Linear Programming Quick Lesson: Class 12 Maths Chapter 12 (NCERT)

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The chapter in one minute

What the chapter is about, and what the exam asks.

Linear programming is about the best choice under limits. You have two amounts, $x$ and $y$ ⁠, a few rules that limit them, and one quantity $Z$ to make as big or as small as possible.

Each rule is a straight line on a graph, and it keeps one side of that line. The part kept by every rule is the feasible region.

The best value of $Z$ ⁠, when there is one, is always at a corner of the region. So you find the corners, put each one into $Z$ ⁠, and pick the largest or the smallest.

The exam gives you the rules and asks you to solve the problem with a graph, or gives you the graph and asks for the rules. Most questions are one of seven kinds, and each kind has a fixed recipe.

The book also shows how to write the rules from a story (the furniture dealer). That step is in the book but not examined: CBSE's syllabus no longer lists it.

7question types

10rule cards

1table that gives the answer:

Z

at every corner

5things your graph must show

Formulas

The rules you must know

There are few formulas in this chapter. These cards are the rules every answer uses.

The objective

Z = ax + by

The profit or cost to make as big or as small as possible. $a$ and $b$ are numbers.

Never negative

x \ge 0,\ y \ge 0

Almost every problem has these. They keep the region in the top-right quarter of the graph.

Where a line meets the axes

\begin{aligned} &ax + by = c \text{ meets} \\ &\text{the axes at } \left(\tfrac{c}{a}, 0\right) \\ &\text{and } \left(0, \tfrac{c}{b}\right) \end{aligned}

Put $y = 0$ for the first point and $x = 0$ for the second. Join the two points. This needs $a$ ⁠, $b$ and $c$ all non-zero; for a line like $x = 3$ or $x - 2y = 0$ ⁠, find another point.

A line from its axis points

\frac{x}{p} + \frac{y}{q} = 1

The line through $(p, 0)$ and $(0, q)$ ⁠. Clear the fractions to get $ax + by = c$ ⁠.

The test point

\begin{aligned} &(0, 0) \text{ in } ax + by \le c \\ &\text{gives } 0 \le c \end{aligned}

True: shade the side with the origin. False: shade the other side. Do the same for $\ge$ ⁠. If $c = 0$ ⁠, the origin is on the line: test another point.

A corner point

\begin{aligned} a_1x + b_1y &= c_1 \\ a_2x + b_2y &= c_2 \end{aligned}

A corner is a point of the region where two boundary lines meet. Solve their two equations together, then check the point obeys every rule.

Bounded region

\begin{aligned} M &= \text{largest corner } Z \\ m &= \text{smallest corner } Z \end{aligned}

Bounded means you can draw a circle round it. Then $M$ is the maximum and $m$ the minimum.

Unbounded: is M the maximum

ax + by \gt M

If this open half-plane has no point in common with the region, $M$ is the maximum. If it has any, there is no maximum. Open means the line itself is left out.

Unbounded: is m the minimum

ax + by \lt m

If this open half-plane has no point in common with the region, $m$ is the minimum. If it has any, there is no minimum.

A tie

Z(P) = Z(Q) = M

If two neighbouring corners $P$ and $Q$ both give the best value, so does every point on the segment $PQ$ ⁠.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one question from the book worked the exam way. In types 1 to 6 you draw a neat graph; in type 7 the graph is given.

1. Find the maximum: lines, shading, corners, table

Spot it: "Maximise $Z$ " with " $\le$ " rules that have only plus signs, like $x + 2y \le 10$ ⁠. The region is then a closed shape with a corner at the origin.

Draw each line through the two points where it meets the axes.
Test $(0, 0)$ in each rule to choose its side. Shade where all the kept sides overlap.
Find every corner. Where two lines cross away from the axes, solve their equations together.
Put each corner into $Z$ in a table. In a bounded region the largest value is the maximum.

