NCERT Solutions for Class 12 Maths Chapter 12: Linear Programming (NCERT 2026–27)

These Class 12 Maths Chapter 12 solutions cover Linear Programming from the NCERT textbook (Reprint 2026–27). Every question of Exercise 12.1 is reproduced verbatim and solved step by step using the Corner Point Method — we set up the feasible region, locate every corner point, evaluate the objective function and read off the optimal value, with each answer cross-checked against the textbook answer key.

Class: 12 Subject: Mathematics Chapter: 12 Chapter Name: Linear Programming Exercises: Exercise 12.1 (10 questions) Session: 2026–27
On this page

Chapter 12 Overview

Chapter 12, Linear Programming, applies the linear inequalities you learnt in Class XI to real-life optimisation problems — maximising profit or minimising cost subject to limited resources. A linear programming problem (LPP) seeks the optimal value of a linear objective function Z = ax + by, where the decision variables x and y are non-negative and satisfy a set of linear constraints. The chapter develops the graphical (Corner Point) method: graph the constraints to find the feasible region, identify its corner points (vertices), evaluate Z at each, and pick the largest (for a maximum) or smallest (for a minimum). It also explains how to handle bounded versus unbounded feasible regions, the special case of multiple optimal solutions (when two corner points give the same value), and problems with no feasible region. The solutions below work through every question of Exercise 12.1.

Key Concepts & Definitions

Objective function: the linear function Z = ax + by (a, b constants) that must be maximised or minimised. Here x and y are the decision variables.

Constraints: the linear inequalities/equations restricting x and y. The conditions x ≥ 0, y ≥ 0 are the non-negative restrictions.

Feasible region: the common region satisfying all the constraints (including x, y ≥ 0). Every point in it is a feasible solution; any point outside is infeasible.

Bounded / unbounded: a feasible region is bounded if it can be enclosed in a circle; otherwise it is unbounded.

Optimal solution: a point of the feasible region that gives the optimal (maximum or minimum) value of Z.

Corner point: a vertex of the feasible region — the intersection of two boundary lines. By the fundamental theorems, the optimal value (if it exists) always occurs at a corner point.

Important Formulas & Method (Chapter 12)

Standard LPP: Optimise Z = ax + by subject to linear constraints and x ≥ 0, y ≥ 0.

Theorem 1: if Z has an optimal value, it occurs at a corner point of the feasible region.

Theorem 2: if the feasible region R is bounded, Z has both a maximum and a minimum, each at a corner point of R.

Corner Point Method: (1) graph the constraints & find the feasible region; (2) find all corner points; (3) evaluate Z = ax + by at each; (4) the largest is M (max), the smallest is m (min) for a bounded region.

Unbounded region: M is the maximum only if ax + by > M has no point in common with the region; m is the minimum only if ax + by < m has no point in common with the region. Otherwise no such optimum exists.

Multiple optimal solutions: if two corner points give the same optimal value, every point on the segment joining them is also optimal.

Exercise 12.1

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original, solved by the Corner Point Method and verified against the answers at the back of the book. (Instruction: solve the following Linear Programming Problems graphically.)

1. Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.

SOLUTION The line x + y = 4 cuts the axes at (4, 0) and (0, 4). With x, y ≥ 0, the feasible region is the triangle OAB with O(0, 0), A(4, 0), B(0, 4); it is bounded. Evaluate Z = 3x + 4y: at O(0, 0), Z = 0; at A(4, 0), Z = 12; at B(0, 4), Z = 16. ∴ Maximum Z = 16 at (0, 4).

2. Minimise Z = − 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

SOLUTION Line x + 2y = 8 meets axes at (8, 0), (0, 4); line 3x + 2y = 12 meets axes at (4, 0), (0, 6). Solving the two lines: subtract to get 2x = 4 ⇒ x = 2, then y = 3, giving the intersection (2, 3). The bounded feasible region OABC has corners O(0, 0), A(4, 0), B(2, 3), C(0, 4). Evaluate Z = −3x + 4y: at O, Z = 0; at A(4, 0), Z = −12; at B(2, 3), Z = −6 + 12 = 6; at C(0, 4), Z = 16. ∴ Minimum Z = −12 at (4, 0).

