Official Class 12 Mathematics Revision Material: CBSE Class 12 Mathematics Linear Programming Notes Set 03
Review targeted revision notes for Class 12 Mathematics with the CBSE Class 12 Mathematics Linear Programming Notes Set 03. Built according to official educational guidelines for the 2026-27 academic year, these downloadable summaries for Chapter 12 Linear Programming support daily study and last-minute exam readiness.
Chapter-wise Concept Summaries: Chapter 12 Linear Programming
Access the complete concept summary PDF for Chapter 12 Linear Programming below. Regular review of these targeted notes builds familiarity with complex Class 12 Mathematics themes and helps secure higher marks in final school evaluations.
Important Notes and Formulas
Introduction
With the help of linear programming models, a decision maker can most efficiently and effectively employ his production factors and limited resources to get maximum profit at minimum cost. He can also take quick decisions, which are important in modern times because any delay or postponement may give an advantage to other organisations.
Linear programming: Linear programming is a method of determining a particular programme or plan of action to optimize (maximize or minimize).
Linear programming problem (L.P.P.): It is a problem of finding the optimal (maximum or minimum) value of a linear function subject to certain restrictions (constraints) determined by a set of linear inequations with variables as non-negative.
General Form of L.P.P.
The general form of a linear programming problem is
Optimize (Maximize or Minimize) \( Z = c_1x_1 + c_2x_2 + \dots + c_nx_n \)
Subject to
\[ \left. \begin{matrix} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n \ (\leq, =, \geq)\ b_1 \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n \ (\leq, =, \geq)\ b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n \ (\leq, =, \geq)\ b_m \end{matrix} \right\rbrace \quad \dots(1) \]
\( x_1, x_2, \dots, x_n \geq 0 \) ...(2)
The definitions of various terms related to the LPP are as follows:
(a) Objective function: The linear function Z which is to be maximized or minimized is the objective function of the LPP.
(b) Decision variables: \( x_1, x_2, \dots, x_n \) are the decision variables.
(c) Constraints: The inequations or equations (1) in the variables of a LPP which describe the conditions under which the optimization (maximization or minimization) is to be accomplished are called constraints.
In the constraints given in the general form of LPP there may be any one of the three signs ≤, =, ≥.
(d) Non-negativity restrictions: These are the constraints (2) which describe that the variables involved in a LPP are non-negative.
(e) Feasible region: The common region determined by all the constraints including the non-negative constraints of a LPP is called the feasible region or solution region.
A feasible region may be (i) bounded (ii) unbounded.
(i) Bounded feasible region:
Constraints are:
x + 3y ≤ 12
3x + y ≤ 12
x ≥ 0, y ≥ 0
(f) Corner points of a feasible region: Corner points of a feasible region are the points of intersection of two boundary lines.
To Solve an LPP Graphically (Corner Point Method)
Working rule:
(a) Formulate the given linear programming problem.
(b) Convert the linear constraints into equalities and then draw their graphs, which will be straight lines.
(c) Find the feasible region of the linear programming problem and determine its corner points by solving the equations of the lines taken two at a time.
(d) Determine the value of the objective function Z at each corner point. Also, find the least value m and the greatest value M of the values of the objective function at the corner points.
(e) If the feasible region R is bounded, then M and m will be the maximum and minimum values of Z, respectively.
Solved Examples
Maximise Z = 22x + 18y
subject to constraints
x + y ≤ 20
360x + 240y ≤ 5,760
x ≥ 0, y ≥ 0.
Answer:
Given
Maximise Z = 22x + 18y ...(1)
subject to constraints
x + y ≤ 20 ...(2)
360x + 240y ≤ 5760 ...(3)
x ≥ 0, y ≥ 0 ...(4)
First we obtained the feasible region determined by the system of constraints (2) to (4).