Exercise 12.1 Q5. Maximise $Z = 3x + 2y$ subject to $x + 2y \le 10$ ⁠, $3x + y \le 15$ ⁠, $x, y \ge 0$ ⁠.

Line $x + 2y = 10$ meets the axes at $(10, 0)$ and $(0, 5)$ ⁠. Line $3x + y = 15$ meets them at $(5, 0)$ and $(0, 15)$ ⁠.

Test $(0, 0)$ ⁠: $0 \le 10$ and $0 \le 15$ are both true, so shade the origin's side of both lines.

The lines cross where $y = 15 - 3x$ and $x + 2(15 - 3x) = 10$ ⁠. So $x = 4$ and $y = 3$ ⁠.

\begin{array}{c|c} \text{Corner} & Z = 3x + 2y \\ \hline (0, 0) & 0 \\ (5, 0) & 15 \\ (4, 3) & 18 \\ (0, 5) & 10 \end{array}

The region is bounded, so the largest value in the table is the maximum.

∴ Maximum $Z = 18$ at $(4, 3)$

See a full exam answer · Go deeper: see it visually · Basics: a corner solves two equations

In past papers (2): click to see

Board 2026 Q30 · Sample 2024-25 Q11

2. A " $\ge$ " rule: shade away from the origin

Spot it: a rule like $x + 2y \ge 10$ ⁠, often with "Minimise $Z$ ". The region no longer touches the origin.

Draw the line from its two axis points, as before.
Test $(0, 0)$ ⁠. Here $0 \ge 10$ is false, so shade the side away from the origin.
Find the corners and make the table. In a bounded region the smallest value is the minimum.

Example 2. Minimise $Z = 200x + 500y$ subject to $x + 2y \ge 10$ ⁠, $3x + 4y \le 24$ ⁠, $x, y \ge 0$ ⁠.

Line $x + 2y = 10$ meets the axes at $(10, 0)$ and $(0, 5)$ ⁠. Test $(0, 0)$ ⁠: $0 \ge 10$ is false, so shade the side away from the origin.

Line $3x + 4y = 24$ meets the axes at $(8, 0)$ and $(0, 6)$ ⁠. Test $(0, 0)$ ⁠: $0 \le 24$ is true, so shade the origin's side.

The two lines meet at $(4, 3)$ ⁠. The region is the triangle with corners $(0, 5)$ ⁠, $(4, 3)$ and $(0, 6)$ ⁠, so it is bounded.

\begin{array}{c|c} \text{Corner} & Z = 200x + 500y \\ \hline (0, 5) & 2500 \\ (4, 3) & 2300 \\ (0, 6) & 3000 \end{array}

∴ Minimum $Z = 2300$ at $(4, 3)$

See a full exam answer · Go deeper: see it visually · Basics: which side of the line

In past papers (7): click to see

Board 2024 Q17 · Board 2024 Q30 · Board 2023 Q29 · Sample 2026-27 Q31(main question) · Sample 2025-26 Q30(main question) · Sample 2022-23 Q30 · Sample 2020-21 Q38(a)

3. Both maximum and minimum, and a tie along an edge

Spot it: "Minimise and maximise $Z$ ". Often one rule, like $x - 2y \ge 0$ ⁠, is a line through the origin.

For a line through the origin, test another point, such as $(10, 0)$ ⁠.
Find the corners and make one table. The smallest value is the minimum and the largest is the maximum.
If two corners share the best value, every point on the segment joining them gives it too. Say so.

Exercise 12.1 Q7. Minimise and maximise $Z = 5x + 10y$ subject to $x + 2y \le 120$ ⁠, $x + y \ge 60$ ⁠, $x - 2y \ge 0$ ⁠, $x, y \ge 0$ ⁠.

The line $x - 2y = 0$ goes through the origin, so test $(10, 0)$ ⁠: $10 \ge 0$ is true. Shade the side with $(10, 0)$ ⁠.