3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

SOLUTION Line 3x + 5y = 15 meets axes at (5, 0), (0, 3); line 5x + 2y = 10 meets axes at (2, 0), (0, 5). Solve the two lines: from them, 15x + 6y = 30 and 6x + 10y = 30 ⇒ multiplying suitably gives the intersection at (20/19, 45/19). The bounded feasible region has corners O(0, 0), A(2, 0), B(20/19, 45/19), C(0, 3). Evaluate Z = 5x + 3y: at O, Z = 0; at A(2, 0), Z = 10; at C(0, 3), Z = 9; at B(20/19, 45/19), Z = 100/19 + 135/19 = 235/19 ≈ 12.37. ∴ Maximum Z = 235/19 at (20/19, 45/19).

4. Minimise Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

SOLUTION Line x + 3y = 3 meets axes at (3, 0), (0, 1); line x + y = 2 meets axes at (2, 0), (0, 2). Solving the two: subtract to get 2y = 1 ⇒ y = 1/2, then x = 3/2, giving (3/2, 1/2). The feasible region is unbounded with corner points A(3, 0), B(3/2, 1/2) and C(0, 2). Evaluate Z = 3x + 5y: at A(3, 0), Z = 9; at B(3/2, 1/2), Z = 9/2 + 5/2 = 7; at C(0, 2), Z = 10. The smallest is 7. Since the region is unbounded, check 3x + 5y < 7 — it has no point in common with the feasible region, so 7 is indeed the minimum. ∴ Minimum Z = 7 at (3/2, 1/2).

5. Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

SOLUTION Line x + 2y = 10 meets axes at (10, 0), (0, 5); line 3x + y = 15 meets axes at (5, 0), (0, 15). Solve the two: from 3x + y = 15 we get y = 15 − 3x; substitute in x + 2y = 10 ⇒ x + 30 − 6x = 10 ⇒ x = 4, y = 3, giving (4, 3). The bounded feasible region OABC has corners O(0, 0), A(5, 0), B(4, 3), C(0, 5). Evaluate Z = 3x + 2y: at O, Z = 0; at A(5, 0), Z = 15; at B(4, 3), Z = 12 + 6 = 18; at C(0, 5), Z = 10. ∴ Maximum Z = 18 at (4, 3).

6. Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. Show that the minimum of Z occurs at more than two points.

SOLUTION Line 2x + y = 3 meets axes at (3/2, 0), (0, 3); line x + 2y = 6 meets axes at (6, 0), (0, 3). The two lines meet at (0, 3). The feasible region is unbounded with corner points A(6, 0) and B(0, 3). Evaluate Z = x + 2y: at A(6, 0), Z = 6; at B(0, 3), Z = 6. Both corner points give Z = 6, so every point on the segment joining (6, 0) and (0, 3) (the line x + 2y = 6) also gives Z = 6. Checking x + 2y < 6 has no common point with the region, so 6 is the minimum. ∴ Minimum Z = 6 at all points on the line segment joining (6, 0) and (0, 3) — hence the minimum occurs at more than two points.

7. Minimise and Maximise Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x, y ≥ 0.

SOLUTION Boundary lines: x + 2y = 120, x + y = 60, x − 2y = 0. Solving pairs: x + 2y = 120 with x − 2y = 0 ⇒ 2x = 120, x = 60, y = 30 ⇒ (60, 30). x + y = 60 with x − 2y = 0 ⇒ (40, 20). x + 2y = 120 meets the x-axis at (120, 0); x + y = 60 meets the x-axis at (60, 0). The bounded feasible region has corners A(60, 0), B(120, 0), C(60, 30), D(40, 20). Evaluate Z = 5x + 10y: at A(60, 0), Z = 300; at B(120, 0), Z = 600; at C(60, 30), Z = 300 + 300 = 600; at D(40, 20), Z = 200 + 200 = 400. Minimum value is 300 at (60, 0). Maximum value is 600, occurring at both B(120, 0) and C(60, 30), so it occurs at every point of segment BC. ∴ Minimum Z = 300 at (60, 0); Maximum Z = 600 at all points on the line segment joining (120, 0) and (60, 30).

