Chapter 12: Linear Programming
Complete Board Exam Focused Notes with Graphical Method & PYQs
Exam Weightage & Blueprint
Total: ~5 MarksLinear Programming focuses purely on graphical solutions of optimization problems. Word problems/formulation has been removed from the latest CBSE syllabus. Focus is entirely on the graphical method.
| Question Type | Marks | Frequency | Focus Topic |
|---|---|---|---|
| MCQ | 1 | High | Identifying Feasible Region, Corner Points |
| Short Answer (2M) | 2 | Medium | Simple Graphical Solution |
| Long Answer (5M) | 5 | Very High | Complete Graphical Solution (Given Formulation) |
⚠️ Important Note: As per latest CBSE syllabus, formulation of LPP from word problems is NOT included. Only graphical method to solve given LPP is required.
⏰ Last 24-Hour Checklist
Key Concepts (Must Know!)
- ☑ Objective Function: Z = ax + by
- ☑ Constraints: Linear inequalities
- ☑ Feasible Region: Common region satisfying all constraints
- ☑ Corner Point Method
- ☑ Bounded vs Unbounded Region
- ☑ Optimal Solution at Corner Points
- ☑ Graphing Linear Inequalities
- ☑ Finding Intersection Points
Important Steps
- ☑ Graph all constraints correctly
- ☑ Identify feasible region
- ☑ Find all corner points
- ☑ Evaluate Z at each corner point
- ☑ Check bounded/unbounded region
- ☑ For unbounded: verify max/min exists
- ☑ State final answer clearly
- ☑ Show all calculations
Key Terminology ★★★★★
Linear Programming Problem (LPP)
A problem concerned with finding the optimal value (maximum or minimum) of a linear function of several variables, subject to conditions that variables are non-negative and satisfy a set of linear inequalities.
Essential Terms
| Term | Definition | Example |
|---|---|---|
| Decision Variables | Variables whose values are to be determined | x (tables), y (chairs) |
| Objective Function | Linear function to be maximized or minimized: Z = ax + by | Z = 250x + 75y (profit) |
| Constraints | Linear inequalities or restrictions on variables | x + y ≤ 60 (storage limit) |
| Non-negative Constraints | Conditions that variables must be ≥ 0 | x ≥ 0, y ≥ 0 |
| Feasible Region | Common region satisfying all constraints | Shaded region in graph |
| Feasible Solution | Any point in feasible region | (10, 20), (5, 30) |
| Corner Point | Vertex where two boundary lines intersect | Intersection of x + y = 60 and 5x + y = 100 |
| Optimal Solution | Feasible solution giving optimal value of Z | Point giving maximum profit |
| Bounded Region | Feasible region that can be enclosed in a circle | Closed polygon |
| Unbounded Region | Feasible region extending indefinitely | Open region |
Memory Aid: Think of LPP as a business optimization problem
• Variables = Your decisions (what to buy/sell)
• Objective = Your goal (maximize profit / minimize cost)
• Constraints = Your limitations (budget, space, time)
• Optimal Solution = Best decision within your limitations
• Variables = Your decisions (what to buy/sell)
• Objective = Your goal (maximize profit / minimize cost)
• Constraints = Your limitations (budget, space, time)
• Optimal Solution = Best decision within your limitations
Understanding LPP Components 🔥🔥🔥
Standard Form of LPP (Given in Exam):
Optimize (Maximize or Minimize) Z = ax + by
Subject to:
• Constraint 1: $a_1x + b_1y$ (≤, ≥, or =) $c_1$
• Constraint 2: $a_2x + b_2y$ (≤, ≥, or =) $c_2$
• ...
• Non-negative constraints: x ≥ 0, y ≥ 0
⚠️ Note: You will be given the complete formulation. No need to formulate from word problems.
Optimize (Maximize or Minimize) Z = ax + by
Subject to:
• Constraint 1: $a_1x + b_1y$ (≤, ≥, or =) $c_1$
• Constraint 2: $a_2x + b_2y$ (≤, ≥, or =) $c_2$
• ...
• Non-negative constraints: x ≥ 0, y ≥ 0
⚠️ Note: You will be given the complete formulation. No need to formulate from word problems.
Example: Complete LPP Given in Exam
Typical Exam Question:
Maximize Z = 250x + 75y
Subject to:
• 5x + y ≤ 100
• x + y ≤ 60
• x ≥ 0, y ≥ 0
Your Task: Solve graphically using corner point method (no need to formulate - it's already given!)
