What happens if the constraints contradict each other?

Then the feasible region is empty and the problem has no solution. Question 10 of Exercise 12.1 is exactly this case: x - y = x + 1 while -x + y <= 0 gives y <= x, and no point can satisfy both. Stating that the feasible region is empty is the complete and correct answer.

Chapter 12 of 13

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CBSE Class 12 Mathematics★ Unit V · Linear Programming, CBSE states no chapter-wise split

Linear Programming

Formulating and solving optimisation problems in two variables by the graphical Corner Point Method, including the bounded and unbounded cases. The chapter has been cut to a single exercise of 10 questions with no Miscellaneous Exercise — the only chapter in the book without one — and every applied word problem has been removed. CBSE Class 12 Mathematics Chapter 12.

⏱ 11 min readEasy difficulty✓ Free · NCERT-aligned

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Key points, glossary and the questions most likely to show up.

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By the end of this chapter you'll be able to…

1Identify the objective function, decision variables, constraints and feasible region of a linear programming problem
2Formulate a real-world optimisation problem as an LPP in two variables with non-negativity constraints
3Draw the feasible region for a system of linear inequalities and locate every corner point
4Apply the Corner Point Method to find the maximum and minimum of a linear objective
5Decide whether a feasible region is bounded or unbounded, and apply the correct extra test in the unbounded case
6Recognise when an optimum is attained along a whole edge, and when a feasible region is empty

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Why this chapter matters

This is the highest marks-per-hour chapter in the paper: one method, ten questions, and no Miscellaneous Exercise. Almost every question is answered by the same four-step Corner Point Method, and the only real difficulty is knowing when an unbounded region invalidates the answer the corners suggest.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Linear inequalities in two variables (Class 11): shading the correct half plane for each inequality is the entire graphical step of this chapter

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Straight lines (Class 11): finding intercepts and solving two linear equations simultaneously is how every corner point is located

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Coordinate geometry (Class 9-10): plotting points and reading a region off a graph accurately

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Simultaneous linear equations (Class 10): each vertex is the solution of the two boundary lines meeting there

Linear Programming

1. Check this before you revise anything

The chapter teaches formulation for six pages, then never tests it. Section 12.2.1 is titled "Mathematical formulation of the problem," and the whole chapter is motivated by a furniture-dealer word problem: a dealer with Rs 50,000 and storage for 60 pieces, buying tables at Rs 2500 and chairs at Rs 500, wanting to maximise profit.

Yet not one of the ten exercise questions asks you to formulate anything. Every question hands you the objective function and the constraints already written out — "Maximise $Z = 3 x + 4 y$ subject to $x + y \leq 4$ ". The skill the chapter spends its first six pages building is never examined.

Section 12.3 has been removed entirely. Older editions carried "Different Types of Linear Programming Problems" — manufacturing problems, diet problems, transportation problems — each with worked word problems, plus Exercise 12.2 and a Miscellaneous Exercise built on them. A full-text search of the current 12-page chapter finds no section 12.3, no "diet problem," and no "manufacturing problem." The two occurrences of "12.3" in the text are both figure labels.

This matches the syllabus line, which covers only "graphical method of solution for problems in two variables... feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints)."

This is the only chapter in the book with no Miscellaneous Exercise. All twelve other chapters have one. Here there is a single Exercise 12.1 with 10 questions, and then the chapter ends.

Exercise	Topic	Questions
12.1	Solve the following LPPs graphically	10

At 10 questions it is the second-smallest chapter in the book, ahead only of Chapter 8 with 9.

The Summary has been cut to a single sentence. The whole end-of-chapter summary now reads as one bullet defining a linear programming problem. Feasible region, optimal solution, the Corner Point Method, and the bounded/unbounded distinction — all taught in the body, all examined — appear nowhere in it. Do not revise from that summary.

