MathsCBSE Class 12 · Linear Programming

Feasible Region and Constraints in Linear Programming (Class 12)

Updated 15 October 2026 · Fact-checked

Each linear constraint ax + by ≤ c (or ≥ c) is a half-plane bounded by the line ax + by = c. Test a point such as the origin to choose the side. The feasible region is the part of the first quadrant that satisfies every constraint; its corner points are where the boundary lines meet.

The feasible region is where all half-planes overlap; its corners (0,0), (5,0), (3,2), (0,4) come from solving pairs of boundary lines.

Illustration drawn for QuestionBuddy.

Understand Feasible Region and Constraints

In a linear programming problem (LPP) the decision variables $x$ and $y$ are the quantities you choose. The constraints are linear inequalities such as $2 x + 3 y \leq 12$ that limit them, and the non-negativity constraints $x \geq 0$ , $y \geq 0$ say that you cannot make a negative number of anything.

One constraint = one half-plane. The line $a x + b y = c$ splits the plane into two halves. To decide which half satisfies $a x + b y \leq c$ , substitute a test point not on the line, usually the origin. For $2 x + 3 y \leq 12$ : $0 \leq 12$ is true, so the region is the side containing the origin. If the line passes through the origin (for example $x - 2 y \geq 0$ ), test another point such as $(1, 0)$ : $1 \geq 0$ is true, so the region is the side containing $(1, 0)$ .

Feasible region. The common region of all the constraints, including $x \geq 0$ and $y \geq 0$ , is the feasible region. Every point in it (boundary included) is a feasible solution; a point outside it is an infeasible solution. If the constraints have no common point, the feasible region is empty and the problem has no solution.

Corner points. A corner (vertex) of the feasible region is where two boundary lines meet. Find each corner by solving the two line equations simultaneously; do not read it off the graph by eye. These corners are what the next step, the corner point method, uses.

Bounded or unbounded. A feasible region is bounded if it can be enclosed in a circle; otherwise it is unbounded (it stretches to infinity in some direction). Regions made only of $\geq$ constraints in the first quadrant, like $x + y \geq 4$ , are unbounded.

Redundant constraints. A constraint that does not change the region (because the other constraints already force it) is redundant. For example, if $x + y \leq 5$ and $x, y \geq 0$ , then $2 x + y \leq 10$ , so the constraint $2 x + y \leq 12$ adds nothing.

Reverse questions. Boards also give a shaded region and ask for its constraints. Find the equation of each boundary line from two points on it, then use a point inside the region to choose $\leq$ or $\geq$ .

The 2026-27 syllabus limits problems to two variables and up to three non-trivial constraints (NCERT Sections 12.2.1 and 12.2.2).

Feasible Region and Constraints: formulas to remember

Half-plane of a constraint: $a x + b y \leq c or a x + b y \geq c, boundary a x + b y = c$; Draw the line, then test a point to pick the side.
Intercepts for drawing a line: $a x + b y = c : (\frac{c}{a}, 0) and (0, \frac{c}{b})$; For a line through the origin, use a second point such as $(b, - a)$ or one read from $y = m x$ .
Origin test: $a (0) + b (0) \leq c ⟺ 0 \leq c$; True: shade towards the origin. False: shade away from it.
Corner point: $a_{1} x + b_{1} y = c_{1}, a_{2} x + b_{2} y = c_{2} \Rightarrow (x, y)$; Solve simultaneously; check it satisfies the other constraints.
Line through two corners: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$; Used to write constraints from a shaded region.
Non-negativity: $x \geq 0, y \geq 0$; Always part of the constraints; keeps the region in the first quadrant.

How to solve Feasible Region and Constraints questions

The routine for drawing a feasible region (the first 1 to 2 marks of every graphical LPP).

1Rewrite each constraint with $x$ and $y$ on the left, for example $y \leq \frac{x}{2}$ as $x - 2 y \geq 0$ .
2For each constraint, draw the line $a x + b y = c$ using its two intercepts (or two convenient points).
3Test the origin (or another point if the line passes through the origin) and mark the side that satisfies the inequality.
4Shade the region common to all constraints inside the first quadrant ( $x \geq 0$ , $y \geq 0$ ).
5Find every corner point by solving the pair of boundary equations that meet there.
6State whether the region is bounded, unbounded or empty, and list the corner points.

Feasible Region and Constraints 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.

