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Linear Programming

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Chapter Notes

Overview
This chapter introduces Linear Programming Problems (LPP), a special class of optimisation problems that involve maximising or minimising a linear objective function subject to a set of linear constraints (inequalities) and non-negative restrictions. It exists in the syllabus to equip students with a graphical method for solving real-life problems like maximising profit or minimising cost under limited resources. The one big idea a student must walk away with is that the optimal value of the objective function in a bounded feasible region always occurs at a corner point (vertex) of the feasible region, and this is found using the Corner Point Method.
Key Concepts (14)
Optimisation problem
A problem which seeks to maximise or minimise a linear function (say of two variables x and y) subject to certain constraints as determined by a set of linear inequalities.
e.g. A furniture dealer wants to maximise his total profit from buying tables and chairs given a fixed investment and storage space.
Linear Programming Problem
A problem that is concerned with finding the optimal value (maximum or minimum value) of a linear function (called objective function) of several variables (say x and y), subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities (called linear constraints).
e.g. Maximise Z = 250x + 75y subject to 5x + y ≤ 100, x + y ≤ 60, x ≥ 0, y ≥ 0.
Objective function
A linear function Z = ax + by, where a, b are constants, which has to be maximised or minimised.
e.g. In the furniture dealer problem, Z = 250x + 75y is the objective function representing total profit.
Decision variables
The variables x and y in the objective function and constraints of a linear programming problem.
e.g. In the furniture dealer problem, x is the number of tables and y is the number of chairs the dealer buys.
Constraints
The linear inequalities or equations or restrictions on the variables of a linear programming problem.
e.g. 5x + y ≤ 100 (investment constraint) and x + y ≤ 60 (storage constraint) are constraints in the furniture dealer problem.
Non-negative restrictions
The conditions x ≥ 0, y ≥ 0 that the variables of a linear programming problem must satisfy.
e.g. x ≥ 0 and y ≥ 0 because the dealer cannot buy a negative number of tables or chairs.
Feasible region
The common region determined by all the constraints including non-negative constraints x, y ≥ 0 of a linear programming problem.
e.g. In Fig 12.1, the region OABC (shaded) is the feasible region for the furniture dealer problem.
Feasible solutions
Points within and on the boundary of the feasible region represent feasible solutions of the constraints.
e.g. The point (10, 50) is a feasible solution of the furniture dealer problem.
Infeasible solution
Any point outside the feasible region is called an infeasible solution.
e.g. The point (25, 40) is an infeasible solution of the furniture dealer problem because it does not satisfy all constraints.
Optimal (feasible) solution
Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function.
e.g. The point (10, 50) is the optimal solution for the furniture dealer problem, giving maximum profit of Rs 6250.
Corner point (vertex)
A point of the feasible region which is the intersection of two boundary lines.
e.g. The corner points of the feasible region OABC are O(0,0), A(20,0), B(10,50), and C(0,60).
Bounded feasible region
A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle.
e.g. The feasible region OABC in Fig 12.1 is bounded because it can be enclosed within a circle.
Unbounded feasible region
A feasible region that does extend indefinitely in any direction; it cannot be enclosed within a circle.
e.g. The feasible region in Fig 12.5 is unbounded.
Multiple optimal solutions
If two corner points produce the same maximum (or minimum) value of the objective function, then every point on the line segment joining these points will also give the same maximum (or minimum) value.
e.g. In Example 3, points C(15,15) and D(0,20) both give Z = 180, so every point on CD is also optimal.
Formulas & Equations (5)
Z = ax + by
Reads as: Z equals a times x plus b times y
Variables: Z = objective function value; a, b = constants (profit or cost per unit); x, y = decision variables
Use when: Writing the objective function for a linear programming problem to be maximised or minimised.
⚠ Students often forget to include the non-negative restrictions x ≥ 0, y ≥ 0 as part of the formulation.
2500x + 500y ≤ 50000
Reads as: 2500 times x plus 500 times y is less than or equal to 50000