8. Minimise and Maximise Z = x + 2y subject to x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

SOLUTION Boundary lines: x + 2y = 100, 2x − y = 0 (i.e. y = 2x), 2x + y = 200. Solving pairs: x + 2y = 100 with y = 2x ⇒ x + 4x = 100, x = 20, y = 40 ⇒ (20, 40). 2x + y = 200 with y = 2x ⇒ 4x = 200, x = 50, y = 100 ⇒ (50, 100). x + 2y = 100 meets the y-axis at (0, 50); 2x + y = 200 meets the y-axis at (0, 200). The bounded feasible region has corners A(0, 50), B(20, 40), C(50, 100), D(0, 200). Evaluate Z = x + 2y: at A(0, 50), Z = 100; at B(20, 40), Z = 20 + 80 = 100; at C(50, 100), Z = 50 + 200 = 250; at D(0, 200), Z = 400. Minimum value is 100, occurring at both A(0, 50) and B(20, 40), hence on all of segment AB. Maximum value is 400 at D(0, 200). ∴ Minimum Z = 100 at all points on the line segment joining (0, 50) and (20, 40); Maximum Z = 400 at (0, 200).

9. Maximise Z = − x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

SOLUTION Boundary lines: x = 3, x + y = 5, x + 2y = 6. Corner points of the (unbounded) feasible region: x = 3 with x + y = 5 ⇒ (3, 2); x + y = 5 with x + 2y = 6 ⇒ (4, 1); and x = 3 on x + 2y = 6 gives (3, 1.5) but x + y = 5 dominates there. The relevant corners are A(6, 0), B(4, 1) and C(3, 2). Evaluate Z = −x + 2y: at A(6, 0), Z = −6; at B(4, 1), Z = −2; at C(3, 2), Z = 1. The region is unbounded (open upward). Z can be made arbitrarily large by increasing y while x = 3 (e.g. at (3, 100), Z = 197). Checking −x + 2y > 1 shows it has points in common with the region, so no finite maximum exists. ∴ Z has no maximum value.

10. Maximise Z = x + y, subject to x − y ≤ −1, −x + y ≤ 0, x, y ≥ 0.

SOLUTION Rewrite: x − y ≤ −1 means y ≥ x + 1, while −x + y ≤ 0 means y ≤ x. For both to hold we would need x + 1 ≤ y ≤ x, i.e. x + 1 ≤ x, which is impossible. No point satisfies all the constraints simultaneously. ∴ there is no feasible region, hence no maximum value of Z.

Miscellaneous Exercise

In the current NCERT 2026–27 edition, Chapter 12 (Linear Programming) contains a single exercise, Exercise 12.1, and no separate Miscellaneous Exercise. After Exercise 12.1 the chapter moves directly to the Summary and a Historical Note. All textbook questions for this chapter are therefore the ten problems solved above.

Common Mistakes to Avoid

Watch out for these

  • Forgetting to find every corner point — always intersect each relevant pair of boundary lines and include the axis-intercept vertices that lie in the feasible region.
  • Treating an unbounded region like a bounded one. For an unbounded region you must test ax + by > M (for a max) or ax + by < m (for a min) to confirm the optimum truly exists.
  • Shading the wrong side of a constraint — use a test point (often the origin) to decide which half-plane satisfies each inequality.
  • Missing multiple optimal solutions: when two corner points give the same optimal value, state that every point on the segment joining them is optimal.
  • Arithmetic slips when evaluating Z at fractional corner points (e.g. (20/19, 45/19)) — keep fractions exact rather than rounding early.
  • Not checking whether the constraints are consistent — if they contradict (as in Q10), there is no feasible region and hence no solution.

Practice MCQs & Assertion–Reason

1. The optimal value of the objective function of an LPP (if it exists) occurs at:

(a) the centre of the feasible region    (b) any interior point    (c) a corner point (vertex) of the feasible region    (d) the origin only

2. A feasible region is said to be bounded if:

(a) it has exactly four corners    (b) it can be enclosed within a circle    (c) it contains the origin    (d) it is a triangle

3. For Maximise Z = 3x + 4y subject to x + y ≤ 4, x, y ≥ 0, the maximum value of Z is:

(a) 12    (b) 16    (c) 4    (d) 0

4. The conditions x ≥ 0, y ≥ 0 in an LPP are called:

(a) objective functions    (b) optimal constraints    (c) non-negative restrictions    (d) feasible solutions

5. If two corner points give the same maximum value of Z, then the maximum:

(a) does not exist    (b) occurs only at the origin    (c) occurs at every point on the segment joining them    (d) occurs at no other point