A budget as a constraint

Suppose you have Rs 600 for stationery. Notebooks cost Rs 40 and pens Rs 20, and you must buy at least 5 notebooks.

Write the constraints for $x$ notebooks and $y$ pens and find the corner points of the feasible region.

Try it first, then show the solutionSolution

Budget: $40 x + 20 y \leq 600$ , that is $2 x + y \leq 30$ .
At least 5 notebooks: $x \geq 5$ ; and $y \geq 0$ .
Corners: $x = 5$ meets $y = 0$ at $(5, 0)$ ; $x = 5$ meets $2 x + y = 30$ at $(5, 20)$ ; $2 x + y = 30$ meets $y = 0$ at $(15, 0)$ .

Answer: $2 x + y \leq 30$ , $x \geq 5$ , $y \geq 0$ ; corner points $(5, 0)$ , $(15, 0)$ and $(5, 20)$ .

Memory hook: Every 'at most' or 'at least' in a story becomes one line on the graph.

A ratio condition

Imagine a student who can study at most 8 hours a day, with $x$ hours of Maths and $y$ hours of Physics. Maths time must be at least Physics time, and Physics needs at least 2 hours.

Write the constraints and find the corner points of the feasible region.

Try it first, then show the solutionSolution

$x + y \leq 8$ , $x \geq y$ (that is $x - y \geq 0$ ), $y \geq 2$ .
$y = 2$ meets $x - y = 0$ at $(2, 2)$ and $x + y = 8$ at $(6, 2)$ .
$x - y = 0$ meets $x + y = 8$ at $(4, 4)$ .

Answer: A triangle with corners $(2, 2)$ , $(6, 2)$ and $(4, 4)$ .

Memory hook: 'At least as much as' becomes a line through the origin.

Tricks to remember Feasible Region and Constraints

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

Zero test, then shade
Put $(0, 0)$ into the inequality. True: shade towards the origin. False: shade away. Line through the origin: test $(1, 0)$ or $(0, 1)$ .
Use it when: Choosing the correct side of each line.
Less-than lies low
For a line with positive $x$ and $y$ coefficients and positive $c$ , $\leq$ is the side towards the origin and $\geq$ the side away from it.
Use it when: A quick check on shading.
Solve, never squint
Corners come from solving two equations, not from reading the graph.
Use it when: Getting exact corner points like $(\frac{8}{3}, \frac{10}{3})$ .
Circle test for boundedness
If you can draw a circle around the whole shaded region, it is bounded.
Use it when: Deciding bounded or unbounded.

Feasible Region and Constraints: solved examples

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

Example 1: Counting corner points · 1 mark · Very short answerBoard level

Find the number of corner points of the feasible region given by $x + y \leq 6$ , $x \leq 4$ , $y \leq 5$ , $x \geq 0$ , $y \geq 0$ .

Try it first, then show the solutionSolution

Corners: $(0, 0)$ , $(4, 0)$ , $x = 4$ with $x + y = 6$ gives $(4, 2)$ , $y = 5$ with $x + y = 6$ gives $(1, 5)$ , and $(0, 5)$ . (½ mark)
Each satisfies all the constraints, so there are 5 corner points. (½ mark)

Answer: 5 corner points: $(0, 0)$ , $(4, 0)$ , $(4, 2)$ , $(1, 5)$ , $(0, 5)$ .

Example 2: Shading a feasible region · 2 marks · Very short answerBoard level

Draw the feasible region for $2 x + 3 y \leq 12$ , $x \leq 3$ , $x \geq 0$ , $y \geq 0$ and write its corner points.

Try it first, then show the solutionSolution

$2 x + 3 y = 12$ passes through $(6, 0)$ and $(0, 4)$ ; origin test $0 \leq 12$ is true, so shade towards the origin. $x = 3$ is a vertical line; shade to its left. (1 mark)
Corners: $(0, 0)$ , $(3, 0)$ , $x = 3$ with $2 x + 3 y = 12$ gives $y = 2$ , so $(3, 2)$ , and $(0, 4)$ . The region is bounded. (1 mark)

Answer: Bounded region with corners $(0, 0)$ , $(3, 0)$ , $(3, 2)$ , $(0, 4)$ .