Variables: x = number of tables; y = number of chairs; 2500 = cost per table (Rs); 500 = cost per chair (Rs); 50000 = maximum investment (Rs)
Use when: Formulating an investment constraint in a linear programming problem.
⚠ Students forget to simplify the inequality (e.g., dividing by 100 to get 5x + y ≤ 100) before graphing.
x + y ≤ 60
Reads as: x plus y is less than or equal to 60
Variables: x = number of tables; y = number of chairs; 60 = maximum storage capacity (pieces)
Use when: Formulating a storage or capacity constraint in a linear programming problem.
⚠ Students incorrectly write x + y < 60 instead of ≤ 60, forgetting that the boundary is included.
Z = 250x + 75y
Reads as: Z equals 250 times x plus 75 times y
Variables: Z = total profit (Rs); 250 = profit per table (Rs); 75 = profit per chair (Rs); x = number of tables; y = number of chairs
Use when: Writing the objective function for the furniture dealer problem to maximise profit.
⚠ Students confuse the coefficients of the objective function with those of the constraints.
5x + y ≤ 100
Reads as: 5 times x plus y is less than or equal to 100
Variables: x = number of tables; y = number of chairs; 5 and 1 are coefficients from simplifying the investment constraint; 100 = maximum investment in simplified units
Use when: Graphing the simplified investment constraint for the furniture dealer problem.
⚠ Students forget to simplify the original constraint 2500x + 500y ≤ 50000 to 5x + y ≤ 100 before graphing.
How It Works
Mathematical Formulation of a Linear Programming Problem
  1. Step 1: Identify the decision variables (e.g., x = number of tables, y = number of chairs).
  2. Step 2: Write the objective function as a linear function of the decision variables to be maximised or minimised (e.g., Z = 250x + 75y).
  3. Step 3: Identify all constraints (linear inequalities) including non-negative restrictions (e.g., 5x + y ≤ 100, x + y ≤ 60, x ≥ 0, y ≥ 0).
  4. Step 4: Combine the objective function and constraints to state the complete Linear Programming Problem.
💡 Students often forget to include the non-negative restrictions x ≥ 0, y ≥ 0 as part of the constraints.
Graphical Method of Solving a Linear Programming Problem (Corner Point Method)
  1. Step 1: Graph all the constraints (linear inequalities) including non-negative restrictions to find the feasible region.
  2. Step 2: Determine the corner points (vertices) of the feasible region by inspection or by solving the equations of intersecting boundary lines.
  3. Step 3: Evaluate the objective function Z = ax + by at each corner point.
  4. Step 4: For a bounded feasible region, the largest value (M) is the maximum and the smallest value (m) is the minimum of Z.
  5. Step 5: For an unbounded feasible region, check if the open half-plane determined by ax + by > M (for maximum) or ax + by < m (for minimum) has any point in common with the feasible region. If not, M or m is the optimal value; otherwise, no optimal value exists.
💡 Students often forget to check the unbounded case — they assume the smallest value at a corner point is the minimum without verifying the open half-plane condition.
Identifying Multiple Optimal Solutions
  1. Step 1: Evaluate the objective function at all corner points of the feasible region.
  2. Step 2: If two corner points give the same maximum (or minimum) value, then every point on the line segment joining these two points is also an optimal solution.
  3. Step 3: State that the problem has multiple optimal solutions.
💡 Students often think only corner points can be optimal solutions, but in the case of multiple optimal solutions, all points on the line segment between the two corner points are also optimal.
Exam Traps — Don't Make These Mistakes
⚠ Forgetting Non-Negative Restrictions
✗ Students write only the given constraints like x + y ≤ 50 and 3x + y ≤ 90, and forget to write x ≥ 0, y ≥ 0.
→ Non-negative restrictions are an essential part of the constraints in any linear programming problem; omitting them makes the formulation incomplete.
✓ Always explicitly write x ≥ 0, y ≥ 0 as part of the constraints.
⚠ Confusing Maximisation and Minimisation in Unbounded Regions
✗ Students write that the smallest value of Z at a corner point of an unbounded region is the minimum value without any further check.
→ For an unbounded feasible region, the smallest value at a corner point may not be the minimum if the open half-plane determined by ax + by < m has points in common with the feasible region.
✓ First find the smallest value m at a corner point, then check if the open half-plane ax + by < m has any point in common with the feasible region. If it does, Z has no minimum value.
⚠ Assuming Only One Optimal Solution Exists
✗ Students state that the maximum (or minimum) value occurs at exactly one corner point.
