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Linear Programming Class 12 Notes CBSE Maths Chapter 12 [Free PDF Download]

Last updated date: 08th Aug 2024
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Revision Notes for CBSE Class 12 Maths Chapter 12 (Linear Programming) - Free PDF Download

Linear programming is an essential topic in mathematics. Today, we will help students learn this topic in our Class 12 Maths Chapter 12 Revision Notes. We will start the Class 12 Maths Linear Programming Revision Notes by Vedantu with the basic definition. After that, we will look at the characteristics, equations, and application of this topic. According to the Revision Notes Class 12 Chapter 12, the main aim of linear programming is to either minimize or maximize a numerical value. To know all about Linear Programming, you can also download the Class 12 Maths Chapter 12 Notes PDF for free.

CBSE Class 12 Maths Revision Notes 2024-25 - Chapter Wise PDF Notes for Free

In the table below we have provided the PDF links of all the chapters of CBSE Class 12 Maths whereby the students are required to revise the chapters by downloading the PDF.

It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

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Find a curated selection of study resources for Class 12 subjects, helping students prepare effectively and excel in their academic pursuits.

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Linear Programming Class 12 Notes Maths - Basic Subjective Questions

Section–A (1 Mark Questions)

1. The feasible solution for a LPP is shown in following figure. Let Z=3x-4y be the objective function. Find where minimum of Z occurs.

Ans.

Hence, the minimum of Z  occurs at (0, 8) and its minimum value is -32.

2. The feasible region for an LPP is shown in the following figure. Let F=3x-4y be the objective function. Then find the Maximum value of F.

Ans. The feasible region as shown in the figure, has objective function F=3x-4y.

Hence, the maximum value of F is 12.

3. The feasible region for an LPP is shown in the following figure. Let F=3x-4y be the objective function. Then find the Minimum value of F.

Ans. The feasible region as shown in the figure, has objective function F=3x-4y.

We have minimum value of F is 16 at   (0, 4).

4. Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F=4x+6y be the objective function. Then find the value of Maximum of F- minimum of F.

Ans.

Maximum of F- Minimum of F=72-12=60.

5. Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z=px+qy where pq>0. Then find the condition on p and q, so that the minimum of Z occurs at (3, 0) and (1, 1).

Ans.

So, condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) is

$q\;+q=3p$

$\Rightarrow 2p=q$

$\Rightarrow p=q/2$

Section–B (2 Marks Questions)

6. Maximize Z=3x+4y, subject to the constraints $x+y\leq 1,x\geq 0,y\geq 0$ .

Ans. We have to maximize Z=3x+4y,

Subject to the constraints $x+y\leq 1$ ; $x\geq 0$ and $y\geq 0$

All these inequalities are plotted as shown below:

The shaded region shown in the figure as OAB is bounded and the coordinates of corner points O, A and B are (0, 0), (1, 0) and (0, 1), respectively.

Hence, the maximum value of Z is 4 at  (0, 1).

7. Maximize the function Z=11x+7y subject to the constraints $x< 3,y\leq 2,x\geq 0$ and $y\geq 0$.

Ans. We have to maximize Z=11x+7y,

Subject to the constraints $x< 3,y\leq 2,x\geq 0$.

All these inequalities are plotted as shown below:

The shaded region OABC is bounded and the coordinates of corner points are (0, 0), (3, 0), (3, 2) and (0, 2), respectively.

Hence, Z is maximum at (3, 2) and its maximum value is 47.

8. Minimize Z=13x-15y subject to the constraints $x+y\leq 7,2x-3y+6\geq 0,x\geq 0$ and $y\geq 0$ .

Ans. We have to minimize Z=13x-15y

Subject to the constraints

$x+y\leq 7$

$2x-3y+6\geq 0$

$x\geq 0$ and $y\geq 0$

All these inequalities are plotted as shown below:

Shaded region shown as OABC is bounded and coordinates of its corner points are (0, 0),

(7, 0), (3, 4) and (0, 2), respectively.

Hence, the minimum value of Z is –30 at (0, 2).

9. Determine the maximum value of Z=3x+4y, if the feasible region (shaded) for a LPP is shown in following figure.

Ans. Two lines in graph are 2x+y=104 and 2x+4y=152

Solving the equations of two lines we have:

$\Rightarrow -3y=-48$

$\Rightarrow y=16 and x=44$

Thus, the point of intersection of two lines is E(44,16).

