Linear Programming Class 12: Notes, Methods and Solved Questions

What Linear Programming is about

Weightage: This chapter is Unit V, Linear Programming, which carries 5 of the 80 theory marks.Source: CBSE Mathematics syllabus 2026-27, Unit V Linear Programming

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5 marks per paper. See the questions

Key concepts

Terminology: objective function, decision variables, constraints

NCERT section 12.2

A linear programming problem (LPP) asks for the optimal (maximum or minimum) value of a linear function subject to linear constraints.

• Decision variables: the unknown quantities, x and y (numbers of items, hours and so on).
• Objective function: $Z=ax+by$, the quantity to maximise (profit, output) or minimise (cost, time).
• Constraints: linear inequalities such as $2x+y\le 10$ from limited resources.
• Non-negativity constraints: $x\ge 0$, $y\ge 0$, because the quantities cannot be negative.

Optimisation means finding the maximum or the minimum. The word "linear" means every term is of degree 1, with no $x^2$ or xy.

Worked example

In "Maximise $Z=3x+2y$ subject to $x+y\le 4$, $x+3y\le 6$, $x,y\ge 0$", identify the objective function and the constraints. [1 mark]

Try it first, then show the solutionSolution

1. Objective function: $Z=3x+2y$ (to be maximised).
2. Constraints: $x+y\le 4$, $x+3y\le 6$; non-negativity: $x\ge 0$, $y\ge 0$.

Answer: As listed.

Mathematical formulation of a problem

NCERT section 12.2.1

To turn a word problem into an LPP:

1. Name the decision variables clearly (x = number of …, y = number of …).
2. Write the objective function from the profit or cost per unit.
3. Write one constraint for each limited resource: (resource used per unit × number of units) summed ≤ amount available. Use ≥ for "at least" conditions.
4. Add $x\ge 0$, $y\ge 0$.

A table (rows for products, columns for resources) keeps the coefficients straight.

Worked example

A workshop makes products P and Q. Each P needs 2 machine-hours and 1 labour-hour; each Q needs 1 machine-hour and 2 labour-hours. At most 10 machine-hours and 8 labour-hours are available daily. Profits are ₹50 per P and ₹40 per Q. Formulate the LPP. [2 marks]

Try it first, then show the solutionSolution

1. Let x = number of P and y = number of Q made per day.
2. Maximise $Z=50x+40y$ subject to $2x+y\le 10$ (machine), $x+2y\le 8$ (labour), $x\ge 0$, $y\ge 0$.

Answer: Maximise $Z=50x+40y$; $2x+y\le 10$, $x+2y\le 8$, $x,y\ge 0$

Feasible region and feasible solutions

NCERT section 12.2.2

Draw each constraint's boundary line (change the inequality to an equation and plot two points, usually the intercepts). Decide which side satisfies the inequality by testing a point not on the line, usually (0, 0). The region common to all constraints, including $x,y\ge 0$ (the first quadrant), is the feasible region.

• A point in the feasible region (including its boundary) is a feasible solution; a point outside is infeasible.
• The region is bounded if it can be enclosed in a circle, and unbounded if it extends indefinitely.
• If no point satisfies all constraints, there is no feasible region and the problem has no solution.

Worked example

For the constraints $2x+y\le 10$, $x+2y\le 8$, $x,y\ge 0$, decide whether (3, 2) and (5, 2) are feasible. [1 mark]

Try it first, then show the solutionSolution

1. (3, 2): 6 + 2 = 8 ≤ 10 and 3 + 4 = 7 ≤ 8: feasible.
2. (5, 2): 10 + 2 = 12 > 10: infeasible.

Answer: (3, 2) is feasible; (5, 2) is infeasible.

Corner point method for a bounded region

NCERT section 12.2.2

Theorem 1: if Z = ax + by has an optimal value on the feasible region R, that value occurs at a corner point (vertex) of R. Theorem 2: if R is bounded, Z has both a maximum and a minimum on R, each at a corner point. (Both theorems are used without proof.)

Method:

1. Draw the feasible region and find all its corner points (read intercepts, or solve the two boundary equations that meet there).
2. Make a table of Z at each corner point.
3. For a bounded region, the largest value is the maximum and the smallest is the minimum.

Worked example

Solve graphically: maximise $Z=50x+40y$ subject to $2x+y\le 10$, $x+2y\le 8$, $x,y\ge 0$. [3 marks]

Try it first, then show the solutionSolution

1. The feasible region is the bounded quadrilateral with corners O(0, 0), A(5, 0), B and C(0, 4); B solves $2x+y=10$, $x+2y=8$: B(4, 2).
2. Z: O → 0, A → 250, B → 200 + 80 = 280, C → 160.
3. Maximum 280 at B(4, 2).