→ If two corner points give the same optimal value, then every point on the line segment joining them is also optimal — the problem has multiple optimal solutions.
✓ Check all corner points. If two give the same optimal value, mention that multiple optimal solutions exist, including all points on the line segment joining those two corner points.
⚠ Not Simplifying Constraints Before Graphing
✗ Students try to graph the constraint 2500x + 500y ≤ 50000 directly without simplifying.
→ Graphing unsimplified constraints is cumbersome and error-prone; dividing by a common factor (e.g., 100) gives a simpler form (5x + y ≤ 100) that is easier to graph.
✓ Always simplify each constraint by dividing by the greatest common divisor before graphing.
⚠ Mistaking Infeasible Region for Feasible Region
✗ Students shade the wrong side of the inequality lines and identify an infeasible region as the feasible region.
→ The feasible region must satisfy all constraints simultaneously; shading the wrong side leads to an incorrect region that does not satisfy the constraints.
✓ Test a point (e.g., (0,0) if it satisfies all constraints) to determine which side of each line to shade. The feasible region is the intersection of all shaded half-planes.
Compare & Contrast
Definition and role in LPP
Feasible solution
Optimal solution
A feasible solution is any point within or on the boundary of the feasible region that satisfies all constraints, while an optimal solution is a feasible solution that gives the maximum or minimum value of the objective function.
Existence of maximum and minimum values
Bounded feasible region
Unbounded feasible region
In a bounded feasible region, the objective function Z has both a maximum and a minimum value, each occurring at a corner point. In an unbounded feasible region, a maximum or minimum may not exist; if it does, it must occur at a corner point, but further verification using open half-planes is required.
Exam Focus
Marks pattern: 1 MCQ (1 mark) on definition or identification, 1 SA (3 marks) on formulation or graphical solution, 1 LA (5 marks) on solving a full LPP graphically including unbounded or multiple optimal solutions.
Most asked: Formulation of LPP from a word problem (identifying objective function and constraints) · Graphical method: finding feasible region and corner points · Evaluating objective function at corner points to find optimal solution · Multiple optimal solutions · Unbounded feasible region and checking for existence of optimal value
Don't skip: Solving a full LPP graphically (5 marks) — includes formulating the problem, graphing constraints, finding feasible region, listing corner points, evaluating objective function, and stating the optimal solution.; Problems with unbounded feasible region or multiple optimal solutions (3-5 marks).
One-Line Revision (read aloud before the exam)
  1. A Linear Programming Problem involves maximising or minimising a linear objective function subject to linear constraints and non-negative restrictions.
  2. The objective function is Z = ax + by, where a and b are constants and x, y are decision variables.
  3. Constraints are linear inequalities or equations that restrict the values of decision variables.
  4. Non-negative restrictions are x ≥ 0 and y ≥ 0.
  5. The feasible region is the common region determined by all constraints including non-negative restrictions.
  6. Feasible solutions are points within and on the boundary of the feasible region.
  7. Infeasible solutions are points outside the feasible region.
  8. An optimal solution is a feasible solution that gives the maximum or minimum value of the objective function.
  9. The optimal value of the objective function for a bounded feasible region always occurs at a corner point (vertex).
  10. Corner points are the intersection points of two boundary lines of the feasible region.
  11. If two corner points give the same optimal value, every point on the line segment joining them is also optimal (multiple optimal solutions).
  12. For an unbounded feasible region, check if the open half-plane ax + by > M (for maximum) or ax + by < m (for minimum) has points in common with the feasible region to decide if an optimal value exists.
  13. The Corner Point Method involves finding the feasible region, listing its corner points, evaluating Z at each, and selecting the optimal value.
  14. A bounded feasible region can be enclosed within a circle; an unbounded one cannot.
  15. Always simplify constraints before graphing (e.g., divide 2500x + 500y ≤ 50000 by 100 to get 5x + y ≤ 100).
  16. The feasible region is always a convex region.
  17. If the feasible region is empty (no point satisfies all constraints), the problem has no feasible solution.

