MathsCBSE Class 12 · Linear Programming

Unbounded Feasible Regions in LPP: When Does an Optimum Exist? (Class 12)

Q: Does an unbounded region always have a minimum?

Not always. With positive coefficients and a region in the first quadrant, a minimum usually exists, but with negative coefficients Z may decrease without limit. The half-plane test decides.

Q: Which NCERT section covers unbounded regions?

Section 12.2.2 (Graphical method of solving linear programming problems) in Chapter 12 of NCERT Mathematics Part II for Class 12, which states the theorems and the open half-plane test.

Q: Is an empty feasible region the same as an unbounded one?

No. An empty region has no feasible point at all, so the LPP has no solution. An unbounded region has infinitely many feasible points; only the optimum may fail to exist.

Updated 15 October 2026 · Fact-checked

On an unbounded feasible region, tabulate Z at the corners as usual. The smallest corner value m is the minimum only if the open half-plane ax + by < m has no point in common with the region. For a maximum M, test ax + by > M. If the half-plane meets the region, that optimum does not exist.

Z = x + 2y: smallest corner value is 5, and the shaded side Z < 5 never touches the region, so 5 is the minimum.

Illustration drawn for QuestionBuddy.

Understand Unbounded Feasible Regions

A feasible region is unbounded when it stretches to infinity in some direction, so no circle can enclose it. This happens mostly with $\geq$ constraints (cost-minimising problems such as 'at least 60 tonnes must be shipped').

What changes. The corner point theorem still says: if $Z$ has an optimal value, it is at a corner. But on an unbounded region the optimum may not exist: $Z$ can grow (or fall) without limit as you move away along the region. So the table of corner values gives only a candidate.

The open half-plane test (NCERT Section 12.2.2). Let $Z = a x + b y$ .

Minimum. Let $m$ be the smallest corner value. Draw the line $a x + b y = m$ . If the open half-plane $a x + b y < m$ has no point in common with the feasible region, then $m$ is the minimum. Otherwise $Z$ has no minimum.
Maximum. Let $M$ be the largest corner value. If the open half-plane $a x + b y > M$ has no point in common with the region, $M$ is the maximum. Otherwise $Z$ has no maximum.

In practice: draw the dotted line $a x + b y = m$ through the best corner and see whether any part of the shaded region lies on the 'better' side.

Typical outcomes.

Minimise cost, region going up and right, positive coefficients: a minimum exists, but there is no maximum (you can always spend more).
Maximise with positive coefficients on such a region: no maximum.
Mixed signs (like $Z = 2 y - 3 x$ ): a maximum may exist even though the region is unbounded, so never assume; always do the test.

Multiple optima. If two adjacent corners share the optimal value and the half-plane test passes, every point on the edge between them is optimal. This was asked as 'show that the minimum occurs at more than two points' (2024-25 sample paper Q30 and 2026 board Q20).

Empty region. If the constraints have no common point, there is no feasible solution at all; this is different from an unbounded region.

The 2026-27 syllabus explicitly lists feasible regions that are 'bounded or unbounded', so this test is examinable every year.

Unbounded Feasible Regions: formulas to remember

Minimum test (unbounded region): $m = min_{i} Z (P_{i}) is the minimum ⟺ {a x + b y < m} \cap R = \emptyset$; $R$ is the feasible region; $P_{i}$ are its corners.
Maximum test (unbounded region): $M = max_{i} Z (P_{i}) is the maximum ⟺ {a x + b y > M} \cap R = \emptyset$; Otherwise $Z$ has no maximum.
Boundary of the test: $a x + b y = m (dotted line through the best corner)$; Dotted because the half-plane is open.
Ray check (equivalent idea): $Z (P + t d) = Z (P) + t (a d_{1} + b d_{2})$; Along an unbounded edge with direction $d = (d_{1}, d_{2})$ , $Z$ grows without limit if $a d_{1} + b d_{2} > 0$ .
Multiple optima: $Z (P) = Z (Q) = m, P Q an edge \Rightarrow Z = m on all of P Q$; After the half-plane test passes.

How to solve Unbounded Feasible Regions questions

The full routine for an LPP whose feasible region turns out to be unbounded.

1Draw and shade the region; note that it is unbounded (open on one side).
2Find all corner points by solving boundary equations, and tabulate $Z$ .
3Pick the candidate: the smallest value $m$ for a minimum, or the largest $M$ for a maximum.
4Draw the dotted line $a x + b y = m$ (or $M$ ) and look at the open half-plane $a x + b y < m$ (or $> M$ ).
5If that half-plane has no point in common with the shaded region, the candidate is the optimum; otherwise state that $Z$ has no minimum (or maximum).
6Write the conclusion, mentioning the half-plane test in one line, because it carries a mark.

