Important Questions Class 12 Maths Chapter 12 Linear Programming with Answers

Linear programming is a method of finding the maximum or minimum value of a linear objective function under given linear constraints. A feasible region contains all points that satisfy every constraint, including non-negative restrictions.

Optimisation questions become easier when students convert words into inequalities before drawing the graph. Important Questions Class 12 Maths Chapter 12 help CBSE 2026 students revise Linear Programming through definitions, graphical method, corner-point tables and word-problem formulation. The NCERT 2026-27 chapter covers objective function, decision variables, constraints, feasible region, feasible solution, optimal solution and graphical solutions of LPPs. It also states that this chapter uses the graphical method only for solving linear programming problems.

Key Takeaways

• Linear Programming Problem: An LPP maximises or minimises a linear objective function under linear constraints.
• Graphical Method: NCERT Class 12 Chapter 12 solves LPPs by graphing feasible regions in two variables.
• Corner Point Rule: The optimal value occurs at a corner point when an optimal value exists.
• CBSE 2026 Focus: Students should practise formulation, feasible region, corner-point tables and unbounded-region tests.

Important Questions Class 12 Maths Chapter 12 Structure 2026

Important Questions Class 12 Maths Chapter 12: CBSE 2026 Exam Focus

Linear Programming connects algebra, inequalities and coordinate geometry. Most school exam questions ask students to graph constraints and test corner points.

CBSE 2026 preparation should follow the NCERT graphical method, not simplex method. The NCERT chapter explains optimisation through maximum profit, minimum cost and resource-use problems.

1. What is Linear Programming in Class 12 Maths?

Linear Programming is a method of finding the optimal value of a linear objective function under linear constraints. The optimal value may be maximum or minimum.

The objective function has the form:

Z = ax + by

The constraints are linear inequalities or equations in x and y.

2. What is an objective function in Linear Programming?

An objective function is the linear function that has to be maximised or minimised. It has the form:

Z = ax + by

Here, a and b are constants.

The variables x and y are called decision variables.

3. What are constraints in Linear Programming?

Constraints are restrictions written as linear equations or inequalities. They limit the possible values of decision variables.

Examples include:

x + y ≤ 60

5x + y ≤ 100

x ≥ 0, y ≥ 0

The last two are non-negative restrictions.

4. What is a feasible region in linear programming?

The feasible region is the common region that satisfies all constraints of an LPP. It includes non-negative restrictions also.

Every point in this region satisfies all given inequalities.

The region may be bounded or unbounded.

5. What is an optimal solution in Linear Programming?

An optimal solution is a feasible point that gives the maximum or minimum value of the objective function. It lies in the feasible region.

For a bounded feasible region, the optimal value occurs at a corner point.

The corner point method uses this rule.

Class 12 Maths Chapter 12 Important Questions with Answers for Linear Programming Basics

Definitions carry direct marks in Class 12 Maths Chapter 12 important questions. Students should write exact meanings and one formula.

The terms objective function, feasible region and corner point appear repeatedly in CBSE 2026 school exams.

6. What is the difference between feasible solution and infeasible solution?

A feasible solution satisfies all constraints, while an infeasible solution violates at least one constraint.

For example, if constraints are:

5x + y ≤ 100
x + y ≤ 60
x ≥ 0, y ≥ 0

Then (10, 50) is feasible.

The point (25, 40) is infeasible because it violates 5x + y ≤ 100.

7. What is the difference between bounded and unbounded feasible region?

A bounded feasible region can fit inside a circle, while an unbounded feasible region extends indefinitely.

A bounded region gives both maximum and minimum values.

An unbounded region may or may not give an optimal value.

Students must test the relevant half-plane in unbounded cases.

8. What is the corner point theorem in Linear Programming?

The corner point theorem says that an optimal value occurs at a corner point of the feasible region.

If Z = ax + by has a maximum or minimum value, test the vertices.

This theorem supports the corner point method.

NCERT uses this theorem for graphical solutions.

9. What are decision variables in Linear Programming?

Decision variables are the unknown quantities whose values decide the solution of an LPP.

For example, if x is the number of tables and y is the number of chairs, then x and y are decision variables.

They must satisfy all constraints.