Maximize Z = 250x + 75y
Subject to:
• 5x + y ≤ 100
• x + y ≤ 60
• x ≥ 0, y ≥ 0
Your Task: Solve graphically using corner point method (no need to formulate - it's already given!)
What You Need to Focus On:
✓ Graphing the constraints correctly
✓ Finding the feasible region
✓ Calculating corner points accurately
✓ Evaluating Z at each corner point
✓ Identifying maximum/minimum value
✗ NOT Required: Formulating LPP from word problems, identifying decision variables from scenarios, writing objective functions from descriptions
✓ Graphing the constraints correctly
✓ Finding the feasible region
✓ Calculating corner points accurately
✓ Evaluating Z at each corner point
✓ Identifying maximum/minimum value
✗ NOT Required: Formulating LPP from word problems, identifying decision variables from scenarios, writing objective functions from descriptions
Graphical Method - Corner Point Method ★★★★★
Fundamental Theorems:
Theorem 1: The optimal value of the objective function occurs at a corner point (vertex) of the feasible region.
Theorem 2: If the feasible region is bounded, both maximum and minimum values exist and occur at corner points.
Theorem 1: The optimal value of the objective function occurs at a corner point (vertex) of the feasible region.
Theorem 2: If the feasible region is bounded, both maximum and minimum values exist and occur at corner points.
Corner Point Method - Complete Steps
Step 1: Graph the Constraints
• Convert each inequality to equation (replace ≤ or ≥ with =)
• Plot the line by finding two points (usually x-intercept and y-intercept)
• Shade the region satisfying the inequality
Step 2: Identify Feasible Region
• Find the common region satisfying ALL constraints
• This is the feasible region (usually shaded)
• Check if it's bounded or unbounded
Step 3: Find Corner Points
• Identify vertices of the feasible region
• Find coordinates by solving simultaneous equations of intersecting lines
• Or read directly from graph
Step 4: Evaluate Objective Function
• Calculate Z = ax + by at each corner point
• Make a table of corner points and Z values
Step 5: Determine Optimal Solution
• For bounded region: Largest value = Maximum, Smallest value = Minimum
• For unbounded region: Additional verification needed (see below)
• Convert each inequality to equation (replace ≤ or ≥ with =)
• Plot the line by finding two points (usually x-intercept and y-intercept)
• Shade the region satisfying the inequality
Step 2: Identify Feasible Region
• Find the common region satisfying ALL constraints
• This is the feasible region (usually shaded)
• Check if it's bounded or unbounded
Step 3: Find Corner Points
• Identify vertices of the feasible region
• Find coordinates by solving simultaneous equations of intersecting lines
• Or read directly from graph
Step 4: Evaluate Objective Function
• Calculate Z = ax + by at each corner point
• Make a table of corner points and Z values
Step 5: Determine Optimal Solution
• For bounded region: Largest value = Maximum, Smallest value = Minimum
• For unbounded region: Additional verification needed (see below)
Special Case: Unbounded Region
For Unbounded Feasible Region:
For Maximum: If M is the largest value of Z at corner points,
• Graph the inequality ax + by > M
• If this half-plane has NO common points with feasible region → M is the maximum
• If it has common points → NO maximum exists
For Minimum: If m is the smallest value of Z at corner points,
• Graph the inequality ax + by < m
• If this half-plane has NO common points with feasible region → m is the minimum
• If it has common points → NO minimum exists
For Maximum: If M is the largest value of Z at corner points,
• Graph the inequality ax + by > M
• If this half-plane has NO common points with feasible region → M is the maximum
• If it has common points → NO maximum exists
For Minimum: If m is the smallest value of Z at corner points,
• Graph the inequality ax + by < m
• If this half-plane has NO common points with feasible region → m is the minimum
• If it has common points → NO minimum exists
⚠️ Common Error: Students often forget to check for unbounded regions! Always verify whether the feasible region is bounded or unbounded before concluding the optimal value.