Unbounded Feasible Regions 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.

Cheapest mix of phone packs

Suppose you need at least 16 GB of data and 900 minutes of calls in a month. Pack A gives 2 GB and 50 minutes for Rs 50; pack B gives 1 GB and 150 minutes for Rs 60.

How many of each pack should you buy to spend the least?

Try it first, then show the solutionSolution

Minimise $Z = 50 x + 60 y$ subject to $2 x + y \geq 16$ , $50 x + 150 y \geq 900$ (that is $x + 3 y \geq 18$ ), $x, y \geq 0$ .
Corners: $(0, 16)$ , $(18, 0)$ , and solving $2 x + y = 16$ with $x + 3 y = 18$ gives $(6, 4)$ . $Z$ : 960, 900, 540.
The open half-plane $50 x + 60 y < 540$ has no point in common with the region, so 540 is the minimum.

Answer: 6 of pack A and 4 of pack B, for Rs 540.

Memory hook: 'At least' needs make the region open upwards: there is a cheapest plan but no 'dearest' plan.

Minimum exists, maximum does not

Imagine a gardener who needs at least 40 kg of compost. Small bags hold 5 kg (Rs 150) and large bags 10 kg (Rs 250), and at least 2 small bags are wanted for easy handling.

Find the cheapest purchase, and say whether there is a most expensive one.

Try it first, then show the solutionSolution

Minimise $Z = 150 x + 250 y$ subject to $5 x + 10 y \geq 40$ (that is $x + 2 y \geq 8$ ), $x \geq 2$ , $y \geq 0$ .
Corners: $(2, 3)$ and $(8, 0)$ ; $Z$ : 1050 and 1200.
$150 x + 250 y < 1050$ has no common point with the region, so the minimum is Rs 1,050. Moving up or right only raises the cost, so there is no maximum.

Answer: 2 small and 3 large bags, Rs 1,050; no maximum cost exists.

Memory hook: Unbounded and positive coefficients: a floor but no ceiling.

Tricks to remember Unbounded Feasible Regions

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

Table, then test
On an open region, the corner table only nominates a candidate; the half-plane test elects it.
Use it when: Never stop at the table for an unbounded region.
Less-than for least
Minimum $m$ : test $Z < m$ . Maximum $M$ : test $Z > M$ . The test half-plane is always the 'better' side.
Use it when: Choosing which half-plane to check.
Dotted line, empty side
Draw $Z = m$ dotted through the best corner; if the better side is empty of shading, you are done.
Use it when: The graph mark for the test.
Signs decide, not shapes
Unbounded does not mean 'no maximum'. With negative coefficients, a maximum can exist.
Use it when: Assertion-reason traps.

Unbounded Feasible Regions: solved examples

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

Example 1: Minimum on an unbounded region · 2 marks · Very short answerBoard level

The feasible region of an LPP is given by $2 x + y \geq 8$ , $2 x + 3 y \geq 16$ , $x \geq 0$ , $y \geq 0$ . Find the minimum value of $Z = 3 x + 2 y$ .

Try it first, then show the solutionSolution

Corners: $(0, 8)$ , $(8, 0)$ , and $2 x + y = 8$ with $2 x + 3 y = 16$ gives $2 y = 8$ , $(2, 4)$ . $Z$ : 16, 24, 14. (1 mark)
The region is unbounded. The open half-plane $3 x + 2 y < 14$ has no point in common with it, so the minimum is 14. (1 mark)

Answer: Minimum $Z = 14$ at $(2, 4)$ .

Example 2: A standard minimisation · 3 marks · Short answerBoard level

Solve graphically: minimise $Z = 5 x + 3 y$ subject to $3 x + y \geq 9$ , $x + 2 y \geq 8$ , $x \geq 0$ , $y \geq 0$ .

Try it first, then show the solutionSolution

$3 x + y = 9$ through $(3, 0)$ , $(0, 9)$ ; $x + 2 y = 8$ through $(8, 0)$ , $(0, 4)$ ; origin fails both, so shade away from the origin. The region is unbounded. (1 mark)
Corners: $(0, 9)$ , $(8, 0)$ , and $y = 9 - 3 x$ in $x + 2 y = 8$ gives $x = 2$ , $y = 3$ : $(2, 3)$ . $Z$ : 27, 40, 19. (1 mark)
The open half-plane $5 x + 3 y < 19$ has no common point with the region, so the minimum is 19 at $(2, 3)$ . (1 mark)

Answer: Minimum $Z = 19$ at $(2, 3)$ .