They usually follow x ≥ 0 and y ≥ 0.

10. What is an optimisation problem in Class 12 Maths?

An optimisation problem seeks the maximum or minimum value of a function under given restrictions.

For example, a shopkeeper may want maximum profit.

A factory may want minimum cost.

Linear programming problems are special optimisation problems.

Linear Programming Class 12 Important Questions on Graphical Method

The graphical method needs accurate inequality shading and corner-point testing. A wrong feasible region gives a wrong answer.

Linear Programming Class 12 important questions usually ask students to solve a given LPP graphically.

11. How do you solve a Linear Programming problem by graphical method?

Solve an LPP graphically by drawing constraints, finding the feasible region and testing corner points.

1. Convert each inequality into a boundary line.
2. Draw the half-plane satisfying each inequality.
3. Mark the common feasible region.
4. Find all corner points.
5. Evaluate Z = ax + by at each corner point.
6. Choose the maximum or minimum value as required.

12. Maximise Z = 3x + 4y subject to x + y ≤ 4, x ≥ 0, y ≥ 0.

The maximum value is 16 at (0, 4).

1. Given Data:
   Z = 3x + 4y
   x + y ≤ 4
   x ≥ 0, y ≥ 0
2. Feasible Region:
   The boundary line is x + y = 4.
   Corner points are (0, 0), (4, 0), (0, 4).
3. Corner-Point Table:

1. 
   Final Result:
   Maximum Z = 16 at (0, 4).

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

The maximum value is 10 at (2, 0).

1. Given Data:
   Z = 5x + 3y
   3x + 5y ≤ 15
   5x + 2y ≤ 10
   x ≥ 0, y ≥ 0
2. Boundary Intercepts:
   3x + 5y = 15 gives (5, 0) and (0, 3).
   5x + 2y = 10 gives (2, 0) and (0, 5).
3. Intersection of lines:
   3x + 5y = 15
   5x + 2y = 10
   Multiply first by 2: 6x + 10y = 30
   Multiply second by 5: 25x + 10y = 50
   Subtract: 19x = 20
   x = 20/19
   Put in 5x + 2y = 10:
   5(20/19) + 2y = 10
   2y = 90/19
   y = 45/19
4. Corner-Point Table:

1. 
   Final Result:
   Maximum Z = 235/19 at (20/19, 45/19).

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

The maximum value is 17 at (4, 3).

1. Given Data:
   Z = 3x + 2y
   x + 2y ≤ 10
   3x + y ≤ 15
   x ≥ 0, y ≥ 0
2. Intersection of lines:
   x + 2y = 10
   3x + y = 15
   From first: x = 10 - 2y
   Put in second:
   3(10 - 2y) + y = 15
   30 - 6y + y = 15
   -5y = -15
   y = 3
   x = 10 - 2(3) = 4
3. Corner-Point Table:

1. 
   Final Result:
   Maximum Z = 18 at (4, 3).

NCERT Solutions Class 12 Maths Chapter 12 Linear Programming Exercise 12.1 Questions

Exercise 12.1 contains graphical LPP questions with maximum, minimum and mixed cases. Students should solve them with corner-point tables.

NCERT Solutions Class 12 Maths Chapter 12 Linear Programming should also cover multiple optimal solutions and no feasible region cases.

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

The minimum value is -12 at (4, 0).

1. Given Data:
   Z = -3x + 4y
   x + 2y ≤ 8
   3x + 2y ≤ 12
   x ≥ 0, y ≥ 0
2. Corner Points:
   (0, 0), (4, 0), (2, 3), (0, 4)
3. Corner-Point Table:

1. 
   Final Result:
   Minimum Z = -12 at (4, 0).

16. Minimise Z = 3x + 5y subject to x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0.

The minimum value is 6 at every point on the segment joining (3/2, 1/2) and (2, 0).

1. Given Data:
   Z = 3x + 5y
   x + 3y ≥ 3
   x + y ≥ 2
   x ≥ 0, y ≥ 0
2. Intersection of boundary lines:
   x + 3y = 3
   x + y = 2
   Subtract: 2y = 1
   y = 1/2
   x + 1/2 = 2
   x = 3/2
3. Corner Points:
   (0, 2), (3/2, 1/2), (2, 0)
4. Corner-Point Table:

1. 
   Final Result:
   Minimum Z = 6 at (2, 0).

17. Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0.