Here the obtained feasible region OABCO is bounded and its corner points are A(16, 0), B(8, 12), C(0, 20) and O(0, 0).
| Corner Points | Z = 22x + 18y |
|---|---|
| O(0, 0) | 0 |
| A(16, 0) | 352 |
| B(8, 12) | 392 |
| C(0, 20) | 360 |
∴ Maximum value of Z = 392 at point B(8, 12)
Maximise Z = x + y
subject to constraints
2x + y ≤ 50
x + 2y ≤ 40
x ≥ 0, y ≥ 0
Answer:
Given
Maximise Z = x + y ...(1)
subject to constraints
2x + y ≤ 50 ...(2)
x + 2y ≤ 40 ...(3)
x ≥ 0, y ≥ 0 ...(4)
First we obtained the feasible region determined by the system of constraints (2) to (4).
Here the obtained feasible region OABCO is bounded and its corner points are A(25, 0), B(20, 10), C(0, 20), O(0, 0).
| Corner Points | Z = x + y |
|---|---|
| O(0, 0) | 0 |
| A(25, 0) | 25 |
| B(20, 10) | 30 |
| C(0, 20) | 20 |
∴ Maximum value of Z = 30 at point B(20, 10)
Maximize Z = 4500x + 5000y
subject to constraints
x + y ≤ 250
25,000x + 40,000y ≤ 70,00,000
5x + 8y ≤ 1400
x ≥ 0, y ≥ 0
Answer:
Given
Maximize Z = 4500x + 5000y ...(1)
subject to constraints
x + y ≤ 250 ...(2)
25,000x + 40,000y ≤ 70,00,000 ...(3)
5x + 8y ≤ 1400 ...(4)
x ≥ 0, y ≥ 0 ...(5)
First we obtained the feasible region determined by the system of constraints (2) to (5).
Here the obtained feasible region ABCOA is bounded and its corner points are A(250, 0), B(200, 50), C(0, 175), O(0, 0).
| Corner Points | Z = 4500x + 5000y |
|---|---|
| O(0, 0) | 0 |
| A(250, 0) | 11,25,000 |
| B(200, 50) | 11,50,000 |
| C(0, 175) | 8,75,000 |
∴ Maximum value of Z = 11,50,000 at point B(200, 50)
Maximise Z = 100x + 120y
subject to constraints
2x + 3y ≤ 30
3x + y ≤ 17
x ≥ 0; y ≥ 0
Answer:
Given
Maximise Z = 100x + 120y ...(1)
subject to constraints
2x + 3y ≤ 30 ...(2)
3x + y ≤ 17 ...(3)
x ≥ 0; y ≥ 0 ...(4)
First we obtained the feasible region determined by the system of constraints (2) to (4).
Here the obtained feasible region OABCO is bounded and its corner points are O(0, 0), \( A\left(\frac{17}{3}, 0\right) \), B(3, 8) and C(0, 10).
| Corner Points | Z = 100x + 120y |
|---|---|
| O(0, 0) | 0 |
| \( A\left(\frac{17}{3}, 0\right) \) | 566.67 |
| B(3, 8) | 1260 |
| C(0, 10) | 1200 |
∴ Maximum value of Z = 1260 at point B(3, 8)
Maximise Z = 80x + 120y
subject to constraints
9x + 12y ≤ 180
x + 3y ≤ 30
x ≥ 0; y ≥ 0
Answer:
Given
Maximise Z = 80x + 120y ...(1)
subject to constraints
9x + 12y ≤ 180 ...(2)
x + 3y ≤ 30 ...(3)
x ≥ 0; y ≥ 0 ...(4)
First we obtained the feasible region determined by the system of constraints (2) to (4).
Here the obtained feasible region OABCO is bounded and its corner points are A(20, 0), B(12, 6), C(0, 10) and O(0, 0).
| Corner Points | Z = 80x + 120y |
|---|---|
| O(0, 0) | 0 |
| A(20, 0) | 1600 |
| B(12, 6) | 1680 |
| C(0, 10) | 1200 |
∴ Maximum value of Z = 1680 at point B(12, 6)
Maximise and Minimise Z = 0.1x + 0.09y
subject to constraints
x ≥ 20,000
y ≥ 10,000
x + y ≤ 50,000
x ≥ y
x ≥ 0, y ≥ 0
Answer:
Given
Maximise and Minimise Z = 0.1x + 0.09y ...(1)
subject to constraints
x ≥ 20,000 ...(2)
y ≥ 10,000 ...(3)
x + y ≤ 50,000 ...(4)
x ≥ y ...(5)
x ≥ 0, y ≥ 0 ...(6)
First we obtained the feasible region determined by the system of constraints (2) to (6).