Textbook section	Topic
12.1	Introduction — optimisation problems and where they arise
12.2	Linear Programming Problem and its Mathematical Formulation
12.2.1	Mathematical formulation of the problem
12.2.2	Graphical method of solving linear programming problems

2. What a Linear Programming Problem Is (Textbook 12.1 to 12.2)

Problems that seek to maximise profit or minimise cost, subject to limits on money, space, time or materials, are called optimisation problems. Linear programming is the special case where everything in sight is linear.

The vocabulary, all of which is examinable:

Term	Meaning
Objective function	The linear expression $Z = a x + b y$ to be maximised or minimised
Decision variables	The quantities $x$ and $y$ being chosen; always non-negative
Constraints	The linear inequalities restricting $x$ and $y$
Non-negative constraints	$x \geq 0$ , $y \geq 0$ , stated separately because quantities cannot be negative
Feasible region	The set of all points satisfying every constraint at once
Feasible solution	Any point inside or on the boundary of the feasible region
Infeasible solution	Any point outside it
Optimal solution	The feasible point giving the best value of $Z$

Why "linear". Both the objective function and every constraint are of the first degree — no $x^{2}$ , no $x y$ , no $x$ . That is what makes the feasible region a polygon and lets the whole problem be solved by inspecting corners.

The furniture dealer, formulated. The dealer's problem becomes: maximise $Z = 250 x + 75 y$ (the profit) subject to $2500 x + 500 y \leq 50000$ (the money limit), $x + y \leq 60$ (the storage limit), and $x, y \geq 0$ . The first inequality simplifies to $5 x + y \leq 100$ .

3. Mathematical Formulation (Textbook 12.2.1)

Although no exercise question tests it, the formulation procedure is the reason the chapter exists, and it is examinable through the syllabus even where this exercise does not reach.

Identify the decision variables and say plainly what each represents, including units.
Write the objective function as a linear expression in those variables, and state whether it is to be maximised or minimised.
Translate each restriction into a linear inequality.
Add the non-negativity constraints $x \geq 0$ , $y \geq 0$ .

The commonest formulation errors are reversing an inequality — a resource limit is $\leq$ , a requirement is $\geq$ — and forgetting the non-negativity constraints, which are what confine the region to the first quadrant.

4. The Graphical Method and the Corner Point Method (Textbook 12.2.2)

Why corners matter. For a linear objective on a polygonal region, the optimum always occurs at a vertex. Moving along any edge changes $Z$ steadily in one direction, so it can never peak in the middle of an edge — it must peak at an end.

The one exception is when an edge happens to lie parallel to the objective line, in which case $Z$ is constant along it and every point of that edge is optimal. That case is the subject of questions 6, 7 and 8.

The Corner Point Method, exactly as the book states it:

Find the feasible region and determine its corner points (vertices), either by inspection or by solving the two equations of the lines meeting there.
Evaluate $Z = a x + b y$ at each corner point. Let $M$ and $m$ be the largest and smallest of these values.
If the feasible region is bounded, $M$ and $m$ are the maximum and minimum of $Z$ .
If the feasible region is unbounded, $M$ and $m$ are only candidates, and each needs a further test described in the next section.

Finding corner points reliably. Each vertex is the intersection of two boundary lines, so solve those two equations simultaneously. Corners on the axes come from setting $x = 0$ or $y = 0$ . Missing a vertex is the most frequent cause of a wrong answer, so count them against the shape of the shaded region before evaluating.

Worked, mirroring the textbook's own Example 1. Maximise $Z = 4 x + y$ subject to $x + y \leq 50$ , $3 x + y \leq 90$ , $x \geq 0$ , $y \geq 0$ .

The feasible region $O A B C$ is bounded, with corners $O (0, 0)$ , $A (30, 0)$ , $B (20, 30)$ and $C (0, 50)$ .

Corner point	$Z = 4 x + y$
$(0, 0)$	$0$
$(30, 0)$	$120$
$(20, 30)$	$110$
$(0, 50)$	$50$

Because the region is bounded, the largest entry is the answer: maximum $Z = 120$ at $(30, 0)$ .