The minimum value is 6 at all points on the line segment x + 2y = 6 within the feasible region.

1. Given Data:
   Z = x + 2y
   2x + y ≥ 3
   x + 2y ≥ 6
   x ≥ 0, y ≥ 0
2. Key Observation:
   The objective function is the same as one constraint.
   Z = x + 2y
   Constraint: x + 2y ≥ 6
3. Feasible Minimum:
   Since x + 2y ≥ 6, minimum Z cannot be below 6.
4. Final Result:
   Minimum Z = 6 at more than two points on x + 2y = 6.

18. Maximise Z = -x + 2y subject to x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

This LPP has no maximum value because the feasible region is unbounded in the direction increasing y.

1. Given Data:
   Z = -x + 2y
   x ≥ 3
   x + y ≥ 5
   x + 2y ≥ 6
   y ≥ 0
2. Feasible Region:
   The region extends upward without bound.
3. Objective Behaviour:
   If y increases while x remains feasible, Z = -x + 2y increases.
4. Final Result:
   No maximum value exists.

19. Maximise Z = x + y subject to x - y ≤ -1, -x + y ≤ 0, x ≥ 0, y ≥ 0.

This LPP has no feasible solution.

1. Given Data:
   x - y ≤ -1 gives x - y + 1 ≤ 0
   This means y ≥ x + 1.
   -x + y ≤ 0 gives y ≤ x.
2. Contradiction:
   y ≥ x + 1 and y ≤ x cannot hold together.
3. Feasible Region:
   No point satisfies both inequalities.
4. Final Result:
   No feasible solution exists.

Corner Point Method Class 12 Questions with Step-by-Step Solutions

Corner points decide the final answer in most LPP questions. Students lose marks when they skip the table.

Corner Point method Class 12 questions should always include feasible region, vertices and objective function values.

20. What is the corner point method in Class 12 Linear Programming?

The corner point method tests the objective function at every vertex of the feasible region.

1. Draw the feasible region.
2. Find the corner points.
3. Substitute each corner point in Z = ax + by.
4. Select the largest value for maximisation.
5. Select the smallest value for minimisation.

21. Solve: Maximise Z = 4x + y subject to x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0.

The maximum value is 120 at (30, 0).

1. Given Data:
   Z = 4x + y
   x + y ≤ 50
   3x + y ≤ 90
   x ≥ 0, y ≥ 0
2. Intersection of lines:
   x + y = 50
   3x + y = 90
   Subtract: 2x = 40
   x = 20
   y = 30
3. Corner Points:
   (0, 0), (30, 0), (20, 30), (0, 50)
4. Corner-Point Table:

1. 
   Final Result:
   Maximum Z = 120 at (30, 0).

22. Solve: Minimise Z = 200x + 500y subject to x + 2y ≥ 10, 3x + 4y ≤ 24, x ≥ 0, y ≥ 0.

The minimum value is 2300 at (4, 3).

1. Given Data:
   Z = 200x + 500y
   x + 2y ≥ 10
   3x + 4y ≤ 24
   x ≥ 0, y ≥ 0
2. Intersection of lines:
   x + 2y = 10
   3x + 4y = 24
   Multiply first by 2: 2x + 4y = 20
   Subtract from second: x = 4
   4 + 2y = 10
   y = 3
3. Corner Points:
   (0, 5), (4, 3), (0, 6)
4. Corner-Point Table:

1. 
   Final Result:
   Minimum Z = 2300 at (4, 3).

23. Solve: Minimise and maximise Z = 3x + 9y subject to x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0.

The minimum value is 60 at (5, 5), and maximum value is 180 on every point of segment joining (15, 15) and (0, 20).

1. Given Data:
   Z = 3x + 9y
   x + 3y ≤ 60
   x + y ≥ 10
   x ≤ y
   x ≥ 0, y ≥ 0
2. Corner Points:
   A(0, 10), B(5, 5), C(15, 15), D(0, 20)
3. Corner-Point Table:

1. 
   Final Result:
   Minimum Z = 60 at (5, 5).
   Maximum Z = 180 on CD.

24. How do multiple optimal solutions occur in Linear Programming?

Multiple optimal solutions occur when two adjacent corner points give the same optimal value.

