This MCQ module is based on: LPP Formulation and Graphical Method
TOPIC 21 OF 25
LPP Formulation and Graphical Method
🎓 Class 12
Mathematics
CBSE
Theory
Ch 12 — Linear Programming
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: LPP Formulation and Graphical Method
Targeting Class 12 level in Linear Programming, with Advanced difficulty.
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12.1 Introduction
In previous classes, solving equations and systems of inequalities was studied. Here we go further: find values of variables subject to certain constraints that optimize (maximize or minimize) a linear function. This is the domain of Linear Programming? — a technique born during World War II for resource-allocation problems and now used in industry, agriculture, transport, diet planning, and finance.
12.2 Linear Programming Problem (LPP) and its Mathematical Formulation
An LPP involves three essentials:
- Decision variables \(x, y\) — quantities to be determined.
- Objective function \(Z = ax + by\) — the linear function to optimize.
- Constraints — system of linear inequalities in \(x, y\), with non-negativity restrictions \(x \ge 0, y \ge 0\).
Definitions
Feasible Region: the set of points \((x,y)\) satisfying all constraints.Feasible Solution: a point inside the feasible region.
Optimal Solution: a feasible solution that maximizes/minimizes \(Z\).
Bounded Region: feasible region can be enclosed in a sufficiently large circle; otherwise unbounded.
12.2.1 Mathematical Formulation — Example: Furniture Factory
A furniture dealer deals in only two items — tables and chairs. He has ₹50,000 to invest and space for at most 60 pieces. A table costs ₹2,500, a chair ₹500; he expects profits of ₹250 per table and ₹75 per chair. Assuming he can sell everything he stocks, how should he invest to maximize profit?
Let \(x\) = number of tables, \(y\) = number of chairs. Then:
Objective: Maximize \(Z = 250x + 75y\).
Constraints:
- \(2500x + 500y \le 50{,}000\) (investment) ⇒ \(5x + y \le 100\)
- \(x + y \le 60\) (space)
- \(x \ge 0,\; y \ge 0\) (non-negativity)
12.3 Graphical Method of Solving LPP
For problems with two decision variables we use a graphical method:
- Graph each constraint; identify the feasible region (intersection of half-planes).
- Find the corner points (vertices) of the region.
- Evaluate \(Z\) at each corner point.
- The largest/smallest value gives the maximum/minimum.
Corner-Point Theorem
If the feasible region of an LPP is bounded, the objective function attains both its maximum and minimum values at corner points (vertices). If unbounded, a maximum/minimum may not exist — but if it does, it occurs at a corner point.
Example 1 — Solving the Furniture Problem Graphically
Fig 12.1: Feasible region OABC for the furniture LPP
Corner points: O(0,0), A(20,0), B(10,50), C(0,60). Evaluate \(Z = 250x + 75y\):
| Corner | x | y | Z = 250x + 75y |
|---|---|---|---|
| O | 0 | 0 | 0 |
| A | 20 | 0 | 5000 |
| B | 10 | 50 | 6250 (max) |
| C | 0 | 60 | 4500 |
Answer: Max profit = ₹6,250 at (x, y) = (10 tables, 50 chairs).
Example 2 — Minimization
Minimize \(Z = 200x + 500y\) subject to \(x + 2y \ge 10,\; 3x + 4y \le 24,\; x \ge 0, y \ge 0\).
Corner points of feasible region: intersection of \(x+2y=10\) with axes and with \(3x+4y=24\).
From \(x+2y=10\) and \(3x+4y=24\): solve — \(x=4, y=3\).
Vertices: A(0, 5), B(4, 3), C(0, 6).
\(Z(A)=0+2500=2500\); \(Z(B)=800+1500=2300\); \(Z(C)=0+3000=3000\).
Min = 2300 at (4, 3).
From \(x+2y=10\) and \(3x+4y=24\): solve — \(x=4, y=3\).
Vertices: A(0, 5), B(4, 3), C(0, 6).
\(Z(A)=0+2500=2500\); \(Z(B)=800+1500=2300\); \(Z(C)=0+3000=3000\).
Min = 2300 at (4, 3).
Example 3 — Diet Problem
A dietician wishes to mix two types of foods F1 and F2 such that the vitamin contents include at least 8 units of vitamin A and 10 units of vitamin C. F1 contains 2 units/kg of A and 1 unit/kg of C; F2 contains 1 unit/kg of A and 2 units/kg of C. F1 costs ₹50/kg, F2 ₹70/kg. Formulate and solve to minimize cost.
Let \(x, y\) kg. Minimize \(Z=50x+70y\).
Constraints: \(2x+y\ge 8,\; x+2y\ge 10,\; x,y\ge 0\).
Corner points: (10,0), (2,4) [from intersection], (0,8).
\(Z(10,0)=500\); \(Z(2,4)=100+280=380\); \(Z(0,8)=560\).
Min = ₹380 at (2, 4). Since feasible region is unbounded, verify \(50x+70y<380\) has no point in feasible region — true ⇒ min attained at (2, 4).
Constraints: \(2x+y\ge 8,\; x+2y\ge 10,\; x,y\ge 0\).
Corner points: (10,0), (2,4) [from intersection], (0,8).
\(Z(10,0)=500\); \(Z(2,4)=100+280=380\); \(Z(0,8)=560\).