The corners: $(60, 0)$ and $(120, 0)$ on the $x$ -axis; $(60, 30)$ ⁠, where $x = 2y$ meets $x + 2y = 120$ ⁠; $(40, 20)$ ⁠, where $x = 2y$ meets $x + y = 60$ ⁠.

\begin{array}{c|c} \text{Corner} & Z = 5x + 10y \\ \hline (60, 0) & 300 \\ (120, 0) & 600 \\ (60, 30) & 600 \\ (40, 20) & 400 \end{array}

The region is bounded. The smallest value is 300, and the largest, 600, comes at two corners.

∴ Minimum $Z = 300$ at $(60, 0)$ ⁠. Maximum $Z = 600$ at every point of the segment joining $(120, 0)$ and $(60, 30)$

See a full exam answer · Go deeper: see it visually · Basics: lines of equal Z

In past papers (13): click to see

Board 2026 Q17 · Board 2025 Q14 · Board 2023 Q17 · Sample 2026-27 Q25(visually impaired version) · Sample 2026-27 Q31(visually impaired version) · Sample 2025-26 Q16(visually impaired version) · Sample 2025-26 Q17(main question) · Sample 2025-26 Q30(visually impaired version) · Sample 2024-25 Q30 · Sample 2023-24 Q7 · Sample 2023-24 Q30(a) · Sample 2022-23 Q11 · Sample 2020-21 Q38(b)

4. An unbounded region: check the minimum with a half-plane

Spot it: all the rules are " $\ge$ ", so the shaded region goes on for ever. The question asks for a minimum.

Find the corners and make the table, as before. Call the smallest value $m$ ⁠.
Draw the line $Z = m$ dotted. The open half-plane $Z \lt m$ is the side of it where $Z$ is smaller.
If it has no point in common with the region, $m$ is the minimum. If it has any, there is no minimum.

Exercise 12.1 Q4. Minimise $Z = 3x + 5y$ such that $x + 3y \ge 3$ ⁠, $x + y \ge 2$ ⁠, $x, y \ge 0$ ⁠.

Line $x + 3y = 3$ meets the axes at $(3, 0)$ and $(0, 1)$ ⁠, and line $x + y = 2$ at $(2, 0)$ and $(0, 2)$ ⁠. Test $(0, 0)$ ⁠: $0 \ge 3$ and $0 \ge 2$ are false, so shade away from the origin. The region is unbounded.

The lines cross where $2y = 1$ (subtract them). So $y = \frac{1}{2}$ and $x = \frac{3}{2}$ ⁠.

\begin{array}{c|c} \text{Corner} & Z = 3x + 5y \\ \hline (3, 0) & 9 \\ \left(\frac{3}{2}, \frac{1}{2}\right) & 7 \\ (0, 2) & 10 \end{array}

The smallest value is 7. The open half-plane $3x + 5y \lt 7$ has no point in common with the region, so 7 is the minimum.

∴ Minimum $Z = 7$ at $\left(\frac{3}{2}, \frac{1}{2}\right)$

See a full exam answer · Go deeper: see it visually · Basics: lines of equal Z

In past papers (3): click to see

Sample 2024-25 Q16 · Sample 2024-25 Q30 · Sample 2022-23 Q11

5. An unbounded region with no maximum

Spot it: the region goes on for ever, and $Z$ keeps growing as you move out into it.

Make the table and call the largest value $M$ ⁠.
Look at the open half-plane $Z \gt M$ ⁠. Find one point in the region on that side.
If you find one, write " $Z$ has no maximum value". The corner value is not the answer.

Exercise 12.1 Q9. Maximise $Z = -x + 2y$ subject to $x \ge 3$ ⁠, $x + y \ge 5$ ⁠, $x + 2y \ge 6$ ⁠, $y \ge 0$ ⁠.

The region goes on for ever, up and to the right: it is unbounded.