If both endpoints of a boundary segment give the same maximum or minimum value, every point on that segment gives the same value.

For example, if C and D both give Z = 180, then every point on CD gives Z = 180.

25. How do you test optimal value in an unbounded feasible region?

Test the open half-plane beyond the corner-point value to decide whether the optimal value exists.

1. Suppose M is the largest corner-point value.
2. Draw ax + by > M for maximisation.
3. If it has no common point with feasible region, M is maximum.
4. If it has common points, no maximum exists.

For minimisation, test ax + by < m.

Linear Programming Word Problems Class 12 with Answers

Word problems need correct variable selection before graphing. Students should write constraints from every resource condition.

Linear Programming word problems Class 12 often involve profit, cost, investment, storage, production or diet restrictions.

26. A dealer buys tables and chairs. How do you formulate the NCERT furniture problem?

The LPP is: maximise Z = 250x + 75y subject to 5x + y ≤ 100, x + y ≤ 60, x ≥ 0, y ≥ 0.

1. Let Variables:
   x = number of tables
   y = number of chairs
2. Investment Constraint:
   Table cost = ₹2500
   Chair cost = ₹500
   Total investment ≤ ₹50000
   2500x + 500y ≤ 50000
   5x + y ≤ 100
3. Storage Constraint:
   x + y ≤ 60
4. Profit Function:
   Z = 250x + 75y
5. Final Formulation:
   Maximise Z = 250x + 75y
   subject to:
   5x + y ≤ 100
   x + y ≤ 60
   x ≥ 0, y ≥ 0

27. Solve the NCERT furniture problem by corner point method.

The maximum profit is ₹6250 when the dealer buys 10 tables and 50 chairs.

1. Given LPP:
   Maximise Z = 250x + 75y
   5x + y ≤ 100
   x + y ≤ 60
   x ≥ 0, y ≥ 0
2. Corner Points:
   O(0, 0), A(20, 0), B(10, 50), C(0, 60)
3. Corner-Point Table:

1. 
   Final Result:
   Maximum profit = ₹6250 at (10, 50).

28. How do you form an investment constraint in Linear Programming?

An investment constraint comes from total money available for buying or producing items.

If item A costs ₹a and item B costs ₹b, and total budget is ₹M, then:

ax + by ≤ M

Here, x and y are quantities of the two items.

29. How do you form a storage constraint in Linear Programming?

A storage constraint limits the total number of items stored.

If a shop can store at most S items, then:

x + y ≤ S

If items use different storage units, then:

ax + by ≤ S

The coefficients show storage used per item.

30. How do you choose variables in Class 12 Linear Programming word problems?

Choose variables for the unknown quantities that the problem asks you to decide.

If the problem asks for tables and chairs, use:

x = number of tables
y = number of chairs

If the problem asks for two products, use:

x = units of first product
y = units of second product

31. How do you write the objective function from profit data?

Multiply each item’s profit by its quantity and add the terms.

If profit per item A is ₹p and item B is ₹q, then:

Z = px + qy

For maximum profit, write:

Maximise Z = px + qy

32. How do you write the objective function from cost data?

Multiply each item’s cost by its quantity and add the terms.

If cost per item A is ₹p and item B is ₹q, then:

Z = px + qy

For minimum cost, write:

Minimise Z = px + qy

Class 12 Maths Chapter 12 Question Answer Writing Tips

Linear Programming answers need graph logic and calculation accuracy. A complete answer shows formulation, graph, vertices and final result.

Class 12 Maths Chapter 12 question answer practice should use copy-friendly equations and clear corner-point tables.

33. How should students write an LPP formulation answer?

Students should define variables first, then write constraints and objective function.

Use this order:

1. Let x and y be decision variables.
2. Write non-negative restrictions.
3. Write resource constraints.
4. Write objective function.
5. State maximise or minimise.

34. How should students write a graphical solution answer?

Students should show boundary lines, feasible region, corner points and objective values.