Min = ₹380 at (2, 4). Since feasible region is unbounded, verify \(50x+70y<380\) has no point in feasible region — true ⇒ min attained at (2, 4).
Unbounded Regions — Caution
If the feasible region is unbounded:• For maximization, \(Z\) may be unbounded ⇒ no finite max. Check whether \(ax+by > M\) has solutions in the region for the suspected max \(M\).
• For minimization, \(Z\) may have a finite min at a corner point only if \(ax+by < m\) has no feasible solution.
Activity: Design Your Own LPP
L6 CreateMaterials: Graph paper, ruler, coloured pens.
- Pick a real-life scenario: making two bakery items with flour and sugar constraints, mixing two fertilisers, or packing two types of books.
- Assign decision variables \(x, y\) (non-negative integers where applicable).
- Write the objective function (profit, cost, nutrients, etc.).
- Write 2–3 linear constraint inequalities that reflect the scenario.
- Plot the feasible region, find corner points, and compute Z at each.
- Compare with a partner: same scenario, different constraints — whose plan gives the better optimum?
Real LPPs often have hundreds of variables and constraints — solved by the Simplex Method (not in syllabus). The graphical method works only for 2 variables but illustrates why the optimum occurs at a corner: linear functions over convex polytopes attain extrema at vertices.
Figure it Out (Exercise 12.1 — Selected)
Q1. Maximize \(Z = 3x + 4y\) subject to \(x + y \le 4, x \ge 0, y \ge 0\).
Corners: (0,0), (4,0), (0,4). Z-values: 0, 12, 16. Max Z = 16 at (0, 4).
Q2. Minimize \(Z = -3x + 4y\) subject to \(x + 2y \le 8, 3x + 2y \le 12, x,y \ge 0\).
Corners: (0,0), (4,0), (2,3) [from intersection of 2 lines], (0,4). Z: 0, -12, -6, 16. Min Z = -12 at (4, 0).
Q3. Maximize \(Z = 5x + 3y\) subject to \(3x + 5y \le 15, 5x + 2y \le 10, x,y \ge 0\).
Intersection: solve \(3x+5y=15\) and \(5x+2y=10\) → \(x=20/19, y=45/19\). Corners: (0,0), (2,0), (20/19, 45/19), (0,3). Z: 0, 10, \(\tfrac{100+135}{19}=\tfrac{235}{19}\), 9. Max Z = 235/19 ≈ 12.37.
Competency-Based Questions
Scenario: A farmer has 10 hectares of land, ₹36,000 capital, and 144 labour-hours. Crop A needs 1 ha, ₹2,000 and 12 hrs/ha, returns ₹3,000/ha profit. Crop B needs 1 ha, ₹6,000 and 12 hrs/ha, returns ₹8,000/ha profit. Let \(x, y\) = hectares of A, B.
Q1. Formulate the LPP (objective + constraints).
L3 ApplyMax \(Z = 3000x + 8000y\). Constraints: \(x+y \le 10\) (land), \(2000x + 6000y \le 36000\) i.e. \(x+3y\le 18\) (capital), \(12x+12y\le 144\) i.e. \(x+y\le 12\) (labour, redundant with land), \(x,y\ge 0\).
Q2. Find corner points of the feasible region.
L3 ApplyFrom \(x+y=10\) and \(x+3y=18\): \(2y=8 ⇒ y=4, x=6\). Corners: (0,0), (10,0), (6,4), (0,6).
Q3. Analyse which corner gives maximum profit and state the amount.
L4 AnalyseZ(0,0)=0; Z(10,0)=30000; Z(6,4)=18000+32000=50000; Z(0,6)=48000. Max = ₹50,000 at (6, 4).
Q4. Evaluate: if the profit per hectare of Crop B drops to ₹5,000, does the optimal plan change? Justify.
L5 EvaluateNew Z=3000x+5000y. Values: Z(10,0)=30000, Z(6,4)=18000+20000=38000, Z(0,6)=30000. Max still at (6,4) = ₹38,000, so plan unchanged — though profit falls.
Assertion–Reason Questions
A: In an LPP with a bounded feasible region, the optimal value occurs at a corner point.
R: A linear function on a convex polygon attains its extrema on vertices.
R: A linear function on a convex polygon attains its extrema on vertices.
(a) Corner-Point Theorem. R explains A.
A: Every LPP has at least one optimal solution.
R: The feasible region of an LPP is always non-empty.
R: The feasible region of an LPP is always non-empty.
(d) A is false — an LPP with an infeasible system (empty feasible region) or an unbounded objective has no optimal solution. R is also false: feasible region can be empty. Actually both false ⇒ no option (a–d) fits cleanly. Standard CBSE answer: (d) with understanding that \"unbounded region with max\" can fail.
Frequently Asked Questions — Linear Programming
What is Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool in NCERT Class 12 Mathematics?
Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool is a key concept covered in NCERT Class 12 Mathematics, Chapter 12: Linear Programming. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool step by step?
To solve problems on Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 12: Linear Programming?
The essential formulas of Chapter 12 (Linear Programming) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool important for the Class 12 board exam?
Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool?
Common mistakes in Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool?
End-of-chapter NCERT exercises for Part 1 — LPP Formulation & Graphical Method | Class 12 Maths Ch 12 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 12, and solve at least one previous-year board paper to consolidate your understanding.
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