Example 3: Writing constraints from a region · 2 marks · Very short answerHarder board level

The feasible region of an LPP is the quadrilateral with corners $O (0, 0)$ , $A (5, 0)$ , $B (3, 4)$ and $C (0, 4)$ . Write the constraints that define it.

Try it first, then show the solutionSolution

Line $A B$ through $(5, 0)$ and $(3, 4)$ has slope $- 2$ : $y = - 2 (x - 5)$ , that is $2 x + y = 10$ . $O$ gives $0 \leq 10$ , so $2 x + y \leq 10$ . (1 mark)
Line $B C$ is $y = 4$ , with $O$ below it: $y \leq 4$ . Together with $x \geq 0$ , $y \geq 0$ . (1 mark)

Answer: $2 x + y \leq 10$ , $y \leq 4$ , $x \geq 0$ , $y \geq 0$ .

Example 4: Formulating constraints and testing plans · 3 marks · Short answerHarder board level

A home bakery makes $x$ cakes and $y$ batches of bread a day. A cake needs 2 hours of oven time and 1 hour of decorating; a bread batch needs 1 hour of oven time and 3 hours of decorating. At most 16 oven hours and 18 decorating hours are available. Write the constraints, find the corner points, and decide whether the plans $(5, 4)$ and $(7, 3)$ are feasible.

Try it first, then show the solutionSolution

Oven: $2 x + y \leq 16$ ; decorating: $x + 3 y \leq 18$ ; $x, y \geq 0$ . (1 mark)
Solving $2 x + y = 16$ and $x + 3 y = 18$ : $x = 18 - 3 y$ , so $36 - 6 y + y = 16$ , $y = 4$ , $x = 6$ . Corners: $(0, 0)$ , $(8, 0)$ , $(6, 4)$ , $(0, 6)$ . (1 mark)
$(5, 4)$ : $14 \leq 16$ and $17 \leq 18$ , feasible. $(7, 3)$ : $17 > 16$ , infeasible (the oven time runs out). (1 mark)

Answer: Corners $(0, 0)$ , $(8, 0)$ , $(6, 4)$ , $(0, 6)$ ; $(5, 4)$ is feasible, $(7, 3)$ is not.

Example 5: Spotting a redundant constraint · 2 marks · Very short answerHOTS / competency

Show that the constraint $2 x + y \leq 12$ is redundant in the system $x + y \leq 5$ , $2 x + y \leq 12$ , $x \geq 0$ , $y \geq 0$ , and write the corner points of the feasible region.

Try it first, then show the solutionSolution

For any feasible point, $2 x + y = (x + y) + x \leq 5 + 5 = 10 < 12$ (since $x \leq x + y \leq 5$ ). So $2 x + y \leq 12$ is automatically true and removing it does not change the region. (1 mark)
The region is the triangle with corners $(0, 0)$ , $(5, 0)$ and $(0, 5)$ . (1 mark)

Answer: Redundant; corners $(0, 0)$ , $(5, 0)$ , $(0, 5)$ .

Example 6: Case study: a canteen's roll plan · 4 marks · Case-basedHOTS / competency

Suppose a school canteen makes $x$ veg rolls and $y$ paneer rolls each morning. It can make at most 60 rolls, the paneer rolls must be at most half the veg rolls, and at least 10 paneer rolls are needed.
(i) Write the constraints. (1 mark)
(ii) Find the corner points of the feasible region. (2 marks)
(iii) Is the region bounded? Is the plan of 25 veg and 15 paneer rolls feasible? (1 mark)

Try it first, then show the solutionSolution

(i) $x + y \leq 60$ , $y \leq \frac{x}{2}$ (that is $x - 2 y \geq 0$ ), $y \geq 10$ , $x \geq 0$ . (1 mark)
(ii) $y = 10$ with $x - 2 y = 0$ : $(20, 10)$ ; $y = 10$ with $x + y = 60$ : $(50, 10)$ ; $x = 2 y$ with $x + y = 60$ : $3 y = 60$ , $(40, 20)$ . (2 marks)
(iii) The region is the triangle with these corners, so it is bounded. For $(25, 15)$ : $x - 2 y = - 5 < 0$ , so it breaks the 'at most half' rule and is infeasible. (1 mark)

Answer: (ii) $(20, 10)$ , $(50, 10)$ , $(40, 20)$ (iii) bounded; $(25, 15)$ is infeasible.