Answer: Maximum Z = 280 at x = 4, y = 2 (make 4 of P and 2 of Q for ₹280 profit).

Unbounded feasible regions

NCERT section 12.2.2

If the feasible region is unbounded, a maximum or minimum may not exist; if it does, it is still at a corner point. After making the corner table:

• Let m be the smallest corner value. Draw (or test) the open half-plane $ax+by<m$. If it has no point in common with the feasible region, m is the minimum; otherwise Z has no minimum.
• Let M be the largest corner value. If $ax+by>M$ has no point in common with the region, M is the maximum; otherwise there is no maximum.

When the region stretches out to the upper right and a, b > 0, a minimum usually exists but a maximum does not.

Worked example

Minimise $Z=2x+3y$ subject to $x+2y\ge 6$, $2x+y\ge 6$, $x,y\ge 0$. [3 marks]

Try it first, then show the solutionSolution

1. The region is unbounded (above both lines). Corners: (0, 6), (6, 0) and the meeting point of $x+2y=6$ and $2x+y=6$, which is (2, 2).
2. Z: (0, 6) → 18, (2, 2) → 10, (6, 0) → 12. Smallest corner value 10.
3. The half-plane $2x+3y<10$ has no point in the region (for instance $2x+3y=\frac{4}{3}(x+2y)+\frac{1}{3}(2x+y)\ge 8+2=10$), so 10 is the minimum.

Answer: Minimum Z = 10 at (2, 2); Z has no maximum.

Multiple optimal solutions

NCERT section 12.2.2

If two adjacent corner points give the same optimal value, then every point on the edge joining them also gives that value: the problem has infinitely many optimal solutions. This happens when the objective function's line is parallel to that edge (its coefficients are proportional to the constraint's).

State the answer as "Z is optimal at every point of the segment joining … and …".

Worked example

Maximise $Z=6x+3y$ subject to $2x+y\le 10$, $x+2y\le 8$, $x,y\ge 0$. [2 marks]

Try it first, then show the solutionSolution

1. Corners O(0, 0), A(5, 0), B(4, 2), C(0, 4): Z = 0, 30, 30, 12.
2. Maximum 30 at both A and B; $6x+3y=3(2x+y)$ is parallel to the edge AB.

Answer: Maximum 30 at every point of the segment joining (4, 2) and (5, 0).

Infeasible problems and checking constraints

NCERT section 12.2.2

Constraints can contradict each other, leaving no feasible region; then the LPP has no feasible solution and no optimal value. Check by shading: if the half-planes do not overlap in the first quadrant, the problem is infeasible.

Board questions also show a feasible region and ask you to write the constraints that produce it: find each boundary line from two points on it, then use a point inside the region to choose ≤ or ≥.

Worked example

A bounded feasible region has corners O(0, 0), A(4, 0), B(3, 2) and C(0, 3). Write the constraints. [2 marks]

Try it first, then show the solutionSolution

1. AB through (4, 0) and (3, 2): $2x+y=8$. BC through (3, 2) and (0, 3): $x+3y=9$.
2. (0, 0) is inside, and $0\le 8$, $0\le 9$, so the constraints are $2x+y\le 8$, $x+3y\le 9$, $x\ge 0$, $y\ge 0$.

Answer: $2x+y\le 8$, $x+3y\le 9$, $x,y\ge 0$

Maths in real life: Linear Programming

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.

Topics in this chapter

• Feasible Region and Constraints
• Corner Point Method
• Unbounded Feasible Regions

Linear Programming formula sheet

Objective function

$Z=ax+by$

Non-negativity constraints

$x\ge 0,y\ge 0$

Corner point principle

$\text{optimal value (if it exists)}=\text{value of }Z\text{ at a corner point}$

Bounded region

$\max Z={\max }_{\text{corners}}Z,\min Z={\min }_{\text{corners}}Z$

Unbounded region, minimum test

$m={\min }_{\text{corners}}Z\text{is the minimum}⟺\{ax+by<m\}\cap R=\varnothing$

Unbounded region, maximum test

$M={\max }_{\text{corners}}Z\text{is the maximum}⟺\{ax+by>M\}\cap R=\varnothing$

Corner point from two boundary lines

$a_{1}x+b_{1}y=c_1,a_{2}x+b_{2}y=c_2\Rightarrow \text{solve simultaneously}$

Multiple optima

$Z(P)=Z(Q)\text{at adjacent corners}\Rightarrow Z\text{ optimal on all of }PQ$

How the board asks Linear Programming

The question patterns to expect, and the method that works for each.

Tricks to remember Linear Programming

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

• Corners are candidates

  The optimum of a linear Z always sits at a vertex: find corners, tabulate, compare.

  Use it when: Every graphical LPP.