The corners: $(6, 0)$ on the $x$ -axis; $(4, 1)$ ⁠, where $x + 2y = 6$ meets $x + y = 5$ ⁠; $(3, 2)$ ⁠, where $x + y = 5$ meets $x = 3$ ⁠.

\begin{array}{c|c} \text{Corner} & Z = -x + 2y \\ \hline (6, 0) & -6 \\ (4, 1) & -2 \\ (3, 2) & 1 \end{array}

The largest value is 1. But the open half-plane $-x + 2y \gt 1$ has points in the region, such as $(3, 5)$ ⁠, where $Z = 7$ ⁠.

∴ $Z$ has no maximum value

See a full exam answer · Go deeper: see it visually · Basics: bounded or not

In past papers (2): click to see

Board 2025 Q27 · Sample 2023-24 Q30(b)

6. No feasible region: the rules disagree

Spot it: the shaded sides of two rules never overlap, so no point obeys every rule.

Rewrite each rule so it starts with $y$ ⁠.
Show that no point can obey both, on the graph or in one line of algebra.
Write "no feasible region", so $Z$ has no maximum or minimum value.

Exercise 12.1 Q10. Maximise $Z = x + y$ subject to $x - y \le -1$ ⁠, $-x + y \le 0$ ⁠, $x, y \ge 0$ ⁠.

Rewrite: $x - y \le -1$ means $y \ge x + 1$ ⁠, and $-x + y \le 0$ means $y \le x$ ⁠.

Both together would need $x + 1 \le x$ ⁠, which is never true. The two shaded parts never overlap.

∴ There is no feasible region, so $Z$ has no maximum value

See a full exam answer · Book Example 5 · Go deeper: see it visually · Basics: where the shadings overlap

7. Read the rules from a graph

Spot it: the graph of the region is given, and the question asks for its constraints, or for the rule that is missing.

For each slanted edge, read where its line meets the axes, say $(p, 0)$ and $(0, q)$ ⁠. Write $\frac{x}{p} + \frac{y}{q} = 1$ ⁠, then clear the fractions. An upright or flat edge is just $x = p$ or $y = q$ ⁠.
Put a point of the region into the left side. Less than the right side gives " $\le$ "; more gives " $\ge$ ".
If the region stays in the top-right quarter, add $x \ge 0$ and $y \ge 0$ ⁠.

Section 12.2, Fig 12.1, read the other way round. The shaded region OABC is closed in by the axes and two lines. One line meets the axes at $(20, 0)$ and $(0, 100)$ ⁠, the other at $(60, 0)$ and $(0, 60)$ ⁠. The origin is in the region. Write its constraints.

\begin{aligned} \frac{x}{20} + \frac{y}{100} &= 1 \\ 5x + y &= 100 \end{aligned}

\begin{aligned} \frac{x}{60} + \frac{y}{60} &= 1 \\ x + y &= 60 \end{aligned}

Put in $(0, 0)$ ⁠: $0 \lt 100$ and $0 \lt 60$ ⁠. The origin is in the region, so both rules are " $\le$ ".

The region stays in the top-right quarter, so add $x \ge 0$ and $y \ge 0$ ⁠.

∴ $5x + y \le 100$ ⁠, $x + y \le 60$ ⁠, $x \ge 0$ ⁠, $y \ge 0$

See the furniture-dealer problem solved in full · Go deeper: see it visually · Basics: where a line meets the axes

In past papers (3): click to see

Sample 2026-27 Q25(main question) · Sample 2025-26 Q16(main question) · Sample 2023-24 Q11

