e. $x - 2 y > 1$ . In the region $x \leq y + 2$ , so $x - 2 y \leq 2 - y \leq 1$ . No point qualifies. (1 mark)
Hence $Z_{m a x} = 2$ at (3, 1). Z has no minimum (along $x = 0$ , $Z = - 4 y \to - \infty$ ). (1 mark)

Answer: $Z_{m a x} = 2$ at (3, 1); no minimum.

Example 4: Proving infeasibility · 2 marks · Short answerHarder board level

Show that the LPP "maximise $Z = x + 2 y$ subject to $x + y \leq 3$ , $2 x + 3 y \geq 12$ , $x, y \geq 0$ " has no feasible solution.

Try it first, then show the solutionSolution

For a feasible point, $2 x + 3 y = 2 (x + y) + y \leq 6 + y$ , and $y \leq x + y \leq 3$ , so $2 x + 3 y \leq 9$ . (1 mark)
This contradicts $2 x + 3 y \geq 12$ , so the region is empty and the LPP is infeasible. (Graphically, $2 x + 3 y = 12$ lies entirely beyond $x + y = 3$ in the first quadrant.) (1 mark)

Answer: Infeasible: no point satisfies all the constraints.

Example 5: Neither maximum nor minimum · 2 marks · Short answerHOTS / competency

On the region $x + y \geq 5$ , $x + 3 y \geq 9$ , $x, y \geq 0$ , show that $Z = x - y$ has neither a maximum nor a minimum.

Try it first, then show the solutionSolution

Along the x-axis beyond (9, 0), $Z = x \to \infty$ : no maximum. (1 mark)
Along the y-axis beyond (0, 5), $Z = - y \to - \infty$ : no minimum. (1 mark)

Answer: Z is unbounded above and below on this region.

Example 6: Cheapest animal feed · 4 marks · Case-basedHOTS / competency

Suppose a farmer mixes feeds P and Q. Feed P costs ₹30 per kg and gives 2 units of vitamin A and 3 units of vitamin B per kg; feed Q costs ₹25 per kg and gives 4 units of A and 1 unit of B per kg. The animals need at least 16 units of A and 9 units of B daily.
(i) Formulate the LPP. (1 mark)
(ii) Find the corner points of the feasible region. (1 mark)
(iii) Find the minimum cost, justifying that it is a minimum. (2 marks)

Try it first, then show the solutionSolution

(i) Minimise $Z = 30 x + 25 y$ subject to $2 x + 4 y \geq 16$ (i.e. $x + 2 y \geq 8$ ), $3 x + y \geq 9$ , $x, y \geq 0$ .
(ii) $x = 8 - 2 y$ in $3 x + y = 9$ : $24 - 5 y = 9$ , $y = 3$ , $x = 2$ . Corners (0, 9), (2, 3), (8, 0).
(iii) Z = 225, 135, 240. Smallest 135 at (2, 3).
The region is unbounded, so test $30 x + 25 y < 135$ ( $6 x + 5 y < 27$ ): its boundary has slope −1.2, between the edge slopes −3 and −0.5, so it meets the region only at (2, 3). Minimum cost ₹135 with 2 kg of P and 3 kg of Q.

Answer: ₹135 a day: 2 kg of P and 3 kg of Q.

Where students lose marks in Unbounded and Infeasible Regions in LPP

The slips examiners see most often, and what to write instead.

Mistake: No half-plane check
Quoting the smallest corner value on an unbounded region without justification loses the final mark.
Do this: Add one line: "the open half-plane $a x + b y < m$ has no point in common with the region".
Usually costs: 1 mark.
Mistake: Solid instead of dashed line
The boundary of the open half-plane is not included; drawing it solid suggests otherwise.
Do this: Draw $a x + b y = m$ dashed and shade the open side lightly.
Usually costs: Up to 1 mark for the graph.
Mistake: Claiming a maximum on a cost region
With "≥" constraints and positive coefficients, Z grows without limit.
Wrong working
"Maximum cost is 240 at (8, 0)."
Right working
No maximum: along the x-axis Z → ∞.
Do this: Check the direction in which the region is open before quoting a maximum.
Usually costs: 1 mark.
Mistake: Not recognising an empty region
Students sometimes pick corner points from lines that never bound a common region.
Do this: Check every chosen corner against all constraints; if none survives, the LPP is infeasible.
Usually costs: 2 marks.

Exam tips

Write "the feasible region is unbounded" in words as soon as you see it.
For minimisation of cost with "≥" constraints, expect an unbounded region and a corner minimum.
A one-line algebraic argument (like $4 x + y \geq 6 + x$ ) is a valid alternative to the graphical half-plane test.
If the LPP is infeasible, say there is no feasible and no optimal solution.