From the graph, corner points are O(0,0), A(52,0), E(44,16) and D(0,38).

Also, given region is bounded.

Here, Z=3x+4y

Hence, Z at (44, 16) is maximum and its maximum value is 196.

10. Feasible region (shaded) for a LPP is shown in following figure. Maximize Z= 5+7y.

Ans. The shaded region is bounded and has coordinates of corner points as (0, 0), (7, 0), (3, 4) and (0, 2). Also, Z=5x+7y.

Hence, the maximum value of Z is 43 at  (3, 4).

11. The feasible region for a LPP is shown in following figure. Find the minimum value of Z = 11x+7y.

Ans. Lines x+3y=9 and x+y=5 intersect at (3, 2).

From the figure, it is clear that feasible region is bounded with corner points as (0, 3), (3, 2) and (0, 5).

Here, Z=11x+7y.

Hence, minimum value of Z is 21 at  (0, 3).

12. The feasible region for a LPP is shown in following figure. Find the maximum value of Z=11x+7y.

Ans. Lines x+3y=9 and x+y=5 intersect at (3, 2).

From the figure, it is clear that feasible region is bounded with corner points as (0, 3), (3, 2) and (0, 5).Here, Z= 11x+7y.

Hence, Z  is maximum at (3, 2) and its maximum value is 47.

13. The feasible region for a LPP is shown in the following figure. Evaluate Z=4x+y at each of the corner points of this region. Find the minimum value of Z if it exists.

Ans.  x+2y=4 and x+y=3 intersect at (2, 1)

From the figure the feasible region is the unbounded region with the corner points A(4,0), B(2,1) and C(0,3) Also, we have Z=4x+y

Here, 3 is the smallest value of Z at the corner point (0, 3). Now as the region is unbounded,

therefore 3 may or may not be the minimum value of Z.

To decide this issue, we graph the inequality $4x+y\leq 3$ and check that the resulting open half plane has no point in common with feasible region otherwise, Z has no minimum value.

From the graph shown above, it is clear that there is no point in common with feasible region and hence Z has minimum value 3 at (0, 3).

PDF Summary - Class 12 Maths Linear Programming Notes (Chapter 12)

Linear Programming & Its Applications

• Linear Programming is a common optimization (maximisation or minimization) approach used in business and everyday life to find the maximum or minimum values necessary of a linear expression in order to meet a set of supplied linear constraints.

• It may entail determining the most profit, the lowest cost, or the least amount of resources used, among other things.

• Industry, commerce, management science, and other fields use it.

Optimal value:

It refers to the Maximum or Minimum value of a linear function.

Objective Function:

• The function that needs to be improved (maximized/minimized)

• Linear function $\text{Z = ax + by }$, where $\text{a, b}$ are constants, which has to be maximised or minimized is called a linear objective function.

• For example, $\text{Z = 250x + 75y }$where variables $\text{x}$ and $\text{y}$ are called decision variables.

Linear Constraints:

• The objective function is to be optimised using a system of linear inequations/equations.

• In a linear programming issue, linear inequalities/equations or limitations on the variables are used.

• Also called Overriding Conditions or Constraints.

• The conditions $\text{x}\ge \text{0, y}\ge \text{0}$are called non-negative restrictions.

Non-Negative Restrictions:

All of the variables used to make decisions are assumed to have non-negative values.

Optimization Problem:

• A problem that seeks to maximize or minimize a linear function (say of two variables $\text{x}$ and $\text{y}$) subject to certain constraints as determined by a set of linear inequalities.

• Linear programming problems (LPP) are a special type of optimization problem.

Note:

• The term "linear" denotes that all of the mathematical relationships in the problem are linear.

• The term "programming" refers to the process of deciding on a specific program or course of action.

Mathematical Formulation of the Problem

• A general LPP can be stated as

$\left( \text{Max / Min} \right)\text{ Z = }{{\text{c}}_{\text{1}}}{{\text{x}}_{\text{1}}}\text{+ }{{\text{c}}_{\text{2}}}{{\text{x}}_{\text{2}}}\text{+ }...\text{ + }{{\text{c}}_{\text{n}}}{{\text{x}}_{\text{n}}}$

(Objective function) subject to constraints and non-negative restrictions.