Where students lose marks in Feasible Region and Constraints

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

Mistake: Shading the wrong side of a line through the origin
The origin lies on the line, so the origin test tells you nothing.
Wrong working
$x - 2 y \geq 0$ : origin gives $0 \geq 0$ , true, so shade towards the origin.
Right working
Test $(1, 0)$ : $1 \geq 0$ is true, so shade the side containing $(1, 0)$ (below the line $y = \frac{x}{2}$ ).
Do this: For lines through the origin, always test a point on an axis.
Usually costs: The graph mark and every corner after it.
Mistake: Forgetting non-negativity
Without $x, y \geq 0$ the region spills into other quadrants and extra corners appear.
Wrong working
Region of $x + y \leq 5$ alone: unbounded, no corners.
Right working
With $x, y \geq 0$ : triangle with corners $(0, 0)$ , $(5, 0)$ , $(0, 5)$ .
Do this: Write $x \geq 0$ , $y \geq 0$ in every formulation.
Usually costs: ½ to 1 mark.
Mistake: Reading corners off the graph
Hand-drawn graphs are not exact; examiners want the solved intersection.
Wrong working
Intersection of $2 x + y = 16$ and $x + 3 y = 18$ 'looks like' $(6.5, 3.5)$ .
Right working
Solving gives $(6, 4)$ .
Do this: Show the two equations and their solution for each corner.
Usually costs: 1 mark, and the optimum later.
Mistake: Wrong inequality sign when writing constraints from a graph
The sign depends on which side the region lies, not on the line alone.
Wrong working
Region lies away from the origin beyond $x + 2 y = 4$ : written $x + 2 y \leq 4$ .
Right working
Test the origin: it is outside the region, so $x + 2 y \geq 4$ .
Do this: Pick a point inside the shaded region and substitute it.
Usually costs: ½ mark per constraint.

Exam tips

Corner-point counting MCQs were asked in 2023 (Q18) and 2024 (Q17); shading and marking corners was a 2-mark question in 2025 (Q24).
Writing constraints from a shaded graph appeared in the 2023-24 sample paper (Q11) and the 2026-27 sample paper (Q25); find each boundary line from two points, then test a point inside.
Label every line with its equation on the graph and mark the corner points with coordinates; marking schemes give a separate mark for the correct shaded region.
Turn 'at most' into $\leq$ , 'at least' into $\geq$ , and 'not more than half of' into an inequality through the origin.
If two constraints cannot hold together, say 'the feasible region is empty, so there is no feasible solution'.