A graph alone is not enough.

The corner-point table proves the final answer.

CBSE 2026 school marking often gives marks for method.

35. What are common mistakes in Class 12 Linear Programming questions?

Common mistakes include wrong inequality signs, missing x ≥ 0, y ≥ 0 and incorrect corner points.

Students also forget to test unbounded regions.

Another common error is choosing the smallest value in a maximisation problem.

The final line must state the point and optimal value.

36. Why is x ≥ 0 and y ≥ 0 always important in Class 12 LPP?

Non-negative restrictions are important because quantities cannot be negative.

A factory cannot produce -5 units.

A shopkeeper cannot buy -3 chairs.

These restrictions keep the feasible region in the first quadrant.

37. What should students write when there is no feasible region?

Students should write that no point satisfies all constraints simultaneously.

They should not calculate objective values.

No feasible region means no feasible solution.

The final answer is:

The LPP has no feasible solution.

38. What should students write when an LPP is unbounded?

Students should test whether the objective value can increase or decrease indefinitely.

A region can be unbounded and still have a minimum.

A region can also have no maximum.

The final answer must mention the relevant half-plane test.

39. How much graph detail is needed in CBSE 2026 Linear Programming answers?

Students should draw all boundary lines clearly and shade the correct feasible region.

They should label axes, intercepts and corner points.

They should also write the corner-point table below the graph.

This keeps the solution checkable.

40. Why is Linear Programming repeatedly asked in CBSE-style papers?

Linear Programming tests modelling, graphing and optimisation together.

It connects inequalities with real-life profit and cost problems.

It also gives step-wise marks for formulation, graph and corner-point calculation.

This makes it suitable for board exam patterns.

Class 12 LPP Questions with Solutions for Extra Practice

Extra practice should mix theory, direct graph questions and formulation problems. These patterns cover most school tests.

Class 12 LPP questions with solutions help students build speed before pre-board exams.

41. Find the feasible region for x + y ≤ 5, x ≥ 0, y ≥ 0.

The feasible region is the triangular region bounded by x = 0, y = 0 and x + y = 5.

1. Boundary Line:
   x + y = 5
2. Intercepts:
   (5, 0), (0, 5)
3. Feasible Region:
   Since x + y ≤ 5, choose the side containing (0, 0).
4. Final Result:
   Feasible region is triangle with vertices (0, 0), (5, 0), (0, 5).

42. Find the maximum of Z = 2x + 3y over x + y ≤ 6, x ≥ 0, y ≥ 0.

The maximum value is 18 at (0, 6).

1. Corner Points:
   (0, 0), (6, 0), (0, 6)
2. Corner-Point Table:

1. 
   Final Result:
   Maximum Z = 18 at (0, 6).

43. Find the minimum of Z = 4x + y over x + y ≥ 3, x ≥ 0, y ≥ 0.

The minimum value is 3 at (0, 3).

1. Feasible Region:
   x + y ≥ 3 is unbounded.
2. Boundary Corner Points on axes:
   (3, 0), (0, 3)
3. Values:
   At (3, 0): Z = 4(3) + 0 = 12
   At (0, 3): Z = 4(0) + 3 = 3
4. Final Result:
   Minimum Z = 3 at (0, 3).

44. Find if (2, 3) is feasible for x + y ≤ 6, 2x + y ≤ 8, x ≥ 0, y ≥ 0.

The point (2, 3) is feasible.

1. Check first constraint:
   x + y = 2 + 3 = 5
   5 ≤ 6
2. Check second constraint:
   2x + y = 2(2) + 3 = 7
   7 ≤ 8
3. Check non-negative restrictions:
   x = 2 ≥ 0
   y = 3 ≥ 0
4. Final Result:
   (2, 3) is a feasible solution.

45. Find if (4, 3) is feasible for x + y ≤ 6, 2x + y ≤ 8, x ≥ 0, y ≥ 0.

The point (4, 3) is not feasible.

1. Check first constraint:
   x + y = 4 + 3 = 7
   7 ≤ 6 is false.
2. Since one constraint fails, no further proof is needed.
3. Final Result:
   (4, 3) is an infeasible solution.

Class 12 Maths Important Links