Example 3: Minimum exists, maximum does not · 3 marks · Short answerHarder board level

Find the minimum of $Z = 2 x + y$ subject to $x + 2 y \geq 4$ , $3 x + y \geq 6$ , $x \geq 0$ , $y \geq 0$ , and show that $Z$ has no maximum.

Try it first, then show the solutionSolution

Corners: $(0, 6)$ , $(4, 0)$ , and $y = 6 - 3 x$ in $x + 2 y = 4$ gives $- 5 x = - 8$ , so $(\frac{8}{5}, \frac{6}{5})$ . $Z$ : 6, 8, $\frac{16}{5} + \frac{6}{5} = \frac{22}{5}$ . (1 mark)
The open half-plane $2 x + y < \frac{22}{5}$ has no common point with the region, so the minimum is $\frac{22}{5}$ . (1 mark)
For the maximum, the largest corner value is 8, but the point $(10, 0)$ is feasible with $Z = 20 > 8$ : the half-plane $2 x + y > 8$ meets the region, so $Z$ has no maximum. (1 mark)

Answer: Minimum $\frac{22}{5}$ at $(\frac{8}{5}, \frac{6}{5})$ ; no maximum.

Example 4: Three constraints, partly closed region · 5 marks · Long answerHarder board level

Minimise $Z = 3 x + 2 y$ subject to $x + y \geq 5$ , $x + 4 y \geq 8$ , $y \leq 4$ , $x \geq 0$ , $y \geq 0$ . Does $Z$ have a maximum?

Try it first, then show the solutionSolution

Lines: $x + y = 5$ , $x + 4 y = 8$ (shade away from the origin) and $y = 4$ (shade below). On $x = 0$ we would need $y \geq 5$ and $y \leq 4$ , so the region does not touch the y-axis. (1 mark)
Corners: $y = 4$ with $x + y = 5$ : $(1, 4)$ ; $x + y = 5$ with $x + 4 y = 8$ : $3 y = 3$ , $(4, 1)$ ; $x + 4 y = 8$ with $y = 0$ : $(8, 0)$ . The region is open to the right, so it is unbounded. (1½ marks)
$Z$ : $3 + 8 = 11$ , $12 + 2 = 14$ , 24. (1 mark)
The open half-plane $3 x + 2 y < 11$ has no common point with the region, so the minimum is 11 at $(1, 4)$ . (1 mark)
No maximum: $(20, 0)$ is feasible with $Z = 60$ , and $Z$ keeps growing to the right. (½ mark)

Answer: Minimum 11 at $(1, 4)$ ; no maximum.

Example 5: A maximum on an unbounded region · 3 marks · Short answerHOTS / competency

Maximise $Z = 2 y - 3 x$ subject to $x + y \geq 4$ , $y - x \leq 2$ , $y \geq 1$ , $x \geq 0$ , $y \geq 0$ . Does $Z$ have a minimum?

Try it first, then show the solutionSolution

Corners: $x + y = 4$ with $y = x + 2$ : $(1, 3)$ ; $x + y = 4$ with $y = 1$ : $(3, 1)$ . The region is unbounded (it opens to the upper right). $Z$ : $6 - 3 = 3$ and $2 - 9 = - 7$ . (1 mark)
Test $2 y - 3 x > 3$ : in the region $y \leq x + 2$ , so $2 y - 3 x \leq 4 - x$ , and $4 - x \leq y \leq x + 2$ forces $x \geq 1$ . Hence $2 y - 3 x \leq 3$ everywhere, the half-plane is empty of feasible points, and the maximum is 3 at $(1, 3)$ . (1½ marks)
No minimum: along $y = 1$ , $Z = 2 - 3 x$ decreases without limit. (½ mark)

Answer: Maximum 3 at $(1, 3)$ ; no minimum.