Note that the optimum is not at the intersection of the two slanted lines. Assuming it always is, without evaluating every corner, is a reliable way to lose the question.

5. Bounded and Unbounded Feasible Regions

A feasible region is bounded if it can be enclosed within some circle, and unbounded if it extends indefinitely. The distinction decides whether the corner values can be trusted.

Bounded regions. Both a maximum and a minimum always exist, and both are found among the corner values. Nothing further is required.

Unbounded regions. The largest corner value $M$ might not be a maximum at all, because the region may run away in a direction where $Z$ keeps growing. The book's test:

$M$ is the maximum of $Z$ only if the open half plane $a x + b y > M$ has no point in common with the feasible region. Otherwise $Z$ has no maximum value.
$m$ is the minimum of $Z$ only if the open half plane $a x + b y < m$ has no point in common with the feasible region. Otherwise $Z$ has no minimum value.

How to apply the test in practice. Sketch the line $a x + b y = M$ and look at the side away from the origin. If any part of the shaded feasible region lies on that side, there is no maximum. Alternatively, find one explicit feasible point that beats $M$ — a single counterexample settles it.

Question 9 is built entirely on this. Maximising $Z = - x + 2 y$ subject to $x \geq 3$ , $x + y \geq 5$ , $x + 2 y \geq 6$ , $y \geq 0$ gives corner values $- 6$ , $- 2$ and $1$ , so $M = 1$ . But the point $(3, 100)$ is feasible and gives $Z = 197$ . Hence $Z$ has no maximum, and answering " $1$ " is exactly the error the question is designed to catch.

Question 10 is a different failure. Its constraints $x - y \leq - 1$ and $- x + y \leq 0$ rearrange to $y \geq x + 1$ and $y \leq x$ , which together demand $x + 1 \leq x$ . No point can satisfy both, so the feasible region is empty and there is nothing to optimise.

When two corners tie. The textbook states that if two corner points produce the same optimal value, every point on the segment joining them gives that same value. Questions 6, 7 and 8 each rely on this, so the correct answer names the whole segment, not one endpoint.

Summary

A linear programming problem maximises or minimises a linear objective $Z = a x + b y$ subject to linear constraints, with the decision variables non-negative.
Formulation means naming the variables, writing the objective, translating each restriction into an inequality, and adding $x \geq 0$ , $y \geq 0$ — resource limits give $\leq$ , requirements give $\geq$ .
The feasible region is the set of points satisfying every constraint; points outside it are infeasible solutions.
The optimum of a linear objective always lies at a corner point, because $Z$ changes steadily along every edge.
Corner Point Method: find the vertices, evaluate $Z$ at each, and take the largest and smallest values as $M$ and $m$ .
For a bounded region, $M$ and $m$ are the maximum and minimum outright.
For an unbounded region, $M$ is the maximum only if $a x + b y > M$ misses the feasible region entirely; otherwise there is no maximum. The mirror test applies to $m$ .
A single feasible point beating $M$ is enough to prove no maximum exists — this is what Question 9 turns on.
If two corner points give the same optimal value, every point on the segment joining them is optimal too.
A feasible region can be empty, in which case the problem has no solution at all.
Section 12.3 (manufacturing, diet and transportation problems), Exercise 12.2 and the Miscellaneous Exercise have all been removed from the current edition — this is the only chapter in the book with no Miscellaneous Exercise.
The chapter's own one-sentence Summary omits the feasible region, the Corner Point Method and the bounded/unbounded distinction; do not revise from it.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Objective function

Z = ax + by, to be maximised or minimised

Always linear — no x^2, xy or square roots, which is what makes the feasible region a polygon

Non-negativity constraints

x >= 0, y >= 0

Stated separately from the other constraints; they confine the region to the first quadrant and are routinely forgotten

Feasible region

The set of all points satisfying every constraint simultaneously

Points inside or on the boundary are feasible solutions; points outside are infeasible