Practice questions on Feasible Region and Constraints

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
Which point lies in the feasible region of $x + 2 y \leq 10$ , $3 x + y \leq 15$ , $x \geq 0$ , $y \geq 0$ ?
(a) $(4, 4)$
(b) $(3, 3)$
(c) $(5, 1)$
(d) $(0, 6)$
Hint
Substitute each point into both constraints.
Answer
(b) $(3, 3)$ : $9 \leq 10$ and $12 \leq 15$ . The others fail: $(4, 4)$ gives $12 > 10$ , $(5, 1)$ gives $16 > 15$ , $(0, 6)$ gives $12 > 10$ .
Q2 · 2 marks · Very short answerBoard level
Find the corner points of the feasible region $x + y \geq 2$ , $x \leq 3$ , $y \leq 3$ , $x \geq 0$ , $y \geq 0$ , and say whether it is bounded.
Hint
The line $x + y = 2$ cuts off a small triangle near the origin from the square.
Answer
Corners $(2, 0)$ , $(3, 0)$ , $(3, 3)$ , $(0, 3)$ , $(0, 2)$ : five corners, and the region is bounded.
Q3 · 3 marks · Short answerHarder board level
The feasible region of an LPP has corners $(0, 2)$ , $(4, 0)$ , $(6, 0)$ , $(6, 3)$ and $(0, 5)$ . Write the constraints.
Hint
Three non-trivial boundary lines: through $(0, 2)$ and $(4, 0)$ ; $x = 6$ ; through $(6, 3)$ and $(0, 5)$ .
Answer
Through $(0, 2)$ , $(4, 0)$ : $x + 2 y = 4$ , and the region is away from the origin, so $x + 2 y \geq 4$ . $x \leq 6$ . Through $(6, 3)$ , $(0, 5)$ : slope $- \frac{1}{3}$ , so $x + 3 y = 15$ , origin side: $x + 3 y \leq 15$ . Also $x \geq 0$ , $y \geq 0$ .
Q4 · 2 marks · Very short answerHarder board level
Show that the constraints $x + y \leq 3$ , $2 x + 3 y \geq 12$ , $x \geq 0$ , $y \geq 0$ have no feasible solution.
Hint
Find the largest value of $2 x + 3 y$ on the triangle $x + y \leq 3$ .
Answer
On the triangle with corners $(0, 0)$ , $(3, 0)$ , $(0, 3)$ , the largest value of $2 x + 3 y$ is $9$ at $(0, 3)$ , since $2 x + 3 y \leq 3 (x + y) \leq 9$ . So $2 x + 3 y \geq 12$ can never hold: the feasible region is empty.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The feasible region given by $x - y \geq 0$ , $x \leq 4$ , $y \geq 0$ is bounded.
Reason (R): A feasible region is bounded if it can be enclosed in a circle.
(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
Sketch $0 \leq y \leq x \leq 4$ .
Answer
(a). The region is the triangle with corners $(0, 0)$ , $(4, 0)$ , $(4, 4)$ , which fits inside a circle; R is the definition used.
Q6 · 3 marks · Short answerHarder board level
Suppose a farmer has 10 acres for wheat ( $x$ acres) and mustard ( $y$ acres). Wheat needs 3 labour-days per acre and mustard 1, and 18 labour-days are available. Write the constraints, find the corner points, and check whether planting 5 acres of wheat and 4 of mustard is possible.
Hint
Two non-trivial constraints: land and labour.
Answer
$x + y \leq 10$ , $3 x + y \leq 18$ , $x, y \geq 0$ . Solving $x + y = 10$ , $3 x + y = 18$ : $x = 4$ , $y = 6$ . Corners $(0, 0)$ , $(6, 0)$ , $(4, 6)$ , $(0, 10)$ . For $(5, 4)$ : $3 (5) + 4 = 19 > 18$ , so the plan is not possible.
Q7 · 3 marks · Short answerHOTS / competency
For which values of $k$ is the feasible region $x + y \leq 4$ , $y \geq k$ , $x \geq 0$ , $y \geq 0$ non-empty? What is the region when $k = 4$ ?
Hint
The largest $y$ on the triangle $x + y \leq 4$ is 4.
Answer
The largest value of $y$ in the triangle $(0, 0)$ , $(4, 0)$ , $(0, 4)$ is 4, so the region is non-empty exactly when $k \leq 4$ (for $k \leq 0$ the constraint $y \geq k$ is redundant). When $k = 4$ the region is the single point $(0, 4)$ .

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 a feasible solution?Show answer
A point that satisfies every constraint, including $x \geq 0$ , $y \geq 0$ .
2.How do you choose the side of $a x + b y \leq c$ ?Show answer
Substitute a test point not on the line (usually the origin); shade the side where the inequality is true.
3.How many corner points does $x + y \geq 4$ , $x, y \geq 0$ have?Show answer
Two, $(4, 0)$ and $(0, 4)$ ; the region is unbounded.
4.When is a feasible region bounded?Show answer
When it can be enclosed in a circle.
5.What is a redundant constraint?Show answer
One that does not change the feasible region because the other constraints already imply it.
6.How many non-trivial constraints can a 2026-27 board LPP have?Show answer
Up to three, in two variables.

Feasible Region and Constraints: frequently asked questions

What is the feasible region in linear programming?

It is the set of points that satisfy all the constraints of the problem, including x ≥ 0 and y ≥ 0. On the graph it is the common shaded region of all the half-planes.

How do I know which side of a constraint line to shade?

Substitute a test point that is not on the line, usually the origin. If the inequality is true there, shade the side containing that point; if false, shade the other side.

Can the feasible region be empty?

Yes. If the constraints contradict each other, such as x + y ≤ 3 with 2x + 3y ≥ 12 in the first quadrant, there is no common point and the problem has no feasible solution.

Which NCERT section covers feasible regions?

Section 12.2 (Linear Programming Problem and its Mathematical Formulation), especially 12.2.1 on formulation and 12.2.2 on the graphical method, in Chapter 12 of NCERT Mathematics Part II for Class 12.

Do I need graph paper for the board exam?

A neat sketch with labelled lines, intercepts, the shaded region and the corner points is enough. Exact scale helps you check, but the corner points must still be found by solving equations.