Graphing Linear Inequalities 🔥🔥
Quick Guide to Shading
| Inequality | Line Type | Region to Shade | Test Point |
|---|---|---|---|
| ax + by ≤ c | Solid line | Below/Left of line | Check (0,0) if possible |
| ax + by ≥ c | Solid line | Above/Right of line | Check (0,0) if possible |
| ax + by < c | Dashed line | Below/Left of line | Check (0,0) if possible |
| ax + by > c | Dashed line | Above/Right of line | Check (0,0) if possible |
| x ≥ 0 | Y-axis (solid) | Right side of Y-axis | - |
| y ≥ 0 | X-axis (solid) | Above X-axis | - |
Steps to Plot a Line
For inequality: ax + by ≤ c
Step 1: Convert to equation: ax + by = c
Step 2: Find x-intercept (put y = 0): x = c/a → Point (c/a, 0)
Step 3: Find y-intercept (put x = 0): y = c/b → Point (0, c/b)
Step 4: Plot these two points and draw the line
Step 5: Test point (0, 0) or any convenient point:
• If point satisfies inequality → shade the side containing that point
• If not → shade the opposite side
Step 1: Convert to equation: ax + by = c
Step 2: Find x-intercept (put y = 0): x = c/a → Point (c/a, 0)
Step 3: Find y-intercept (put x = 0): y = c/b → Point (0, c/b)
Step 4: Plot these two points and draw the line
Step 5: Test point (0, 0) or any convenient point:
• If point satisfies inequality → shade the side containing that point
• If not → shade the opposite side
Quick Test for Origin:
If the line does NOT pass through origin (0,0):
• Substitute (0,0) in the inequality
• If TRUE → shade the region containing origin
• If FALSE → shade the region NOT containing origin
Example: For x + y ≤ 60
Test (0,0): 0 + 0 ≤ 60 ✓ TRUE
So shade the region containing origin
If the line does NOT pass through origin (0,0):
• Substitute (0,0) in the inequality
• If TRUE → shade the region containing origin
• If FALSE → shade the region NOT containing origin
Example: For x + y ≤ 60
Test (0,0): 0 + 0 ≤ 60 ✓ TRUE
So shade the region containing origin
Visual Examples of Graphing
Example 1: Graphing x + y ≤ 50
The shaded region shows all points satisfying x + y ≤ 50 with x ≥ 0, y ≥ 0
Example 2: Graphing 2x + y ≥ 100
For ≥ inequalities, the feasible region is ABOVE/RIGHT of the line
Solved Examples (Board Marking Scheme)
Example 1: Maximize Z = 4x + y (6 Marks)
Subject to: x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0
Graph with Feasible Region:
Step 1: Graph the constraints 1.5 Marks
Line 1: x + y = 50 → Points: (50, 0) and (0, 50)
Line 2: 3x + y = 90 → Points: (30, 0) and (0, 90)
Also plot x = 0 (Y-axis) and y = 0 (X-axis)
Shade the feasible region OABC (bounded)
Step 2: Find corner points 1.5 Marks
O: (0, 0) - Origin
A: (30, 0) - Intersection of 3x + y = 90 and y = 0
B: Solve x + y = 50 and 3x + y = 90 simultaneously:
Subtracting: 2x = 40 → x = 20
From x + y = 50: y = 30
So B = (20, 30)
C: (0, 50) - Intersection of x + y = 50 and x = 0
Step 3: Evaluate Z at corner points 2 Marks
| Corner Point | Z = 4x + y |
|---|---|
| O (0, 0) | 0 |
| A (30, 0) | 120 ← Maximum |
| B (20, 30) | 110 |
| C (0, 50) | 50 |
Step 4: Conclusion 1 Mark
The feasible region is bounded. Maximum value of Z is 120 at point (30, 0)
Answer: x = 30, y = 0, Maximum Z = 120
Example 2: Minimize Z = 200x + 500y (6 Marks)
Subject to: x + 2y ≥ 10, 3x + 4y ≤ 24, x ≥ 0, y ≥ 0
Graph with Feasible Region:
Step 1: Graph the constraints 1.5 Marks
Line 1: x + 2y = 10 → Points: (10, 0) and (0, 5)
For x + 2y ≥ 10, shade above the line
Line 2: 3x + 4y = 24 → Points: (8, 0) and (0, 6)
For 3x + 4y ≤ 24, shade below the line
Feasible region is ABC (bounded)
Step 2: Find corner points 1.5 Marks
A: (0, 5) - Intersection of x + 2y = 10 and x = 0
B: Solve x + 2y = 10 and 3x + 4y = 24:
From first: x = 10 - 2y
Substitute: 3(10 - 2y) + 4y = 24
30 - 6y + 4y = 24 → -2y = -6 → y = 3
x = 10 - 6 = 4
So B = (4, 3)
C: (0, 6) - Intersection of 3x + 4y = 24 and x = 0
Step 3: Evaluate Z 2 Marks