Corner Point Method, step 1

Find the feasible region and determine its corner points by inspection or by solving the two intersecting line equations

Missing a vertex is the most frequent cause of a wrong answer — count them against the shape of the shaded region

Corner Point Method, step 2

Evaluate Z = ax + by at each corner point; let M be the largest value and m the smallest

Tabulate them; the optimum is often not at the intersection of the two slanted lines

Bounded feasible region

If the region is bounded, M is the maximum of Z and m is the minimum

A region is bounded if it can be enclosed inside some circle; no further test is needed

Unbounded region, maximum test

M is the maximum only if the open half plane ax + by > M has no point in common with the feasible region; otherwise Z has no maximum

One explicit feasible point beating M is enough to prove no maximum exists

Unbounded region, minimum test

m is the minimum only if the open half plane ax + by < m has no point in common with the feasible region; otherwise Z has no minimum

The mirror image of the maximum test; check the side of the line towards the origin

Optimum along an edge

If two corner points give the same optimal value, every point on the segment joining them gives that value too

The correct answer then names the whole segment, not one endpoint — questions 6, 7 and 8 all turn on this

Empty feasible region

If the constraints are mutually contradictory, no feasible point exists and the problem has no solution

Rearrange each constraint to isolate y and compare — e.g. y >= x+1 with y <= x is impossible

Why the optimum sits at a corner

A linear Z changes steadily along every edge, so it cannot peak in the middle of one

The exception is an edge parallel to the objective line, where Z is constant along it

Formulation procedure

Name the decision variables, write Z, translate each restriction into a linear inequality, then add x >= 0 and y >= 0

Resource limits give <=, minimum requirements give >= — reversing one is the commonest formulation error

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Applying the corner values directly when the feasible region is unbounded

✓ For an unbounded region the largest corner value M is only a candidate. Test whether ax + by > M meets the region; if it does, Z has no maximum at all.

WATCH OUT

✗ Assuming the optimum lies at the intersection of the two slanted constraint lines

✓ Evaluate Z at every corner including the axis points. In the textbook's own Example 1 the maximum is at (30,0), not at the intersection (20,30).

WATCH OUT

✗ Missing one or more corner points of the feasible region

✓ Each vertex is where two boundary lines meet, including the axes. Count the vertices against the shape of the shaded polygon before evaluating anything.

WATCH OUT

✗ Reporting a single point when two corners give the same optimal value

✓ When two corner points tie, every point on the segment joining them is optimal — state the whole segment.

WATCH OUT

✗ Reversing an inequality when formulating from a word problem

✓ A resource that is limited gives <=; a requirement that must be met gives >=. Read whether the phrase is 'at most' or 'at least'.

WATCH OUT

✗ Forgetting the non-negativity constraints x >= 0 and y >= 0

✓ They are separate constraints and they confine the region to the first quadrant; without them the feasible region is the wrong shape entirely.

WATCH OUT

✗ Shading the wrong side of a constraint line

✓ Substitute a test point (usually the origin) into the inequality. If it satisfies the inequality, shade the side containing it; if not, shade the other side.

WATCH OUT

✗ Assuming every LPP has a solution

✓ The feasible region can be empty when constraints contradict each other, and an unbounded region can leave Z with no maximum. Both outcomes are valid answers.

WATCH OUT

✗ Revising from the chapter's printed Summary

✓ The current edition's Summary is a single sentence and omits the feasible region, the Corner Point Method and the bounded/unbounded distinction — all of which are examined.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 12.1

Exercise 12.1

Solve linear programming problems graphically, including bounded and unbounded regions, ties along an edge, and one empty feasible region

