Class 12 Maths · Chapter 12 – swipe through all 6 cards to understand the whole chapter.
What is Linear Programming?
Optimising (max or min) a linear quantity under linear constraints.
Z = objective function
- Used for profit/cost problems
- All relations are linear
- x, y ≥ 0 (non-negative)
Constraints & Feasible Region
Inequalities that limit the solution.
Feasible region = common area
- Constraints = linear inequalities
- Feasible region: all satisfied
- Corners = corner points
Corner Point Method
The optimum lies at a corner of the feasible region.
Pick the max / min value
- Find all corner points
- Compute Z at each
- Largest = max, smallest = min
Bounded vs Unbounded
The feasible region may be closed or open.
Unbounded → may not exist
- Bounded: enclosed region
- Unbounded: open region
- Check open regions carefully
Optimal Solution
The corner point that gives the best value of Z.
May occur at two corners
- Unique or multiple optima
- On an edge ⇒ infinite optima
- Always at the boundary
Graphing Tips
How to plot and read the feasible region.
x ≥ 0, y ≥ 0 → first quadrant
- Convert ≤ / ≥ to lines
- Test origin to pick the side
- Mark all corner coordinates
Linear Programming (Class 12 Maths) is a method to maximise or minimise a linear objective function subject to linear constraints. You write the objective Z = ax + by, list the constraints as inequalities, and plot them to get a feasible region (a convex polygon). The optimal value of Z always occurs at a corner (vertex) of this feasible region, so you evaluate Z at every corner point and pick the largest or smallest value. Typical problems are manufacturing, diet and transportation problems, solved by the corner-point method.
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📥 Download PDF (Free)Linear Programming (LP) is an optimisation technique used to find the best outcome (maximum profit or minimum cost) given constraints. This chapter carries 5 marks in CBSE boards and is one of the most scoring chapters - once you master the graphical method, you can solve any LP problem!
Key Concepts
1. What is Linear Programming?
It is a method to achieve the best outcome (maximum or minimum) in a mathematical model whose requirements are represented by linear relationships.
| Term | Meaning | Example |
|---|---|---|
| Objective Function | The function to be maximised or minimised | Z = 3x + 5y (profit function) |
| Decision Variables | Variables we control | x = units of A, y = units of B |
| Constraints | Limitations/restrictions (linear inequalities) | 2x + y ≤ 100 (machine hours) |
| Non-negativity | Variables can’t be negative | x ≥ 0, y ≥ 0 |
| Feasible Region | Set of all points satisfying all constraints | Shaded region on graph |
| Feasible Solution | Any point in the feasible region | (10, 20) |
| Optimal Solution | Feasible solution that optimises Z | The point giving max/min Z |
2. Corner Point Method (Steps)
Step 2: Graph each constraint as an equation (line), shade the feasible region
Step 3: Find all corner points (vertices) of the feasible region
Step 4: Calculate Z at each corner point
Step 5: The corner point giving max/min Z is the optimal solution
3. Types of Feasible Regions
| Type | Description | Optimal Solution |
|---|---|---|
| Bounded | Enclosed region (finite area) | Always exists (both max and min) |
| Unbounded | Region extends infinitely | May or may not exist - need verification |
4. Types of LP Problems in CBSE
| Type | Objective | Common Scenario |
|---|---|---|
| Manufacturing | Maximise profit | Units of products A and B |
| Diet Problem | Minimise cost | Amount of food items for nutrition |
| Transportation | Minimise cost | Shipping goods from factories to warehouses |
| Allocation | Maximise output | Distributing resources |
5. Graphing Tips
- Convert inequality to equation to plot the line
- Use test point (0,0) to determine which side to shade
- Find intersection points by solving pairs of equations simultaneously
- Don’t forget non-negativity constraints (x ≥ 0, y ≥ 0) - this restricts the region to the first quadrant
Important Definitions
| Term | Definition |
|---|---|
| Linear Programming | Optimisation technique with linear objective function and linear constraints |
| Feasible Region | Set of all points satisfying all constraints simultaneously |
| Optimal Solution | A feasible solution at which the objective function has max or min value |
| Corner Point | Vertex of the feasible region where constraint boundaries intersect |
| Bounded Region | Feasible region that can be enclosed in a circle |
Solved Examples - NCERT-Based
Example 1: Manufacturing Problem
A factory makes two products A and B. Each A requires 2 hrs on machine I and 1 hr on machine II. Each B requires 1 hr on machine I and 3 hrs on machine II. Machine I is available for 10 hrs, machine II for 12 hrs. Profit on A = ₹40, on B = ₹60. Find the maximum profit.