\left\{ \begin{align} & {{\text{a}}_{\text{11}}}{{\text{x}}_{\text{1}}}\text{+ }{{\text{a}}_{\text{12}}}{{\text{x}}_{\text{2}}}\text{+ }...\text{ + }{{\text{a}}_{\text{1n}}}{{\text{x}}_{\text{n}}}\left( \le \text{ = }\ge \right){{\text{b}}_{\text{1}}} \\ & {{\text{a}}_{\text{21}}}{{\text{x}}_{\text{1}}}\text{+ }{{\text{a}}_{\text{22}}}{{\text{x}}_{\text{2}}}\text{+ }...\text{ + }{{\text{a}}_{\text{2n}}}{{\text{x}}_{\text{n}}}\left( \le \text{ = }\ge \right){{\text{b}}_{\text{2}}} \\ & \text{.} \\ & \text{.} \\ & \text{.} \\ & \text{.} \\ & {{\text{a}}_{\text{m1}}}{{\text{x}}_{\text{1}}}\text{+ }{{\text{a}}_{\text{m2}}}{{\text{x}}_{\text{2}}}\text{+ }...\text{ + }{{\text{a}}_{\text{mn}}}{{\text{x}}_{\text{n}}}\left( \le \text{ = }\ge \right){{\text{b}}_{\text{m}}} \\ \end{align} \right\}

• ${{\text{x}}_{\text{1}}}\text{, }{{\text{x}}_{\text{2}}}\text{, }.....\text{, }{{\text{x}}_{\text{n}}}\ge 0$ where ${{\text{a}}_{\text{11}}}\text{, }{{\text{a}}_{\text{12}}}\text{, }....\text{, }{{\text{a}}_{\text{mn}}}$;

• ${{\text{b}}_{\text{1}}}\text{, }{{\text{b}}_{\text{2}}}\text{, }....\text{, }{{\text{b}}_{\text{m}}}$ and ${{\text{c}}_{\text{1}}}\text{, }{{\text{c}}_{\text{2}}}\text{, }....\text{, }{{\text{c}}_{\text{n}}}$ are constants and

• ${{\text{x}}_{\text{1}}}\text{, }{{\text{x}}_{\text{2}}}\text{, }.....\text{, }{{\text{x}}_{\text{n}}}$ are variables.

Terminologies

Solution of an LPP:

A set of values of the variables ${{\text{x}}_{\text{1}}}\text{, }{{\text{x}}_{\text{2}}}\text{, }.....\text{, }{{\text{x}}_{\text{n}}}$ that satisfy the restrictions of an LPP.

Feasible Solution of an LPP:

• A set of values of the variables ${{\text{x}}_{\text{1}}}\text{, }{{\text{x}}_{\text{2}}}\text{, }.....\text{, }{{\text{x}}_{\text{n}}}$ that satisfy the restrictions and non-negative restrictions of an LPP.

• Possible solutions to the restrictions are represented as points within and on the boundary of the feasible zone.

Feasible Region:

The common region determined by all the constraints including non-negative constraints $\text{x, y}\ge \text{0}$ of a linear programming problem, is called the feasible region (or solution region) for the problem.

Feasible Choice:

Each point is in the feasible region.

Infeasible Region:

The region outside the feasible region.

Infeasible Solution:

Any point outside the feasible region.

Optimal Solution of an LPP:

A feasible solution of an LPP is said to be optimal (or optimum) if it also optimizes the objective function of the problem.

Graphical Solution of an LPP:

The solution of an LPP is obtained by the graphical method, that is by drawing the graphs corresponding to the constraints and the non-negative restrictions.

Unbounded Solution:

Such solutions exist if the value of the objective function can be increased or decreased forever.

Example: Graph the constraints stated as linear inequalities:

• $\text{5x + y }\le \text{ 100 }...\text{ }\left( 1 \right)$

• $\text{x + y }\le \text{ 60 }......\text{ }\left( 2 \right)$

• $\text{x}\ge \text{ 0 }.................\text{ }\left( 3 \right)$

• $\text{y}\ge \text{ 0 }.................\text{ }\left( 4 \right)$

Ans:

For Plotting the Equation (1),

• Let $\text{x = 0}$. Hence we get the point $\text{y = 100}$

• Let $\text{y = 0}$. Hence we get the point $\text{x = }\dfrac{\text{100}}{5}=20$

• The equation $\left( 1 \right)$ is obtained by joining the points $\left( 20,100 \right)$

For Plotting the Equation $\left( 2 \right)$,

• Let $\text{x = 0}$. Hence we get the point $\text{y = 60}$

• Let $\text{y = 0}$. Hence we get the point $\text{x = 60}$

• The equation $\left( 2 \right)$ is obtained by joining the points $\left( 60,60 \right)$

• From Equation $\left( 3 \right)$ and Equation $\left( 4 \right)$, we know both $\text{x}$ and $\text{y}$ are more significant than $\text{0}$.