| Corner Point | Z = 200x + 500y |
|---|---|
| A (0, 5) | 2500 |
| B (4, 3) | 2300 ← Minimum |
| C (0, 6) | 3000 |
Step 4: Conclusion 1 Mark
Minimum value of Z is 2300 at point (4, 3)
Example 3: Min & Max Z = 3x + 9y (6 Marks)
Subject to: x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0
Solution: 6 Marks
Graph all constraints and find feasible region ABCD (bounded)
Corner Points:
A (0, 10): Intersection of x = 0 and x + y = 10
B (5, 5): Solve x + y = 10 and x = y → 2x = 10 → x = 5
C (15, 15): Solve x + 3y = 60 and x = y → 4y = 60 → y = 15
D (0, 20): Intersection of x = 0 and x + 3y = 60
| Corner Point | Z = 3x + 9y |
|---|---|
| A (0, 10) | 90 |
| B (5, 5) | 60 ← Minimum |
| C (15, 15) | 180 ← Maximum |
| D (0, 20) | 180 ← Maximum |
Answer:
Minimum Z = 60 at (5, 5)
Maximum Z = 180 at both C and D (Multiple optimal solutions exist on line segment CD)
Example 4: Unbounded Region - Minimize Z = -50x + 20y (6 Marks)
Subject to: 2x - y ≥ -5, 3x + y ≥ 3, 2x - 3y ≤ 12, x ≥ 0, y ≥ 0
Graph with Unbounded Feasible Region:
Solution: 6 Marks
Graph constraints - feasible region is unbounded (extends indefinitely)
Corner Points: (0, 5), (0, 3), (1, 0), (6, 0)
| Corner Point | Z = -50x + 20y |
|---|---|
| (0, 5) | 100 |
| (0, 3) | 60 |
| (1, 0) | -50 |
| (6, 0) | -300 ← Smallest |
Check for unbounded region:
Since region is unbounded, check if -300 is actually the minimum.
Graph the half-plane: -50x + 20y < -300
Simplify: -5x + 2y < -30
This half-plane HAS common points with the feasible region (extends to infinity)
Conclusion: Z has NO minimum value
Previous Year Questions (PYQs)
2023 (1 Mark MCQ): The corner points of the feasible region determined by the system x + y ≤ 6, 2x + y ≤ 8, x ≥ 0, y ≥ 0 are:
(A) (0,0), (4,0), (2,4), (0,6)
(B) (0,0), (2,0), (4,2), (0,8)
(C) (0,0), (4,0), (3,2), (0,6)
(D) (0,0), (0,6), (2,4), (4,0)
Answer: (A)
Solution: Graph the constraints. Find intersections: (0,0), (4,0) from 2x+y=8 & y=0, (2,4) from x+y=6 & 2x+y=8, (0,6) from x+y=6 & x=0
(A) (0,0), (4,0), (2,4), (0,6)
(B) (0,0), (2,0), (4,2), (0,8)
(C) (0,0), (4,0), (3,2), (0,6)
(D) (0,0), (0,6), (2,4), (4,0)
Answer: (A)
Solution: Graph the constraints. Find intersections: (0,0), (4,0) from 2x+y=8 & y=0, (2,4) from x+y=6 & 2x+y=8, (0,6) from x+y=6 & x=0
2023 (5 Marks): Solve graphically:
Maximize Z = 30x + 40y
Subject to: 2x + y ≤ 20, x + 2y ≤ 16, x ≥ 0, y ≥ 0
Solution:
Corner points: (0,0), (10,0), (8,4), (0,8)
Z values: 0, 300, 400, 320
Answer: Maximum Z = 400 at (8,4)
Maximize Z = 30x + 40y
Subject to: 2x + y ≤ 20, x + 2y ≤ 16, x ≥ 0, y ≥ 0
Solution:
Corner points: (0,0), (10,0), (8,4), (0,8)
Z values: 0, 300, 400, 320
Answer: Maximum Z = 400 at (8,4)
2022 (5 Marks): Solve graphically:
Minimize Z = 50x + 70y
Subject to: 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Solution:
Corner points: (0,10), (2,4), (8,0)
Z values: 700, 380, 400
Answer: Minimum Z = 380 at (2,4)
Minimize Z = 50x + 70y
Subject to: 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Solution:
Corner points: (0,10), (2,4), (8,0)
Z values: 700, 380, 400
Answer: Minimum Z = 380 at (2,4)
2021 (5 Marks): Minimize and Maximize Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0.
Solution:
Graph all constraints to find feasible region (bounded)
Corner points: (60,0), (120,0), (60,30), (40,20)
Z at (60,0) = 300
Z at (120,0) = 600
Z at (60,30) = 600
Z at (40,20) = 400
Answer:
Minimum Z = 300 at (60,0)
Maximum Z = 600 at (120,0) and (60,30) - multiple solutions
Solution:
Graph all constraints to find feasible region (bounded)
Corner points: (60,0), (120,0), (60,30), (40,20)
Z at (60,0) = 300
Z at (120,0) = 600
Z at (60,30) = 600
Z at (40,20) = 400
Answer:
Minimum Z = 300 at (60,0)
Maximum Z = 600 at (120,0) and (60,30) - multiple solutions
2020 (5 Marks): Solve graphically: Maximize Z = x + y subject to x - y ≤ -1, -x + y ≤ 0, x ≥ 0, y ≥ 0.