Solution:
Let x = units of A, y = units of B
Maximise Z = 40x + 60y
Subject to: 2x + y ≤ 10, x + 3y ≤ 12, x ≥ 0, y ≥ 0
Corner points: (0,0), (5,0), (0,4), and intersection of 2x+y=10 and x+3y=12
Solving: 2x+y=10 → y=10−2x → x+3(10−2x)=12 → x+30−6x=12 → −5x=−18 → x=18/5=3.6, y=2.8
Corner points and Z values:
| Corner Point | Z = 40x + 60y |
|---|---|
| (0, 0) | 0 |
| (5, 0) | 200 |
| (3.6, 2.8) | 144 + 168 = 312 |
| (0, 4) | 240 |
Maximum profit = ₹312 at (3.6, 2.8)
Example 2: Diet Problem
A diet requires at least 10 units of vitamin A and 12 units of vitamin B. Food I costs ₹6/unit and provides 2 units of A, 1 unit of B. Food II costs ₹10/unit and provides 1 unit of A, 4 units of B. Find minimum cost.
Solution:
Let x = units of Food I, y = units of Food II
Minimise Z = 6x + 10y
Subject to: 2x + y ≥ 10, x + 4y ≥ 12, x ≥ 0, y ≥ 0
Corner points: (0, 10), (12, 0), intersection of 2x+y=10 and x+4y=12
Solving: y = 10−2x → x + 4(10−2x) = 12 → x + 40−8x = 12 → −7x = −28 → x=4, y=2
| Corner Point | Z = 6x + 10y |
|---|---|
| (0, 10) | 100 |
| (4, 2) | 44 |
| (12, 0) | 72 |
Minimum cost = ₹44 at x = 4 units of Food I, y = 2 units of Food II
Example 3: Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Solution:
Corner points: (0,0), (2,0), (0,3)
Intersection: 3x+5y=15 and 5x+2y=10
From 5x+2y=10: y = (10−5x)/2. Substituting: 3x + 5(10−5x)/2 = 15 → 6x + 50−25x = 30 → −19x = −20 → x = 20/19, y = 45/19
| Corner Point | Z = 5x + 3y |
|---|---|
| (0, 0) | 0 |
| (2, 0) | 10 |
| (20/19, 45/19) | 100/19 + 135/19 = 235/19 ≈ 12.37 |
| (0, 3) | 9 |
Maximum Z = 235/19 ≈ 12.37 at (20/19, 45/19)
Example 4: Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0
Solution:
Corner points of unbounded feasible region: (6, 0), (0, 3)
Intersection: 2x+y=3 and x+2y=6 → y=3−2x → x+2(3−2x)=6 → x+6−4x=6 → x=0, y=3
Also check: (6, 0)
| Corner Point | Z = x + 2y |
|---|---|
| (6, 0) | 6 |
| (0, 3) | 6 |
Minimum Z = 6 (achieved at both points and all points on the line segment joining them)
Important Questions for Board Exams
1 Mark Questions
- Define a feasible region in an LP problem.
- What is the corner point theorem?
2 Mark Questions
- Draw the feasible region for: x + y ≤ 4, x ≥ 0, y ≥ 0 and find its corner points.
- If the feasible region is bounded with vertices (0,0), (4,0), (0,3), find max Z = 7x + 4y.
5 Mark Questions
- Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, y ≥ 0. Graph the constraints and find optimal solution.
- A dealer deals in two products A and B. He has ₹30,000 to invest and space for 20 items. Product A costs ₹2,500, B costs ₹500. He can sell A at ₹500 profit and B at ₹150 profit. Formulate the LP problem and solve graphically for maximum profit.
Quick Revision Points
- LP = optimise a linear function subject to linear constraints
- Always check corner points - optimal solution is always at a vertex
- Steps: Formulate → Graph → Find corners → Evaluate Z → Pick best
- Bounded region → max and min both exist
- Unbounded region → may need extra verification
- Non-negativity (x ≥ 0, y ≥ 0) restricts to first quadrant
- If Z has same value at two corners → all points on that edge are optimal
- CBSE mostly asks 5-mark graphical problems - practice drawing accurate graphs!
Chapter Navigation
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Frequently Asked Questions
A linear programming problem asks you to maximise or minimise a linear objective function Z = ax + by subject to a set of linear inequalities called constraints, along with non-negativity conditions x, y greater than or equal to 0. It models real allocation problems like manufacturing, diet and transport.
The feasible region is the set of all points that satisfy every constraint; graphically it is the common shaded area, usually a convex polygon. The corner-point method says the optimal value of Z occurs at a vertex (corner) of this region, so you evaluate Z at each corner and choose the maximum or minimum.
A feasible solution is any point inside or on the boundary of the feasible region that satisfies all constraints. An optimal solution is the feasible point that gives the best (maximum or minimum) value of the objective function Z, and it always lies at a corner point of the feasible region.
If the feasible region is unbounded, a maximum or minimum may not exist. You evaluate Z at the corner points and then check, using an open half-plane test, whether Z can be made larger or smaller without limit; if it can, there is no optimal value.
Common Class 12 linear programming problems are manufacturing problems (maximise profit), diet problems (minimise cost while meeting nutrition needs), and transportation problems (minimise shipping cost). All are set up with a linear objective and linear constraints.