Example 6: Case study: shipping at least 60 tonnes · 4 marks · Case-basedHOTS / competency

Suppose a company must ship at least 60 tonnes of cement using trucks (10 tonnes each, Rs 3,000 a trip) and tempos (4 tonnes each, Rs 1,500 a trip). To cover all delivery routes, at least 9 vehicles must be sent.
(i) Write the LPP to minimise cost. (1 mark)
(ii) Find the corner points; is the region bounded? (1 mark)
(iii) Find the minimum cost, justifying it. (1 mark)
(iv) Is there a maximum cost? (1 mark)

Try it first, then show the solutionSolution

(i) Minimise $Z = 3000 x + 1500 y$ subject to $10 x + 4 y \geq 60$ (that is $5 x + 2 y \geq 30$ ), $x + y \geq 9$ , $x, y \geq 0$ . (1 mark)
(ii) Corners: $(0, 15)$ , $(9, 0)$ , and $5 x + 2 y = 30$ with $x + y = 9$ : $3 x = 12$ , $(4, 5)$ . The region is unbounded. (1 mark)
(iii) $Z$ : 22500, 27000, 19500. The open half-plane $3000 x + 1500 y < 19500$ (that is $2 x + y < 13$ ) has no common point with the region, so the minimum cost is Rs 19,500 with 4 trucks and 5 tempos. (1 mark)
(iv) No: sending more vehicles is always feasible and always costs more, so $Z$ has no maximum. (1 mark)

Answer: (ii) $(0, 15)$ , $(4, 5)$ , $(9, 0)$ ; unbounded (iii) Rs 19,500 with 4 trucks and 5 tempos (iv) no maximum.

Where students lose marks in Unbounded Feasible Regions

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

Mistake: Skipping the half-plane test
On an unbounded region the corner table gives only a candidate.
Wrong working
Corners give 19, 27, 40, so the minimum is 19. (No test.)
Right working
'The open half-plane $5 x + 3 y < 19$ has no point in common with the region, so the minimum is 19.'
Do this: Write the test sentence and draw the dotted line every time the region is unbounded.
Usually costs: 1 mark.
Mistake: Reporting the largest corner value as the maximum
On an unbounded region with positive coefficients, Z grows without limit.
Wrong working
Maximum $Z = 40$ at $(8, 0)$ .
Right working
$(20, 0)$ is feasible with $Z = 100 > 40$ , so Z has no maximum.
Do this: Test $Z > M$ before declaring a maximum.
Usually costs: The conclusion mark.
Mistake: Using the wrong side for the test
For a minimum you must check the side where Z is smaller.
Wrong working
Minimum 14: 'the half-plane $3 x + 2 y > 14$ meets the region, so there is no minimum'.
Right working
Check $3 x + 2 y < 14$ ; it has no common point, so 14 is the minimum.
Do this: Minimum: test $<$ . Maximum: test $>$ .
Usually costs: The final answer.
Mistake: Assuming 'unbounded' means 'no maximum'
With negative coefficients the maximum can still exist.
Wrong working
Region unbounded, so $Z = 2 y - 3 x$ has no maximum.
Right working
Maximum 3 at $(1, 3)$ , confirmed by the half-plane test.
Do this: Always test; never decide from the shape alone.
Usually costs: 1 to 2 marks.

Exam tips

Unbounded-region minimisation was the 3-mark LPP in 2025 (Q30), and 2025 Q27 asked a maximisation on an unbounded region.
Multiple minima on an unbounded region were asked in the 2024-25 sample paper (Q30) and as the 2026 board assertion-reason (Q20).
Shade the region open on one side and write 'the feasible region is unbounded' before the table.
For the half-plane test, draw $a x + b y = m$ as a dotted line and say clearly whether the open half-plane meets the region.
If an optimum does not exist, say so in words: 'Z has no maximum value.' That is a complete answer.