• Origin test for shading

  Put (0, 0) into each inequality: true → shade the origin side; false → the other side.

  Use it when: Drawing the feasible region.

• Bounded gives both

  A closed region always has a max and a min; an open one may lack one of them.

  Use it when: Deciding whether to apply the half-plane test.

• Tie means a whole edge

  Two neighbouring corners with the same optimum → every point between them is optimal.

  Use it when: Multiple optimal solutions.

• ≤ is under, ≥ is over

  For lines with positive intercepts, ≤ keeps the side towards the origin and ≥ the side away from it.

  Use it when: Quick shading.

Linear Programming: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Linear Programming

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

• ✗Mistake: Missing a corner point

  Forgetting corners on the axes or the meeting point of two slanted lines.

  ✓Do this: Walk round the boundary of the region and list every vertex; solve equations for each meeting point.

  Usually costs: 1-2 marks, and often the wrong answer.

• ✗Mistake: Declaring a maximum on an unbounded region without the test

  Reading the largest corner value as the maximum.

  Wrong working

  Region x + y ≥ 2, x − y ≤ 1: corners give Z = 3x + 2y values 4 and 5.5, so "maximum 5.5".

  Right working

  The half-plane 3x + 2y > 5.5 meets the region (for example at (2, 2)), so Z has no maximum.

  ✓Do this: For unbounded regions always check the open half-plane.

  Usually costs: 1 mark.

• ✗Mistake: Shading the wrong side of a line

  Guessing the direction of ≤ or ≥.

  ✓Do this: Test a point (usually the origin) in each inequality.

  Usually costs: Most of the marks.

• ✗Mistake: No graph or unlabelled graph

  Writing only the table.

  ✓Do this: Draw the lines with intercepts labelled, shade the feasible region and name the corners.

  Usually costs: 1 mark for the figure.

• ✗Mistake: Giving only one point when two corners tie

  Stopping at the first optimum found.

  ✓Do this: Say the optimum holds at every point of the segment joining the two corners.

  Usually costs: 1 mark.

NCERT exercise guide: Linear Programming

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Linear Programming: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5 marks per paper; 19 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Linear Programming

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Linear Programming question, by year

Practice questions on Linear Programming

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

1. Q1 · 1 mark · MCQBoard level

   The objective function of a linear programming problem is:
   (a) a constant
   (b) a linear function to be optimised
   (c) an inequality
   (d) a quadratic expression

   Hint

   Think of Z = ax + by.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The number of corner points of the feasible region given by x + y ≤ 4, x ≥ 0, y ≥ 0 is:
   (a) 2
   (b) 3
   (c) 4
   (d) infinitely many

   Hint

   It is a triangle.

   Answer

   (b) 3

3. Q3 · 1 mark · MCQBoard level

   The corner points of an unbounded feasible region are (0, 8), (2, 3) and (8, 0), and 2x + 3y < 13 has no point in common with the region. The minimum of Z = 2x + 3y is:
   (a) 24
   (b) 16
   (c) 13
   (d) no minimum

   Hint

   Tabulate, then use the given test.

   Answer

   (c) 13 at (2, 3)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): (3, 3) is a feasible solution of x + y ≤ 5, x ≥ 0, y ≥ 0.
   Reason (R): A feasible solution satisfies every constraint of the problem.
   (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

   3 + 3 = ?

   Answer

   (d). 3 + 3 = 6 > 5, so A is false; R is the correct definition.

5. Q5 · 2 marks · Very short answerHarder board level

   For the feasible region $x+y\ge 3$, $x,y\ge 0$, does $Z=x+y$ have a minimum? A maximum?

   Hint

   Corners (3, 0) and (0, 3).

   Answer

   Minimum 3, attained at every point of the segment joining (3, 0) and (0, 3); no maximum, since the region is unbounded and Z grows without limit.

6. Q6 · 3 marks · Short answerHarder board level

   Solve graphically: find the maximum and minimum of $Z=3x+2y$ subject to $x+2y\le 10$, $x+y\ge 3$, $x-y\le 2$, $x,y\ge 0$.

   Hint

   Five lines bound the region; four corners survive.

   Answer

   Corners (0, 3), (0, 5), $\left(\frac{14}{3},\frac{8}{3}\right)$, $\left(\frac{5}{2},\frac{1}{2}\right)$; Z = 6, 10, $\frac{58}{3}$, $\frac{17}{2}$. Maximum $\frac{58}{3}$ at $\left(\frac{14}{3},\frac{8}{3}\right)$; minimum 6 at (0, 3).

7. Q7 · 3 marks · Short answerHarder board level

   Minimise $Z=4x+3y$ subject to $x+y\ge 5$, $2x+y\ge 6$, $x,y\ge 0$.