MCQs — 9 Questions with Answers

Q1. In a linear programming problem, what is the feasible region?

  • A. The region that satisfies all constraints including non-negative restrictions ✓
  • B. The region that violates at least one constraint
  • C. The area where the objective function is zero
  • D. The region containing only the corner points of the graph

Answer: A — The feasible region is the common area determined by all constraints where each point satisfies every inequality and restriction.

Q2. A company manufactures two products X and Y. Let the profit from X be Rs 40 per unit and from Y be Rs 50 per unit. If the company can produce at most 100 units total, which statement correctly represents the objective function and one constraint?

  • A. Maximize Z = 40x + 50y; constraint: x + y ≤ 100 ✓
  • B. Minimize Z = 40x + 50y; constraint: x + y ≥ 100
  • C. Maximize Z = 100x + 100y; constraint: 40x + 50y ≤ 100
  • D. Minimize Z = 40 + 50; constraint: x + y = 100

Answer: A — Profit is maximized with objective function Z = 40x + 50y, and the production limit 'at most 100 units' translates to constraint x + y ≤ 100.

Q3. Which of the following statements is correct about the optimal solution of a linear programming problem?

  • A. It always occurs at the origin
  • B. It occurs at one of the corner points of the feasible region ✓
  • C. It occurs at the center of the feasible region
  • D. It occurs wherever the objective function equals zero

Answer: B — By the Corner Point Theorem, the optimal value of a linear objective function over a feasible region always occurs at a vertex (corner point) of that region.

Q4. A problem has constraints: x + y ≤ 8, 2x + y ≤ 12, x ≥ 0, y ≥ 0. If we want to maximize Z = 3x + 2y, which corner point will NOT be a vertex of the feasible region?

  • A. (0, 0)
  • B. (6, 0)
  • C. (4, 4) ✓
  • D. (0, 8)

Answer: C — Checking each point: (0,0) ✓, (0,8) ✓, (6,0) satisfies 2x+y=12 and x+y≤8 ✓, (4,4) ✓—all are feasible; this tests careful constraint verification.

Q5. Which of the following is NOT a requirement for formulating a linear programming problem?

  • A. All constraints must be linear inequalities or equations
  • B. Decision variables must be non-negative
  • C. The objective function must be a quadratic expression ✓
  • D. There must be an objective function to maximize or minimize

Answer: C — Linear programming requires linear objective functions of the form Z = ax + by, NOT quadratic expressions like Z = x² + y.

Q6. A company produces chairs and tables. The profit is Z = 100x + 150y. The constraints are: 2x + 3y ≤ 180 (labor hours), x + y ≤ 70 (storage), x ≥ 0, y ≥ 0. The corner points of the feasible region are (0, 0), (70, 0), (0, 60), and (45, 25). What is the maximum profit?

  • A. Rs 7000
  • B. Rs 9000
  • C. Rs 10500 ✓
  • D. Rs 11250

Answer: C — Evaluate the objective function Z = 100x + 150y at all corner points; the largest value is the maximum profit.

Q7. For a linear programming problem, which statement is true about the feasible region and the objective function?

  • A. The feasible region must always be bounded
  • B. The feasible region can be unbounded, and in that case no optimal solution exists ✓
  • C. The objective function cannot be evaluated at corner points if the feasible region is unbounded
  • D. An unbounded feasible region guarantees multiple optimal solutions

Answer: B — If the feasible region is unbounded (extends to infinity), the objective function may not have a finite maximum or minimum value, resulting in no optimal solution or unbounded solutions.

Q8. In a linear programming problem, if two corner points of the feasible region give the same maximum value for the objective function, which of the following is true?

  • A. The problem has no optimal solution
  • B. There are exactly two optimal solutions
  • C. All points on the line segment joining these two corner points are optimal solutions ✓
  • D. Only one of these corner points is actually optimal

Answer: C — When two corner points yield the same objective function value, the entire edge between them (all points on the line segment) represents optimal solutions due to the linearity of the objective function.

Q9. A factory has constraints: x + 2y ≤ 10 and 3x + y ≤ 15, with x ≥ 0, y ≥ 0. Which of the following corner points does NOT lie on the boundary of the feasible region?

  • A. (0, 0)
  • B. (5, 0)
  • C. (0, 5)
  • D. (3, 3) ✓

Answer: D — Point (3,3) lies in the interior of the feasible region since 3+2(3)=9<10 and 3(3)+3=12<15; corner points must lie on at least one constraint boundary line as an equality.

Flashcards

What is the feasible region in a linear programming problem?

The feasible region is the common area determined by all constraints (including non-negative restrictions) where every point represents a valid solution satisfying all conditions.

Define an objective function in linear programming.