Practice questions on Unbounded and Infeasible Regions in LPP

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For an unbounded feasible region, the smallest corner value m of Z = ax + by is the minimum if:
(a) the region contains the origin
(b) the half-plane $a x + b y < m$ has no point in common with the region
(c) the half-plane $a x + b y > m$ has no point in common with the region
(d) always
Hint
Smaller values of Z live where?
Answer
(b).
Q2 · 2 marks · Short answerBoard level
Is the region $x \geq 2$ , $y \geq 1$ , $x + y \leq 6$ bounded? Give its corners.
Hint
The "≤" closes it off.
Answer
Bounded: a triangle with corners (2, 1), (5, 1), (2, 4).
Q3 · 3 marks · Short answerHarder board level
Minimise $Z = 4 x + y$ subject to $x + y \geq 4$ , $3 x + y \geq 6$ , $x, y \geq 0$ .
Hint
Corners (0, 6), (1, 3), (4, 0).
Answer
Z = 6, 7, 16. In the region $y \geq 6 - 3 x$ , so $4 x + y \geq 6 + x \geq 6$ : no point has $Z < 6$ . Minimum 6 at (0, 6).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An LPP whose feasible region is unbounded never has a maximum.
Reason (R): On an unbounded region a minimisation problem may still have an optimal solution.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Recall the example maximising 2x − 4y.
Answer
(d).
Q5 · 1 mark · Very short answerHarder board level
Show that the constraints $x - y \geq 2$ and $y - x \geq 1$ (with $x, y \geq 0$ ) have no common solution.
Hint
Add them.
Answer
Adding gives $0 \geq 3$ , which is false: infeasible.
Q6 · 3 marks · Short answerHOTS / competency
On the region $x - y \leq 1$ , $x + y \geq 3$ , $x, y \geq 0$ , find the minimum of $Z = y - x$ and describe all the points where it occurs. Does Z have a maximum?
Hint
The edge $x - y = 1$ is unbounded and parallel to the objective lines.
Answer
Corners (0, 3) and (2, 1): Z = 3, −1. Since $x - y \leq 1$ , $Z = y - x \geq - 1$ everywhere, so $Z_{m i n} = - 1$ , attained at every point of the ray $x - y = 1$ with $x \geq 2$ . No maximum (Z = y → ∞ along the y-axis).
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a company must move at least 40 tonnes of goods. A type A truck carries 8 tonnes and costs ₹5000 a trip; a type B truck carries 4 tonnes and costs ₹3500. At least 2 type B trucks must be used.
(i) Formulate the LPP to minimise cost. (1 mark)
(ii) Find the corners of the feasible region. (1 mark)
(iii) Find the minimum cost. (1 mark)
(iv) Explain why this is a true minimum although the region is unbounded. (1 mark)
Hint
$8 x + 4 y \geq 40$ is $2 x + y \geq 10$ .
Answer
(i) Min $Z = 5000 x + 3500 y$ , $2 x + y \geq 10$ , $y \geq 2$ , $x \geq 0$ . (ii) (0, 10), (4, 2). (iii) Z = 35 000, 27 000: ₹27 000 with 4 A and 2 B. (iv) On the edge $2 x + y = 10$ , $Z = 35000 - 2000 x$ falls until $y = 2$ ; beyond, along $y = 2$ or upwards, Z rises; so $5000 x + 3500 y < 27000$ meets no feasible point.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is an infeasible LPP?Show answer
One whose constraints have no common solution: the feasible region is empty.
2.On an unbounded region, if an optimum exists, where is it?Show answer
At a corner point.
3.Which open half-plane do you test for a maximum M?Show answer
$a x + b y > M$ .
4.Is the region $x + y \geq 2$ , $x, y \geq 0$ bounded?Show answer
No, it is unbounded.
5.Constraints $x + y \leq 1$ and $x + y \geq 2$ : feasible?Show answer
No: infeasible.

Unbounded and Infeasible Regions in LPP: frequently asked questions

Does an unbounded region always mean there is no solution?

No. A minimum (or sometimes a maximum) can still exist. You find the best corner value and then check, with the open half-plane, that no feasible point does better.

What is the difference between unbounded and infeasible?

An unbounded region has infinitely many feasible points stretching without end. An infeasible problem has no feasible points at all.

Why is the boundary line dashed in the half-plane test?

The test is about points with Z strictly less than (or greater than) the corner value. The line itself has Z equal to that value, so it is excluded from the half-plane.

Is it enough to say "Z increases without limit"?

Yes, if you show it: give a direction along the region (such as the x-axis) and how Z behaves along it.

How is this examined in ISC?

The 2027 syllabus explicitly lists bounded, unbounded and infeasible regions and optimal solutions that may or may not exist, so expect a justification step inside the LPP question or a 1-mark reasoning item.