Solution:
Graph the constraints
Observe that feasible region has no points satisfying all constraints simultaneously
Answer: No feasible solution exists
Solution:
Graph the constraints
Observe that feasible region has no points satisfying all constraints simultaneously
Answer: No feasible solution exists
2019 (5 Marks): Maximize Z = 4x + y subject to x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0
Solution:
Corner points: O(0,0), A(30,0), B(20,30), C(0,50)
Z values: 0, 120, 110, 50
Answer: Maximum Z = 120 at (30,0)
Solution:
Corner points: O(0,0), A(30,0), B(20,30), C(0,50)
Z values: 0, 120, 110, 50
Answer: Maximum Z = 120 at (30,0)
Previous Year Questions (PYQs)
2023 (1 Mark MCQ): The corner points of the feasible region determined by the system x + y ≤ 6, 2x + y ≤ 8, x ≥ 0, y ≥ 0 are:
(A) (0,0), (4,0), (2,4), (0,6)
(B) (0,0), (2,0), (4,2), (0,8)
(C) (0,0), (4,0), (3,2), (0,6)
(D) (0,0), (0,6), (2,4), (4,0)
Answer: (A)
Solution: Graph the constraints. Find intersections: (0,0), (4,0) from 2x+y=8 & y=0, (2,4) from x+y=6 & 2x+y=8, (0,6) from x+y=6 & x=0
(A) (0,0), (4,0), (2,4), (0,6)
(B) (0,0), (2,0), (4,2), (0,8)
(C) (0,0), (4,0), (3,2), (0,6)
(D) (0,0), (0,6), (2,4), (4,0)
Answer: (A)
Solution: Graph the constraints. Find intersections: (0,0), (4,0) from 2x+y=8 & y=0, (2,4) from x+y=6 & 2x+y=8, (0,6) from x+y=6 & x=0
2023 (6 Marks): A manufacturer produces two types of products A and B. Each unit of A requires 2 hours on machine I and 1 hour on machine II. Each unit of B requires 1 hour on machine I and 2 hours on machine II. Machine I is available for 20 hours and machine II for 16 hours. Profit on product A is Rs 30 per unit and on B is Rs 40 per unit. Formulate and solve graphically.
Solution:
Let x = units of A, y = units of B
Maximize Z = 30x + 40y
Subject to: 2x + y ≤ 20 (Machine I)
x + 2y ≤ 16 (Machine II)
x ≥ 0, y ≥ 0
Corner points: (0,0), (10,0), (8,4), (0,8)
Z values: 0, 300, 400, 320
Answer: Maximum profit Rs 400 at x=8, y=4
Solution:
Let x = units of A, y = units of B
Maximize Z = 30x + 40y
Subject to: 2x + y ≤ 20 (Machine I)
x + 2y ≤ 16 (Machine II)
x ≥ 0, y ≥ 0
Corner points: (0,0), (10,0), (8,4), (0,8)
Z values: 0, 300, 400, 320
Answer: Maximum profit Rs 400 at x=8, y=4
2022 (6 Marks): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 50 per kg to purchase Food I and Rs 70 per kg to purchase Food II. Formulate and solve to minimize cost.
Solution:
Let x = kg of Food I, y = kg of Food II
Minimize Z = 50x + 70y
Subject to: 2x + y ≥ 8 (Vitamin A)
x + 2y ≥ 10 (Vitamin C)
x ≥ 0, y ≥ 0
Corner points: (0,10), (2,4), (8,0)
Z values: 700, 380, 400
Answer: Minimum cost Rs 380 at x=2, y=4
Solution:
Let x = kg of Food I, y = kg of Food II
Minimize Z = 50x + 70y
Subject to: 2x + y ≥ 8 (Vitamin A)
x + 2y ≥ 10 (Vitamin C)
x ≥ 0, y ≥ 0
Corner points: (0,10), (2,4), (8,0)
Z values: 700, 380, 400
Answer: Minimum cost Rs 380 at x=2, y=4
2021 (6 Marks): Minimize and Maximize Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0.