An objective function Z = ax + by is a linear function of decision variables that must be maximized or minimized according to the problem requirement.

What are non-negative constraints and why are they important?

Non-negative constraints (x ≥ 0, y ≥ 0) ensure that decision variables represent real quantities like number of items, which cannot be negative in practical problems.

State the Corner Point Theorem in linear programming.

The optimal value of the objective function in a linear programming problem occurs at one of the corner points (vertices) of the feasible region.

What is the difference between feasible and infeasible regions?

The feasible region consists of all points satisfying every constraint, while the infeasible region contains points that violate at least one constraint.

How do you formulate a linear programming problem from a word problem?

Identify decision variables, write the objective function (what to maximize/minimize), list all constraint inequalities, and add non-negative restrictions.

Why must the objective function be linear in linear programming?

A linear objective function ensures the optimal solution always occurs at a corner point of the feasible region, making the graphical solution method reliable.

What does the feasible solution represent in a linear programming problem?

Each feasible solution is a point (x, y) within or on the boundary of the feasible region that satisfies all constraints and represents a valid strategy.

In the graphical method, why do we evaluate the objective function only at corner points?

For a linear objective function over a convex feasible region, the maximum or minimum always occurs at a vertex (corner point), not in the interior.

What are decision variables and what do they represent in linear programming?

Decision variables (like x and y) are the quantities we can control and optimize (e.g., number of tables and chairs) to achieve the objective function.

Important Board Questions

Define the feasible region in a linear programming problem and state the Corner Point Theorem. [2 marks]

State that feasible region is the intersection of all constraint half-planes; then define Corner Point Theorem: optimal value occurs at a vertex of the feasible region.

A manufacturer produces two products A and B. Product A requires 2 hours of labor and yields profit Rs 80, while product B requires 3 hours of labor and yields profit Rs 120. If 180 labor hours are available per week and the manufacturer can produce at most 70 units total, formulate the linear programming problem to maximize profit. Then, identify the objective function and list all constraints including non-negative restrictions. [5 marks]

Let x = units of A, y = units of B. Objective: Maximize Z = 80x + 120y. Constraints: 2x + 3y ≤ 180 (labor), x + y ≤ 70 (production limit), x ≥ 0, y ≥ 0. Show each constraint's origin clearly.

Solve the linear programming problem graphically: Maximize Z = 5x + 4y subject to x + 2y ≤ 8, 3x + y ≤ 9, x ≥ 0, y ≥ 0. Find all corner points of the feasible region, evaluate the objective function at each, and determine the optimal solution with the maximum value of Z. [6 marks]

Plot lines x + 2y = 8 and 3x + y = 9; find intersections with axes and each other: (0,0), (3,0), (0,4), and solve x+2y=8 with 3x+y=9 simultaneously to get (2,3). Verify each point satisfies both inequalities. Evaluate Z = 5x + 4y at all four corners: (0,0)→0, (3,0)→15, (0,4)→16, (2,3)→22. Maximum is Z = 22 at (2,3).

Assertion-Reason Questions for Linear Programming

Q1. Assertion (A):

The feasible region in a linear programming problem is the set of all points that satisfy all the constraints including non-negativity conditions.

Reason (R):

The feasible region is constructed by finding the intersection of all half-planes determined by the linear constraints.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why the feasible region is defined as stated. The intersection of half-planes from all constraints creates the feasible region.

Q2. Assertion (A):

In a linear programming problem, the optimal value of the objective function always occurs at a vertex of the feasible region.

Reason (R):

The feasible region is a convex polygon, and linear functions achieve their maximum and minimum values at boundary points.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—the optimal solution occurs at a vertex. However, the reason is incomplete; it should specifically mention that the optimum occurs at vertices (not just boundary points) due to the Fundamental Theorem of Linear Programming.

Q3. Assertion (A):

If the feasible region is empty, then the linear programming problem has no solution.

Reason (R):

An empty feasible region results from inconsistent constraints that cannot be satisfied simultaneously by any non-negative values of the variables.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are accurate. When constraints are mutually exclusive (e.g., x ≥ 5 and x ≤ 2), no feasible solution exists, making the problem unsolvable.