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Linear Programming?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Objective function Z = ax + by is linear, and so is every constraint — that is what makes the feasible region a polygon
•Decision variables are always non-negative; x >= 0 and y >= 0 are separate constraints
•Feasible region: all points satisfying every constraint at once; feasible solutions lie inside or on its boundary
•The optimum of a linear objective always occurs at a corner point of the feasible region
•Corner Point Method: find the vertices, evaluate Z at each, take the largest M and smallest m
•Bounded region: M and m are the maximum and minimum outright, with no further check
•Unbounded region: M is the maximum only if ax + by > M misses the feasible region entirely
•Unbounded region: m is the minimum only if ax + by < m misses the feasible region entirely
•A single feasible point beating M proves that no maximum exists
•If two corner points tie at the optimum, every point on the joining segment is optimal
•A feasible region can be empty when constraints contradict one another, leaving no solution at all
•Resource limits translate to <= and minimum requirements to >= when formulating from a word problem
•Shade each half plane by testing the origin in the inequality
•Section 12.3, Exercise 12.2 and the Miscellaneous Exercise have been removed from the current edition
•This is the only chapter in the book with no Miscellaneous Exercise, and at 10 questions the second-smallest overall

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit V: Linear Programming, no chapter-wise split published by CBSE

Question type	Marks each	Typical count	What it tests
Corner Point Method and Feasible Region	5	1	Drawing the feasible region, locating every corner point, tabulating Z and identifying the optimum
Bounded and Unbounded Regions, and Formulation	3-5	1	Applying the unbounded-region test, recognising ties along an edge or an empty region, and formulating an LPP from a word problem

Prep strategy

Always state whether the feasible region is bounded or unbounded before quoting an answer — it decides whether the corner values can be trusted
Tabulate Z at every corner point, including the axis intercepts; never assume the optimum is at the intersection of the slanted lines
For unbounded regions, either sketch the line ax + by = M or produce one explicit feasible point that beats M
If two corners give the same optimal value, write the answer as the whole segment joining them
Do not revise from the chapter's printed Summary — it is one sentence and omits almost everything examined

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Production planning in manufacturing

Deciding how many units of two products to make given limited machine hours, labour and raw material is exactly the furniture-dealer problem that opens this chapter.

Diet and nutrition planning

Choosing quantities of two foods to meet minimum nutritional requirements at least cost is a classic minimisation LPP, described by the economist G. Stigler in 1945.

Transportation and logistics

The first linear programming problem ever formulated, by Kantorovich and Hitchcock in 1941, was a transportation problem — minimising the cost of shipping goods between locations.

Portfolio and budget allocation

Splitting a fixed budget between two investments to maximise return, subject to risk limits, is a two-variable LPP solved by the same Corner Point Method.

Crop planning in agriculture

Allocating limited land, water and fertiliser between two crops to maximise yield or profit reduces to the same graphical method.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Draw the graph on the answer sheet with a ruler and label every constraint line — the diagram itself carries marks

State explicitly in words whether the feasible region is bounded or unbounded before quoting any answer

Present the corner values in a table with two columns, corner point and value of Z, exactly as the textbook does

Solve simultaneously for each intersection rather than reading coordinates off the graph, which is rarely accurate enough

For an unbounded region, write out the half-plane test as a sentence and give the counterexample point if one exists

If two corners tie, write 'Z is optimal at every point on the segment joining A and B' rather than naming one point

Check that all corner points actually satisfy every constraint before evaluating — a point on one boundary can still violate another

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

The simplex method, suggested by G. B. Dantzig in 1947, solves linear programming problems in any number of variables by moving between adjacent vertices, and is what makes large real-world problems tractable

STRETCH

Duality theory pairs every maximisation problem with a minimisation problem whose optimal value is identical, giving a way to certify that a solution really is optimal

STRETCH

Integer linear programming, where the decision variables must be whole numbers, is dramatically harder than the continuous case and underlies scheduling and routing problems

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainUnbounded feasible regionTesting whether the largest corner value is genuinely a maximum

Maximise $Z = - x + 2 y$ subject to $x \geq 3$ , $x + y \geq 5$ , $x + 2 y \geq 6$ , $y \geq 0$ .

Stuck? Show the approach

Compute the corner values, then apply the unbounded-region test rather than reporting the largest one.