Solution:
Graph all constraints to find feasible region (bounded)
Corner points: (60,0), (120,0), (60,30), (40,20)
Z at (60,0) = 300
Z at (120,0) = 600
Z at (60,30) = 600
Z at (40,20) = 400
Answer:
Minimum Z = 300 at (60,0)
Maximum Z = 600 at (120,0) and (60,30) - multiple solutions
Solution:
Graph all constraints to find feasible region (bounded)
Corner points: (60,0), (120,0), (60,30), (40,20)
Z at (60,0) = 300
Z at (120,0) = 600
Z at (60,30) = 600
Z at (40,20) = 400
Answer:
Minimum Z = 300 at (60,0)
Maximum Z = 600 at (120,0) and (60,30) - multiple solutions
2020 (6 Marks): Solve graphically: Maximize Z = x + y subject to x - y ≤ -1, -x + y ≤ 0, x ≥ 0, y ≥ 0.
Solution:
Graph the constraints
Observe that feasible region has no points satisfying all constraints simultaneously
Answer: No feasible solution exists
Solution:
Graph the constraints
Observe that feasible region has no points satisfying all constraints simultaneously
Answer: No feasible solution exists
Exam Strategy & Mistake Bank
Common Mistakes 🚨
Mistake 1: Incorrect shading of inequalities. Always test a point (usually origin) to verify which side to shade.
Mistake 2: Forgetting non-negative constraints x ≥ 0, y ≥ 0. These are ALWAYS required unless stated otherwise.
Mistake 3: Not checking if region is bounded/unbounded. This affects whether maximum/minimum exists.
Mistake 4: Wrong calculation of corner points. Always solve simultaneous equations carefully.
Mistake 5: Forgetting to state units in word problems. "Maximum profit is 500" vs "Maximum profit is Rs 500"
Mistake 6: Evaluating Z at wrong points. Only corner points of feasible region should be tested.
Scoring Tips 🏆
Tip 1: Draw neat, labeled graphs. Mark axes, lines, corner points clearly. Use ruler!
Tip 2: Show all working for corner points. Write the two equations being solved.
Tip 3: Make a proper table for evaluating Z at corner points. This makes checking easy.
Tip 4: Write conclusion clearly: "Maximum value of Z is ___ at point (___,___)"
Tip 5: For word problems, clearly define variables at the start: "Let x = number of tables"
Tip 6: If graph paper is not provided, draw a rough but clear sketch showing the feasible region.
Quick Reference Formulas & Facts
Standard Form
General LPP Format:Optimize (Max/Min) Z = ax + by
Subject to:
• $a_1x + b_1y$ (≤/≥/=) $c_1$
• $a_2x + b_2y$ (≤/≥/=) $c_2$
• x ≥ 0, y ≥ 0
Key Theorems
Theorem 1: Optimal value occurs at a corner pointTheorem 2: For bounded region, both max and min exist at corner points
Important Points
• Feasible region is always convex• Corner points are intersections of boundary lines
• If two corner points give same optimal value, all points on line segment joining them are also optimal
• For unbounded region, max/min may not exist - needs verification
Practice Problems (Self-Assessment)
Level 1: Basic (2 Marks Each)
Q1. Maximize Z = 3x + 4y subject to x + y ≤ 4, x ≥ 0, y ≥ 0
Hint: Corner points are (0,0), (4,0), (0,4). Max at (0,4) = 16
Q2. Minimize Z = -3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Hint: Graph and find corner points. Min at (4,0) = -12
Q3. Maximize Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Hint: Find intersection of the two constraint lines
Level 2: Intermediate (6 Marks Each)
Q4. Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0. Show that minimum occurs at more than one point.
Hint: Multiple optimal solutions on a line segment
Q5. Minimize and Maximize Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0
Hint: Find all four corner points of the bounded region
Q6. Maximize Z = -x + 2y subject to x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
Hint: Feasible region is unbounded. Check if maximum exists.
Level 3: Advanced (5 Marks Each)
Q7. Maximize Z = -x + 2y subject to x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
Hint: Feasible region is unbounded. Check if maximum exists.