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Your graph must show five things: (1) both axes with a scale, (2) each line with its equation written on it, (3) the points where each line meets the axes, (4) the feasible region shaded, (5) every corner with its coordinates.
To draw a line, put $x = 0$ ⁠, then $y = 0$ ⁠. For $3x + 4y = 24$ that gives $(0, 6)$ and $(8, 0)$ ⁠.
Choose a scale that fits the largest intercept. For numbers like 120, let one square be 10 or 20.
With a positive number on the right, " $\le$ " keeps the origin's side and " $\ge$ " keeps the other side.
A rule like $x \le y$ has the line $x - y = 0$ ⁠, through the origin. Test a point such as $(1, 0)$ instead.
Check every corner in every rule before you put it in the table. A crossing point that breaks a rule is not a corner.
Always make the table, even when one corner looks best. The table is the working the examiner reads.
Two corners with the same best value: the answer is every point on the segment joining them.
Told that the maximum comes at two corners, such as $(4, 0)$ and $(0, 3)$ ⁠? Then $Z$ is equal there: $4a = 3b$ for $Z = ax + by$ ⁠.
Unbounded region: never stop at the table. Draw the half-plane's line dotted on your graph, and do the open half-plane check for each answer you give.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Shading the wrong side → test $(0, 0)$ in the rule, and shade the side where it is true.
Testing $(0, 0)$ on a line through the origin → it lies on the line and tells you nothing; test $(1, 0)$ ⁠.
Using a crossing point that is outside the region, like $(0, 60)$ in Exercise 12.1 Q7 → check each corner in every rule.
Reading a corner such as $\left(\frac{20}{19}, \frac{45}{19}\right)$ in Exercise 12.1 Q3 off the graph → solve the two equations together.
Giving the smallest corner value of an unbounded region as the minimum at once → do the open half-plane check first.
Leaving out $x \ge 0$ and $y \ge 0$ when you list the rules → write them every time.
Giving one point when two corners tie → say every point on the segment joining them.

Marks

How the examiner gives marks

From CBSE's own papers and marking rules. Write these lines and the marks follow.

Linear programming carries 5 of the 80 marks in the theory paper.
CBSE asks for neat figures. Your graph, with the region shaded and the corners marked, is part of the answer. In CBSE's sample marking schemes the graph earns the biggest share of a "solve graphically" question.
Examiners mark by steps. Keep each one on its own line: the graph, the corner points, the table of $Z$ ⁠, and the answer.
Finish with a conclusion in words, such as "Maximum $Z = 18$ at $(4, 3)$ ". For an unbounded region, write the half-plane check as well.
A slip is penalised only once, so keep going. An MCQ earns 1 mark for the right option only.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 12.1 Q1 one line and the axes: a triangle with three corners
Example 1 the maximum of a bounded region; then read its rules back from the graph
Exercise 12.1 Q2 a minimum, with a minus sign in $Z$
Section 12.2 the furniture dealer: solve it with a corner table
Exercise 12.1 Q3 a corner with fractions: solve the two lines together
Example 3 minimum and maximum, with a tie at the maximum
Exercise 12.1 Q8 a " $\ge$ " rule and a line through the origin
Exercise 12.1 Q6 unbounded, with the minimum along a whole edge
Example 4 unbounded, and the smallest corner value is not a minimum
Example 5 no feasible region

Seven kinds of question, one method. Draw the lines, shade the region, find the corners and make the table. Work through the practice list and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Linear programming, the quick way

The chapter in one minute

The rules you must know

The question types and how to crack each one

1. Find the maximum: lines, shading, corners, table

2. A " $\ge$ " rule: shade away from the origin

3. Both maximum and minimum, and a tie along an edge

4. An unbounded region: check the minimum with a half-plane

5. An unbounded region with no maximum

6. No feasible region: the rules disagree

7. Read the rules from a graph

Shortcuts and quick tips

Mistakes that cost marks

How the examiner gives marks

Practise these first

You can do this

Linear programming, the quick way

The chapter in one minute

The rules you must know

The question types and how to crack each one

1. Find the maximum: lines, shading, corners, table

2. A "≥\ge≥" rule: shade away from the origin

3. Both maximum and minimum, and a tie along an edge

4. An unbounded region: check the minimum with a half-plane

5. An unbounded region with no maximum

6. No feasible region: the rules disagree

7. Read the rules from a graph

Shortcuts and quick tips

Mistakes that cost marks

How the examiner gives marks

Practise these first

You can do this

2. A " $\ge$ " rule: shade away from the origin