(Graph is drawn with the help of Geogebra and Paint)

From the graph above,

• Possible solutions to the constraints are represented by points within and on the edge of the feasible zone.

• Here, every point within and on the boundary of the feasible region $\text{OABC}$ represents a possible solution to the problem.

• For example, point $\left( 10,50 \right)$ is a feasible solution to the problem, and so are the points $\left( 0,60 \right),\left( 20,0 \right)$, etc.

• Any point outside the possible region is called an infeasible solution. For example, point $\left( 25,40 \right)$ is an infeasible solution to the problem.

• Now, we can see that every point in the feasible region $\text{OABC}$ satisfies all the constraints given in $\left( 1 \right)$to $\left( 4 \right)$. As there are endless points, it is unclear how to discover a position that yields the objective function's most significant value  $\text{z = 250x + 75y}$.

 Vertex of  the Feasible Region The Corresponding Value of Z (in Rs) $\text{O }\left( 0,0 \right)$ $0$ $\text{A }\left( 0,60 \right)$ $4500$ $\text{B }\left( 10,50 \right)$ $6250$ (Maximum) $\text{C }\left( 20,0 \right)$ $5000$

Theorem $\text{1}$

• Let $\text{R}$ be the feasible region (convex polygon) for a linear programming problem.

• Let $\text{Z = ax + by }$ be the objective function.

• When the variables x and y are subject to constraints specified by linear inequalities, the optimal value Z (maximum or minimum) must occur at a corner point* (vertex) of the feasible region.

Theorem $\text{2}$

• Let $\text{R}$ be the feasible region for a linear programming problem.

• Let $\text{Z = ax + by }$ be the objective function.

• If an objective function $\text{R}$ is bounded, then the objective function $\text{Z}$ has both a maximum and a minimum value on $\text{R}$ and each of these occurs at a corner point (vertex) of $\text{R}$.

Remark:

The objective function may not have a maximum or minimum value if $\text{R}$ is unbounded. It must, however, occur at a corner point of $\text{R}$ if it exists.

A Graphical Approach to Solving a Linear Programming Issue

It can be solved by using below methods, they are as follows;

1. Corner point method

2. Iso-profit or Iso-cost method

Corner Point Method:

• Based on the principle of the extreme point theorem.

• Procedure to Solve an LPP Graphically by Corner Point Method

• Consider each constraint as an equation.

• Plot each equation on a graph, as each one will geometrically represent a straight line.

• The common region thus satisfies all the constraints, and the non-negative restrictions are called the feasible region. It is a convex polygon.

• Determine the vertices (corner points) of the convex polygon. These vertices are known as the feasible region's extreme points of corners.

• Find the objective function's values at each of the extreme points.

• The optimal solution of the given LPP is the point where the value of the objective function is optimum (maximum or minimum).

Isom-Profit or Iso-Cost Method:

Iso-profit or Iso-cost Method for Graphically Solving an LPP:

• Consider each constraint to be a mathematical equation.

• Draw each equation on a graph, as each one will represent a straight line geometrically.

• The polygonal region reached by meeting all constraints and non-negative limits is the feasible region, a convex set of all viable solutions of a given LPP.

• Determine the viable region's extreme points.

• Give the objective function $\text{Z}$ a handy value k and draw the matching straight line in the $\text{xy}$-plane.

• If the problem is maximization, draw parallel lines to $\text{Z = k}$ and find the line farthest from the origin and has at least one point in common with the viable zone.

• If the problem is of minimisation, then draw lines parallel to the line $\text{Z = k}$ that is closest to the origin and has minimum one point in common with the feasible zone.

• The optimal solution of the given LPP is the common point so achieved.

Working Rule for Marking Feasible Region

Consider the constraint $\text{ax + by}\le \text{c}$, where $\text{c 0}$.

• To begin, make a straight line $\text{ax + by}=\text{c}$ y connecting any two points on it.

• Find two points that satisfy this equation as a starting point.