Practice questions on Unbounded Feasible Regions

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
The feasible region of an LPP is unbounded. Which statement is always true?
(a) Z has a maximum.
(b) Z has a minimum.
(c) If Z has an optimal value, it occurs at a corner point.
(d) Z has neither a maximum nor a minimum.
Hint
Which theorem does not need boundedness?
Answer
(c). The corner point theorem holds for any feasible region; existence is what boundedness guarantees.
Q2 · 2 marks · Very short answerBoard level
For the feasible region $x + y \geq 4$ , $x \geq 0$ , $y \geq 0$ , find the minimum of $Z = 3 x + 2 y$ . Does $Z$ have a maximum?
Hint
Two corners only.
Answer
Corners $(4, 0)$ and $(0, 4)$ give 12 and 8. Since $3 x + 2 y = 2 (x + y) + x \geq 8$ on the region, $3 x + 2 y < 8$ has no common point: minimum 8 at $(0, 4)$ . No maximum, as $Z$ grows without limit.
Q3 · 3 marks · Short answerHarder board level
Minimise $Z = 6 x + 4 y$ subject to $2 x + y \geq 8$ , $x + 2 y \geq 7$ , $x \geq 0$ , $y \geq 0$ .
Hint
The middle corner comes from $y = 8 - 2 x$ .
Answer
Corners $(0, 8)$ , $(3, 2)$ , $(7, 0)$ ; $Z$ : 32, 26, 42. The half-plane $6 x + 4 y < 26$ has no common point with the region, so the minimum is 26 at $(3, 2)$ .
Q4 · 3 marks · Short answerHarder board level
Maximise $Z = x - 2 y$ subject to $y \geq 1$ , $x - y \leq 3$ , $x \geq 0$ , $y \geq 0$ .
Hint
In the region $x \leq y + 3$ .
Answer
Corners $(0, 1)$ and $(4, 1)$ ; $Z$ : $- 2$ and 2. In the region $x - 2 y \leq (y + 3) - 2 y = 3 - y \leq 2$ , so $x - 2 y > 2$ has no common point: the maximum is 2 at $(4, 1)$ , although the region is unbounded.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The LPP 'maximise $Z = x + y$ subject to $x + y \geq 3$ , $x, y \geq 0$ ' has no maximum.
Reason (R): No objective function has a maximum on an unbounded feasible region.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Think of $Z = x - 2 y$ on an unbounded region.
Answer
(c). A is true since $Z$ grows without limit. R is false: for example $Z = x - 2 y$ has maximum 2 on the unbounded region $y \geq 1$ , $x - y \leq 3$ .
Q6 · 3 marks · Short answerHarder board level
Minimise $Z = 3 x + 3 y$ subject to $x + y \geq 5$ , $x + 3 y \geq 9$ , $x \geq 0$ , $y \geq 0$ . Show that the minimum occurs at more than two points.
Hint
$Z = 3 (x + y)$ .
Answer
Corners $(0, 5)$ , $(3, 2)$ , $(9, 0)$ ; $Z$ : 15, 15, 27. The half-plane $3 x + 3 y < 15$ , that is $x + y < 5$ , has no common point with the region, so the minimum is 15. It occurs at both $(0, 5)$ and $(3, 2)$ and so at every point of the segment joining them, such as $(1.5, 3.5)$ .
Q7 · 3 marks · Short answerHOTS / competency
For which real values of $p$ does $Z = p x + y$ have a minimum on the region $x + y \geq 3$ , $x \geq 0$ , $y \geq 0$ ? Find the minimum in terms of $p$ .
Hint
Move along the ray $y = 0$ , $x \to \infty$ .
Answer
Along $y = 0$ , $Z = p x$ , which falls without limit if $p < 0$ , so no minimum then. For $p \geq 0$ the corners $(3, 0)$ and $(0, 3)$ give $3 p$ and 3, and $Z$ does not decrease along any unbounded direction, so the minimum is $min (3 p, 3)$ : $3 p$ when $0 \leq p \leq 1$ , and 3 when $p \geq 1$ .

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.On an unbounded region, how do you confirm a minimum $m$ ?Show answer
Check that the open half-plane $a x + b y < m$ has no point in common with the region.
2.On an unbounded region, how do you confirm a maximum $M$ ?Show answer
Check that the open half-plane $a x + b y > M$ has no point in common with the region.
3.Does every objective function have a minimum on an unbounded region?Show answer
No; the optimum may or may not exist, so the test is needed.
4.Region $x + y \geq 2$ , $x, y \geq 0$ : maximum of $Z = x + y$ ?Show answer
None; $Z$ grows without limit.
5.Why is the test line drawn dotted?Show answer
The half-plane is open (strict inequality), so the line itself is not included.
6.Can an unbounded region give a maximum?Show answer
Yes, for example when some coefficients are negative, such as $Z = x - 2 y$ on $y \geq 1$ , $x - y \leq 3$ .

Unbounded Feasible Regions: frequently asked questions

What is an unbounded feasible region?

It is a feasible region that extends without limit in some direction, so it cannot be enclosed in a circle. It usually comes from 'at least' (≥) constraints.

How do I check whether the minimum exists on an unbounded region?

Find the smallest corner value m. Draw the line ax + by = m. If the open half-plane ax + by < m has no point in common with the feasible region, m is the minimum; otherwise there is no minimum.

Does an unbounded region always have a minimum?

Not always. With positive coefficients and a region in the first quadrant, a minimum usually exists, but with negative coefficients Z may decrease without limit. The half-plane test decides.

Which NCERT section covers unbounded regions?

Section 12.2.2 (Graphical method of solving linear programming problems) in Chapter 12 of NCERT Mathematics Part II for Class 12, which states the theorems and the open half-plane test.

Is an empty feasible region the same as an unbounded one?

No. An empty region has no feasible point at all, so the LPP has no solution. An unbounded region has infinitely many feasible points; only the optimum may fail to exist.