   Hint

   Unbounded region; check the half-plane.

   Answer

   Corners (0, 6), (1, 4), (5, 0); Z = 18, 16, 20. Since $4x+3y=2(x+y)+(2x+y)\ge 10+6=16$, the minimum is 16 at (1, 4).

8. Q8 · 3 marks · Short answerHarder board level

   Show that $Z=3x+2y$ has no maximum subject to $x+y\ge 2$, $x-y\le 1$, $x,y\ge 0$, and find its minimum.

   Hint

   Corners (0, 2) and $\left(\frac{3}{2},\frac{1}{2}\right)$.

   Answer

   Z = 4 at (0, 2) and $\frac{11}{2}$ at $\left(\frac{3}{2},\frac{1}{2}\right)$. The point (2, 2) is feasible with Z = 10 > $\frac{11}{2}$, so there is no maximum. Since $3x+2y\ge 2(x+y)\ge 4$, the minimum is 4 at (0, 2).

9. Q9 · 2 marks · Very short answerHarder board level

   A company makes x units of A and y units of B. Each A uses 3 machine-hours and each B 2 machine-hours, with 120 hours available, and at least 10 units of A must be made. Profits are ₹30 per A and ₹25 per B. Formulate the LPP.

   Hint

   One resource constraint and one "at least" constraint.

   Answer

   Maximise $Z=30x+25y$ subject to $3x+2y\le 120$, $x\ge 10$, $y\ge 0$.

10. Q10 · 5 marks · Long answerHarder board level

    A tailor makes shirts (x) and kurtas (y). A shirt uses 2 m of cloth and 2 hours; a kurta uses 3 m and 1 hour. There are 24 m of cloth and 16 hours, and at least 2 kurtas must be made. Profits are ₹60 per shirt and ₹70 per kurta. Formulate and solve graphically.

    Hint

    Three constraints besides x ≥ 0.

    Answer

    Maximise $Z=60x+70y$; $2x+3y\le 24$, $2x+y\le 16$, $y\ge 2$, $x\ge 0$. Corners (0, 2), (7, 2), (6, 4), (0, 8); Z = 140, 560, 640, 560. Maximum ₹640 for 6 shirts and 4 kurtas.

11. Q11 · 3 marks · Short answerHOTS / competency

    A bounded feasible region has corners (0, 0), (5, 0), (4, 3) and (0, 5). For which values of k does Z = kx + y attain its maximum only at (4, 3)?

    Hint

    Compare Z(4, 3) = 4k + 3 with the other corners.

    Answer

    Need $4k+3>5k$ (so k < 3), $4k+3>5$ (so $k>\frac{1}{2}$) and $4k+3>0$. Hence $\frac{1}{2}<k<3$.

12. Q12 · 2 marks · Very short answerHOTS / competency

    Show that the LPP "maximise $Z=x+y$ subject to $2x+y\le 4$, $x+2y\ge 10$, $x,y\ge 0$" has no feasible solution.

    Hint

    Bound x + 2y using the first constraint.

    Answer

    From $2x+y\le 4$: $y\le 4-2x$, so $x+2y\le x+8-4x=8-3x\le 8<10$ for $x\ge 0$. No point satisfies both constraints, so the feasible region is empty.

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. 1.What is the degree of the objective function of an LPP?Show answer

   1 (it is linear).

2. 2.Where does the optimal value of Z occur?Show answer

   At a corner point of the feasible region.

3. 3.When is a feasible region bounded?Show answer

   When it can be enclosed within a circle.

4. 4.Does Z always have a maximum on an unbounded region?Show answer

   No; it may not exist. Use the half-plane test.

5. 5.What are non-negativity constraints?Show answer

   x ≥ 0 and y ≥ 0.

6. 6.What does it mean if two adjacent corners give the same optimal Z?Show answer

   Every point on the segment joining them is optimal.

7. 7.Number of corner points of x + y ≤ 4, x ≥ 0, y ≥ 0?Show answer

   3: (0, 0), (4, 0), (0, 4).

8. 8.What is an infeasible solution?Show answer

   A point that fails at least one constraint.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you identify the objective function and constraints in a problem?
• ☐Can you formulate an LPP from a word problem?
• ☐Can you draw the boundary lines and shade the correct half-planes?
• ☐Can you find every corner point, including those found by solving two equations?
• ☐Can you tabulate Z and pick the maximum and minimum on a bounded region?
• ☐Can you apply the half-plane test on an unbounded region?
• ☐Can you recognise multiple optimal solutions?
• ☐Can you recognise an infeasible problem?
• ☐Can you write the constraints of a given feasible region?
• ☐Can you find a condition on p and q for Z = px + qy to be optimal at two given corners?

Linear Programming: frequently asked questions