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Case-Based Questions (CBQ) for Linear Programming

Case Study:

TechGear Manufacturing produces two types of smartwatches: the FitPro and the HealthMax. The FitPro generates a profit of Rs 1,200 per unit, while the HealthMax generates Rs 1,500 per unit. Production is constrained by machine hours and labour availability. The FitPro requires 2 hours of machine time and 1 hour of labour, whereas the HealthMax requires 1 hour of machine time and 2 hours of labour. The company has a maximum of 100 machine hours and 120 labour hours available per week. The company aims to maximise weekly profit while meeting these constraints.

Q1. Let F = number of FitPro units and H = number of HealthMax units. Which linear constraint represents the machine hour limitation?

  • A. 2F + H ≤ 100 ✓
  • B. F + 2H ≤ 100
  • C. 2F + H ≤ 120
  • D. F + 2H ≤ 120

Answer: A

Q2. What is the objective function that TechGear should maximise?

  • A. Z = 1,200F + 1,500H ✓
  • B. Z = 1,500F + 1,200H
  • C. Z = 100F + 120H
  • D. Z = 2F + H + F + 2H

Answer: A

Q3. If the feasible region has vertices at (0, 0), (50, 0), (0, 60), and (40, 20), at which vertex is profit maximised?

  • A. (50, 0) with profit Rs 60,000
  • B. (0, 60) with profit Rs 90,000
  • C. (40, 20) with profit Rs 78,000
  • D. (40, 20) with profit Rs 84,000 ✓

Answer: D

Q4. Which of the following production plans is NOT feasible for TechGear?

  • A. 30 FitPro and 40 HealthMax units
  • B. 25 FitPro and 50 HealthMax units
  • C. 45 FitPro and 10 HealthMax units
  • D. 20 FitPro and 50 HealthMax units ✓

Answer: D

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Previous Year Questions (PYQ) for Linear Programming

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

The feasible region of a linear programming problem with objective function Z = 5x + 7y is shown below. The maximum value of Z – minimum value of Z is

  • 8
  • 29
  • 35
  • 43

CBSE 2026 · Set 65/1/1 · 1 mark

The degree of an objective function of a linear programming problem is

  • 0
  • 1
  • 2
  • Any natural number

CBSE 2026 · Set 65/1/1 · 3 marks

Solve the following linear programming problem graphically: Minimize Z = 13x – 15y subject to constraints x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0.

CBSE 2026 · Set 65/1/2 · 3 marks

Solve the following linear programming problem graphically: Minimize Z = 13x – 15y Subject to constraints: x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0

CBSE 2026 · Set 65/1/3 · 3 marks

Solve the following linear programming problem graphically: Minimize Z = 13x – 15y Subject to constraints x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0

CBSE 2026 · Set 65/2/1 · 1 mark

In a linear programming problem, the linear function which has to be maximized or minimized is called

  • a feasible function
  • an objective function
  • an optimal function
  • a constraint

CBSE 2026 · Set 65/2/1 · 1 mark

For the feasible region shown below, the non-trivial constraints of the linear programming problem are

  • x + y ≤ 5, x + 3y ≤ 9
  • x + y ≥ 5, x + 3y ≤ 9
  • x + y ≤ 5, x + 3y ≥ 9
  • x + y ≤ 5, 3x + y ≤ 9

CBSE 2026 · Set 65/2/1 · 3 marks

Solve the following linear programming problem graphically: Maximize Z = 10500x + 9000y Subject to constraints: x + y ≤ 50, 2x + y ≤ 80, x, y ≥ 0

CBSE 2026 · Set 65/2/2 · 3 marks

Solve the following linear programming problem graphically: Maximize Z = 8000x + 12000y Subject to constraints 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

The corner points of the feasible region for an LPP are (0, 2), (6, 0), (6, 8) and (0, 5). Let Z = 4x + 6y be the objective function. (Maximum of Z) – (Minimum of Z) is equal to :

  • 60
  • 48
  • 42
  • 18

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

In an LPP, if the objective function has the same minimum value on two corner points of the feasible region, then the number of points at which the minimum value of objective function occurs is :

  • 0
  • 2
  • Infinite
  • Finite

CBSE 2026 · Set 65(B) R Mathematics · 3 marks

The corner points of the feasible region determined by the system of linear constraints in an LPP are (0, 10), (5, 5), (15, 15) and (0, 20). Let Z = px + qy, p, q > 0, be the objective function, then find the relation between p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20).

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Class 12 Maths Formula Sheet

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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