Q8. Minimize and Maximize Z = 3x + 9y subject to x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0
Hint: Find all corner points of quadrilateral feasible region
Q9. Maximize Z = x + y subject to x - y ≤ -1, -x + y ≤ 0, x ≥ 0, y ≥ 0
Hint: Check if feasible region exists
Step-by-Step Guide for 5-Mark Questions
Mark Distribution (Total 5 Marks)
1. Graphing Constraints (1.5 Marks)
• Convert inequalities to equations
• Find intercepts and plot lines
• Shade feasible region correctly
• Label axes and lines clearly
2. Finding Corner Points (1.5 Marks)
• Identify all vertices of feasible region
• Show calculation for intersections (solve simultaneous equations)
• Write coordinates clearly
3. Evaluating Objective Function (1.5 Marks)
• Make a table with corner points
• Calculate Z at each corner point
• Identify max/min value
4. Conclusion (0.5 Mark)
• State optimal value clearly
• State optimal point
• For unbounded regions, verify if max/min exists
Time Management for 5-Mark Question:
Total Time: 8-10 minutes
• Graphing: 3-4 minutes
• Corner points: 2 minutes
• Evaluation: 2 minutes
• Checking & writing answer: 1-2 minutes
Total Time: 8-10 minutes
• Graphing: 3-4 minutes
• Corner points: 2 minutes
• Evaluation: 2 minutes
• Checking & writing answer: 1-2 minutes
⚠️ Important: Since formulation is not required, all 5 marks are for the graphical solution. Ensure your graph is neat and all calculations are clearly shown!
Special Cases & Exceptions
Case 1: No Feasible Solution
When: Constraints contradict each other
Example: x + y ≥ 8 and 3x + 5y ≤ 15 with x ≥ 0, y ≥ 0
Result: No common region exists
Answer: "The problem has no feasible solution"
Example: x + y ≥ 8 and 3x + 5y ≤ 15 with x ≥ 0, y ≥ 0
Result: No common region exists
Answer: "The problem has no feasible solution"
Case 2: Multiple Optimal Solutions
When: Two corner points give same optimal value
Example: Z = 180 at both (15, 15) and (0, 20)
Result: All points on line segment joining these points are optimal
Answer: "Maximum Z = 180 at infinite points on line segment joining (15,15) and (0,20)"
Example: Z = 180 at both (15, 15) and (0, 20)
Result: All points on line segment joining these points are optimal
Answer: "Maximum Z = 180 at infinite points on line segment joining (15,15) and (0,20)"
Case 3: Unbounded Feasible Region
When: Region extends indefinitely
Check Required:
• For max: Graph ax + by > M (where M is largest value at corners)
• For min: Graph ax + by < m (where m is smallest value at corners)
If half-plane has common points with feasible region: Max/min does NOT exist
If NO common points: M is max / m is min
Check Required:
• For max: Graph ax + by > M (where M is largest value at corners)
• For min: Graph ax + by < m (where m is smallest value at corners)
If half-plane has common points with feasible region: Max/min does NOT exist
If NO common points: M is max / m is min
Case 4: Redundant Constraints
Definition: A constraint that doesn't affect the feasible region
Example: If x + y ≤ 100 and x + y ≤ 50, the first is redundant
Action: Identify but still show in graph (doesn't affect answer)
Example: If x + y ≤ 100 and x + y ≤ 50, the first is redundant
Action: Identify but still show in graph (doesn't affect answer)
Final Checklist (Print & Keep)
✓ Before Entering Exam Hall
Concepts to Recall:
- Objective function (given in exam)
- Types of constraints
- Feasible vs infeasible region
- Corner point theorem
- Bounded vs unbounded region
- How to shade inequalities
- Finding intersection points
Techniques to Remember:
- How to find x and y intercepts
- Solving 2 linear equations
- Test for bounded region
- Verification for unbounded
- Inequality direction rules
- Making evaluation tables
✓ During Exam
- ✓ Read the given LPP carefully (it's already formulated)
- ✓ Draw neat graph with ruler and label everything
- ✓ Show all calculations for corner points
- ✓ Make table for evaluating Z at each corner
- ✓ Check if region is bounded/unbounded
- ✓ For unbounded: verify if max/min exists
- ✓ Write complete conclusion clearly
- ✓ Double-check arithmetic calculations
Golden Rules for Full Marks in LPP
1. Graph neatly - Use ruler, label axes and lines clearly2. Show calculations - Don't just read corner points from graph
3. Make a table - Organized Z evaluation gets you marks
4. Write conclusion - State maximum/minimum value and point
5. Check bounded/unbounded - Critical for correctness
6. Shade correctly - Test with origin if needed
7. Practice graphing - Speed and accuracy come from practice
8. No formulation needed - Focus entirely on graphical solution!
Quick Revision Summary
5-Minute Quick Revision
What is LPP?Finding optimal value of linear function subject to linear constraints with non-negative variables.