Here the obtained feasible region ABCDA is bounded and its corner points are A(20000, 10000), B(20000, 20000), C(25000, 25000) and D(40000, 10000).
| Corner Points | Z = 0.1x + 0.09y |
|---|---|
| A(20000, 10000) | 2900 |
| B(20000, 20000) | 3800 |
| C(25000, 25000) | 4750 |
| D(40000, 10000) | 4900 |
∴ Maximum value of Z = 4900 at point D(40000, 10000) and minimum value of Z = 2900 at point A(20000, 10000)
Maximize Z = 0.7x + 1y
subject to the constraints
4x + 6y ≤ 240
6x + 3y ≤ 240
x, y ≥ 0
Answer:
Given, we have to
Maximize Z = 0.7x + 1y ...(1)
subject to the constraints
4x + 6y ≤ 240 ...(2)
6x + 3y ≤ 240 ...(3)
x, y ≥ 0 ...(4)
First we obtained the feasible region determined by the system of constraints (2) to (4).
Here the obtained feasible region ABCOA is bounded and its corner points are A(40, 0), B(30, 20), C(0, 40) and O(0, 0).
| Corner Points | Z = 0.7x + 1y |
|---|---|
| A(40, 0) | 28 |
| B(30, 20) | 41 |
| C(0, 40) | 40 |
| O(0, 0) | 0 |
∴ Maximum value of Z = 41 at point B(30, 20)
Maximise Z = 40x + 50y
subject to constraints
3x + y ≤ 9
x + 2y ≤ 8
x ≥ 0, y ≥ 0
Answer:
Given
Maximise Z = 40x + 50y ...(1)
subject to constraints
3x + y ≤ 9 ...(2)
x + 2y ≤ 8 ...(3)
x ≥ 0, y ≥ 0 ...(4)
First we obtained the feasible region determined by the system of constraints (2) to (4).
Here the obtained feasible region OABCO is bounded and its corner points are A(3, 0), B(2, 3), C(0, 4), O(0, 0).
| Corner Points | Z = 40x + 50y |
|---|---|
| O(0, 0) | 0 |
| A(3, 0) | 120 |
| B(2, 3) | 230 |
| C(0, 4) | 200 |
∴ Maximum value of Z = 230 at point B(2, 3)
Previous Year Questions
Very Short Answer Questions (1 Mark)
Question 1. The corner points of the feasible region of an LPP are (0, 0), (0, 8), (2, 7), (5, 4) and (6, 0). The maximum profit P = 3x + 2y occurs at the point ________ . [CBSE 2020, 1M]
Answer:
Given that the corner points are (0, 0), (0, 8), (2, 7), (5, 4) and (6, 0)
and P = 3x + 2y
at (0, 8), P = 16; at (5, 4), P = 23
at (2, 7), P = 20; at (6, 0), P = 18
at (0, 0), P = 0
\( P_{max} \) = 23 at (5, 4) [1]
Long Answer Questions I (4 Marks)
Maximise Z = 34x + 45y
under the following constraints
x + y ≤ 300
2x + 3y ≤ 70
x ≥ 0, y ≥ 0.
Answer:
[3]
| Corner Points | Z = 34x + 45y |
|---|---|
| O(0, 0) | 0 |
| A(35, 0) | 1190 |
| \( B\left(0, \frac{70}{3}\right) \) | 1050 |
Since the feasible region is bounded,
hence the maximum value of Z = 1190 at A(35, 0). [1]
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Study Material and Revision Notes for Class 12 Mathematics Chapter 12 Linear Programming
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