6. For an unbounded feasible region, the value M is the maximum of Z = ax + by only if:

(a) ax + by > M has no point in common with the region    (b) ax + by < M is empty    (c) the region is a polygon    (d) M = 0

7. In Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0, the minimum value of Z is:

(a) 3    (b) 6    (c) 9    (d) 12

8. A point lying within and on the boundary of the feasible region is called a:

(a) corner point    (b) feasible solution    (c) optimal solution    (d) infeasible point

9. For Maximise Z = x + y subject to x − y ≤ −1, −x + y ≤ 0, x, y ≥ 0:

(a) maximum is 1    (b) maximum is 0    (c) there is no feasible region    (d) maximum is unbounded but exists

10. The graphical method of solving an LPP studied in Class 12 is the:

(a) simplex method    (b) corner point method    (c) transportation method    (d) substitution method

Answer key: 1-(c), 2-(b), 3-(b), 4-(c), 5-(c), 6-(a), 7-(b), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: For an LPP with a bounded feasible region, both a maximum and a minimum of Z always exist.

Reason: If the feasible region is bounded, the objective function attains its optimum at a corner point of the region.

A-R 2. Assertion: In Maximise Z = 3x + 4y subject to x + y ≤ 4, x, y ≥ 0, the maximum is 16 at (0, 4).

Reason: The optimum of an LPP always occurs at an interior point of the feasible region.

A-R 3. Assertion: The LPP Maximise Z = x + y subject to x − y ≤ −1, −x + y ≤ 0, x, y ≥ 0 has no solution.

Reason: The constraints require y ≥ x + 1 and y ≤ x simultaneously, which is impossible, so the feasible region is empty.

A-R 4. Assertion: For Maximise Z = −x + 2y subject to x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0, Z has no maximum value.

Reason: The feasible region is unbounded and ax + by > M has points in common with it for every M.

A-R 5. Assertion: When the minimum of Z occurs at two corner points, it also occurs at every point of the segment joining them.

Reason: The objective function is linear, so it is constant along the line through two points giving the same value.

Answer key: 1-(A), 2-(C), 3-(A), 4-(A), 5-(A).

Quick Revision Summary

  • An LPP optimises a linear objective function Z = ax + by subject to linear constraints and x, y ≥ 0.
  • The feasible region is the common solution of all constraints; its vertices are the corner points.
  • By the fundamental theorems, the optimum (if it exists) occurs at a corner point; a bounded region always has both a max and a min.
  • Corner Point Method: graph constraints → find all corner points → evaluate Z at each → pick largest (max) / smallest (min).
  • For an unbounded region, confirm the optimum with the ax + by > M (or < m) test; it may not exist.
  • Equal Z at two corner points ⇒ multiple optimal solutions along the joining segment; contradictory constraints ⇒ no feasible region.

How to score full marks in this chapter

Draw a neat, labelled graph and shade the feasible region using a test point. List every corner point with its coordinates, then make a clear table of Z-values — markers reward the table. For unbounded regions, always add the line stating whether ax + by > M (or < m) meets the region, and explicitly conclude “maximum/minimum exists / does not exist”. When two corner points tie, state the full segment of optimal solutions to bag the last mark.

Frequently Asked Questions

What is Class 12 Maths Chapter 12 Linear Programming about?

Chapter 12 teaches how to solve optimisation problems graphically. You formulate a linear objective function Z = ax + by with linear constraints, graph the feasible region, and use the Corner Point Method to find the maximum or minimum value of Z.

How many exercises are there in Class 12 Maths Chapter 12?

The NCERT 2026–27 edition has a single exercise, Exercise 12.1, with 10 graphical LPP questions. There is no separate Miscellaneous Exercise in this chapter — all questions are solved on this page.

What is the Corner Point Method?

It is the method of evaluating the objective function Z at every corner point (vertex) of the feasible region. The largest value gives the maximum and the smallest gives the minimum, since the optimum of an LPP always occurs at a corner point.

Are these Class 12 Maths Chapter 12 solutions free?

Yes. All solutions are free and follow the official NCERT textbook for the 2026–27 session, solved step by step and verified against the book’s answer key.

← Chapter 11: Three Dimensional Geometry All Class 12 Maths Chapters Chapter 13: Probability →
Scroll to Top