Standard Form:
Max/Min Z = ax + by
Subject to: Linear inequalities + x ≥ 0, y ≥ 0
Solution Method (Corner Point):
1. Graph constraints → 2. Find feasible region → 3. Locate corner points → 4. Evaluate Z → 5. Choose optimal
Key Remember:
• Optimal at corner point (Theorem 1)
• Bounded region → max & min both exist
• Unbounded → need to verify
• Multiple solutions → on line segment
• No feasible region → no solution
Common Question Types:
1. Manufacturing (maximize profit)
2. Diet (minimize cost)
3. Investment (maximize return)
4. Allocation (minimize cost)
Graphing Tips:
• Find intercepts: put y=0 for x-intercept, x=0 for y-intercept
• Test (0,0) to determine shading
• Use ruler for neat lines
• Label everything clearly
Last-Minute Memory Triggers:
FLOC - Four steps of LPP
• Formulate (variables, objective, constraints)
• Lines (graph the constraints)
• Optimal points (find corners)
• Calculate Z (evaluate and conclude)
ICE - For graphing lines
• Intercepts (find both)
• Connect (draw the line)
• Evaluate (test point for shading)
FLOC - Four steps of LPP
• Formulate (variables, objective, constraints)
• Lines (graph the constraints)
• Optimal points (find corners)
• Calculate Z (evaluate and conclude)
ICE - For graphing lines
• Intercepts (find both)
• Connect (draw the line)
• Evaluate (test point for shading)
Syllabus Focus & Applications
As Per Latest CBSE Syllabus:
✓ INCLUDED:
• Introduction to Linear Programming
• Related terminology (objective function, constraints, feasible region, etc.)
• Graphical method of solving LPP in two variables
• Corner point method
• Bounded and unbounded feasible regions
✗ NOT INCLUDED (Removed):
• Mathematical formulation from word problems
• Application problems requiring formulation
• Real-world scenario to LPP conversion
✓ INCLUDED:
• Introduction to Linear Programming
• Related terminology (objective function, constraints, feasible region, etc.)
• Graphical method of solving LPP in two variables
• Corner point method
• Bounded and unbounded feasible regions
✗ NOT INCLUDED (Removed):
• Mathematical formulation from word problems
• Application problems requiring formulation
• Real-world scenario to LPP conversion
What to Expect in Exams
Typical Question Format:
The LPP will be completely given to you in this form:
"Solve graphically: Maximize/Minimize Z = ax + by
Subject to:
• Constraint 1
• Constraint 2
• x ≥ 0, y ≥ 0"
You just need to:
1. Graph the constraints
2. Find feasible region
3. Calculate corner points
4. Evaluate Z
5. Find optimal solution
The LPP will be completely given to you in this form:
"Solve graphically: Maximize/Minimize Z = ax + by
Subject to:
• Constraint 1
• Constraint 2
• x ≥ 0, y ≥ 0"
You just need to:
1. Graph the constraints
2. Find feasible region
3. Calculate corner points
4. Evaluate Z
5. Find optimal solution
Historical Context (For Knowledge Only)
Historical Background:
Linear Programming was developed during World War II to optimize military operations. The first LPP was formulated in 1941 by Russian mathematician L. Kantorovich and American economist F. L. Hitchcock (transportation problem).
In 1947, American economist G. B. Dantzig developed the Simplex Method, an efficient algorithm to solve LPPs with multiple variables.
Nobel Prize: L. Kantorovich and T. C. Koopmans received the Nobel Prize in Economics (1975) for their pioneering work in linear programming.
Note: This is for general knowledge only - not required for exam.
Linear Programming was developed during World War II to optimize military operations. The first LPP was formulated in 1941 by Russian mathematician L. Kantorovich and American economist F. L. Hitchcock (transportation problem).
In 1947, American economist G. B. Dantzig developed the Simplex Method, an efficient algorithm to solve LPPs with multiple variables.
Nobel Prize: L. Kantorovich and T. C. Koopmans received the Nobel Prize in Economics (1975) for their pioneering work in linear programming.
Note: This is for general knowledge only - not required for exam.
Real-World Applications (Knowledge)
Business & Industry:
- Production planning
- Inventory management
- Resource allocation
- Supply chain optimization
- Portfolio management
Other Fields:
- Military operations
- Transportation & logistics
- Diet & nutrition planning
- Agriculture planning
- Telecommunications
Beyond Class 12:
While we study only graphical method (2 variables), real-world problems involve hundreds or thousands of variables, solved using:
• Simplex Method
• Interior Point Methods
• Computer Software (Excel Solver, MATLAB, Python)
• Integer Programming
This information is not required for board exams.
While we study only graphical method (2 variables), real-world problems involve hundreds or thousands of variables, solved using:
• Simplex Method
• Interior Point Methods
• Computer Software (Excel Solver, MATLAB, Python)
• Integer Programming
This information is not required for board exams.