Show the full solution

The corner points are $(6, 0)$ , $(4, 1)$ and $(3, 2)$ , giving $Z = - 6$ , $- 2$ and $1$ , so $M = 1$ . Since every constraint is of $\geq$ type the region is unbounded, and the test asks whether $- x + 2 y > 1$ meets it. The point $(3, 100)$ satisfies all four constraints and gives $Z = 197$ . The half plane therefore intersects the feasible region.

Answer: Z has no maximum value

The trap

Reporting $Z = 1$ at $(3, 2)$ — the largest corner value — without running the unbounded test. This is the single most common error in the chapter.

JEE MainOptimum attained along an edgeRecognising a tie between two corner points

Minimise and Maximise $Z = 5 x + 10 y$ subject to $x + 2 y \leq 120$ , $x + y \geq 60$ , $x - 2 y \geq 0$ , $x, y \geq 0$ .

Stuck? Show the approach

Evaluate all four corners and read the table twice, watching for repeated values.

Show the full solution

The corners are $(60, 0)$ , $(120, 0)$ , $(60, 30)$ and $(40, 20)$ , giving $Z = 300$ , $600$ , $600$ and $400$ . The region is bounded, so both extremes exist. The minimum is $300$ at $(60, 0)$ . The maximum $600$ occurs at two distinct corners, so every point on the segment joining them is optimal.

Answer: Minimum 300 at (60,0); Maximum 600 along the segment joining (120,0) and (60,30)

The trap

Naming only $(120, 0)$ for the maximum. When two corners tie, the optimal set is the entire connecting segment.

JEE AdvancedInconsistent constraintsDetecting an empty feasible region before plotting

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

Stuck? Show the approach

Rearrange both constraints to isolate $y$ and compare them directly.

Show the full solution

The first gives $y \geq x + 1$ and the second gives $y \leq x$ . Together they require $x + 1 \leq y \leq x$ , hence $x + 1 \leq x$ , that is $1 \leq 0$ . This is impossible, so no point satisfies both. Graphically the two half planes lie on opposite sides of $y = x$ and never overlap.

Answer: No feasible region exists, so Z has no maximum

The trap

Plotting carelessly and inventing a small overlap region near the origin. Rearranging algebraically first settles the question with certainty.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

A region is bounded if it can be enclosed inside some circle. In practice, if every constraint is of <= type with positive coefficients, the region is closed off and bounded. If the constraints are mostly of >= type, the region usually extends indefinitely and is unbounded. The distinction matters because for an unbounded region the largest corner value may not be the maximum at all.

Because the region runs off to infinity, and Z may keep growing in that direction. The corner values only tell you the best value among the vertices. The test is: if the open half plane ax + by > M contains any feasible point, then Z has no maximum. Finding one explicit point that beats M is the quickest way to settle it.

Yes. If two corner points give the same value of Z, then every point on the line segment joining them gives that same value, so there are infinitely many optimal solutions. The textbook states this explicitly, and questions 6, 7 and 8 of Exercise 12.1 are all built on it. Your answer should name the segment, not a single point.

Section 12.2.1 is titled 'Mathematical formulation of the problem' and the whole chapter is motivated by the furniture-dealer example, but all ten exercise questions supply the objective and constraints ready-made. The applied word problems that used to test formulation lived in the removed section 12.3 and Exercise 12.2. Formulation remains in the syllabus, so it is still worth practising from the worked examples.

Correct. Every other chapter in the Class 12 Mathematics book has one; this chapter has only Exercise 12.1 with 10 questions and then ends with a one-sentence Summary and a Historical Note. Nothing is missing from your copy of the book.

Then the feasible region is empty and the problem has no solution. Question 10 of Exercise 12.1 is exactly this case: x - y <= -1 gives y >= x + 1 while -x + y <= 0 gives y <= x, and no point can satisfy both. Stating that the feasible region is empty is the complete and correct answer.

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