• This straight-line divides the $\text{xy}$ -plane into two parts.

• The inequation $\text{ax + by c}$ will represent that part of the $\text{xy}$ -plane which lies to that side of the line $\text{ax + by}=\text{c}$ in which the origin lies.

Again, consider the constraint $\text{ax + by}\ge \text{c}$, where $\text{c 0}$.

• Draw the straight line $\text{ax + by}=\text{c}$ by joining any two points on it.

• This straight-line divides the $\text{xy}$-plane into two parts.

• The inequation $\text{ax + by}\ge \text{c}$ will represent that part of the $\text{xy}$ -plane, which lies to that side of the line $\text{ax + by}=\text{c}$ in which the origin does not lie.

Important Points to be Remembered

1. Basic Feasible Solution:

A fundamental solution that also meets the non-negativity constraints is known as a BFS.

1. Optimum Basic Feasible Solution:

A BFS is said to be optimum if it also optimizes (Max or min) the objective function.

Example: Solve the following linear programming problem graphically:

Maximize $\text{Z = 4x + y }...\text{ }\left( 1 \right)$

Subject to the constraints:

• $\text{x + y }\le \text{ 50 }...\text{ }\left( 2 \right)$

• $\text{3x + y }\le \text{ 90 }...\text{ }\left( 3 \right)$

• $\text{x}\ge \text{ 0 , y}\ge \text{ 0 }...\text{ }\left( 4 \right)$

Ans: For Plotting the Equation $\left( 2 \right)$ ,

• Let $\text{x = 0}$, Hence we get the point $\text{y = 50}$

• Let $\text{y = 0}$. Hence we get the point $\text{x = 50}$

• Equation $\left( 2 \right)$ is obtained by joining the points $\left( 50,50 \right)$

For Plotting the Equation $\left( 3 \right)$ ,

• Let $\text{x = 0}$. Hence we get the point $\text{y = 90}$

• Let $\text{y = 0}$. Hence we get the point $\text{x = 30}$

• The equation (3) is obtained by joining the points $\left( 30,90 \right)$

From Equation $\left( 4 \right)$, we know both $\text{x}$ and $\text{y}$ are greater than $\text{0}$.

As a result, the points are $\left( \text{0,0} \right)\text{, }\left( \text{50,50} \right)\text{, }\left( \text{0,50} \right)\text{, }\left( \text{30,0} \right)\text{, }\left( \text{30,90} \right)$.

The viable region in the graph is colored, as determined by the system of constraints $\left( 2 \right)$ to $\left( 4 \right)$.

The viable region $\text{OAEC}$ is bounded, as shown below;

By replacing the vertices of the bounded region for the vertices of the bounded region, the maximum value of $\text{Z}$ may be determined using the Corner Point Method.

As a result, the highest value of $\text{Z}$ at the position is $\text{120}$ at the point $\left( 30,0 \right)$.

 Corner Point The Corresponding Value of $\text{Z}$ $\left( 0,0 \right)$ $0$ $\left( 30,0 \right)$ $120$ (Maximum) $\left( 20,30 \right)$ $110$ $\left( 0,50 \right)$ $50$

Example: Determine the minimum value of the objective function graphically.,

$\text{Z = -50x + 20y }...\text{ }\left( 1 \right)$

Subject to the constraints:

• $\text{2x - y }\ge \text{ -5 }...\text{ }\left( 2 \right)$

• $\text{3x + y }\ge \text{ 3 }...\text{ }\left( 3 \right)$

• $\text{2x - 3y }\le \text{ 12 }...\text{ }\left( 4 \right)$

• $\text{x}\ge \text{ 0 , y}\ge \text{ 0 }...\text{ }\left( 5 \right)$

Ans: We need to graph the feasible region of the system of inequalities $\left( 2 \right)$ to $\left( 5 \right)$. The visible shaded area is shown in the graph.

 Corner Point $\text{Z = -50x + 20y}$ $\left( 0,5 \right)$ $100$ $\left( 0,3 \right)$ $60$ $\left( 1,0 \right)$ $-50$ $\left( 6,0 \right)$ $-300$ (Smallest)

By Observation that the feasible region is unbounded.

At the corner points, we now examine $\text{Z}$ .

From this table, we found that $-300$ is the smallest value of $\text{Z}$ at the corner point $\left( 6,0 \right)$.

Since the region would have been bounded, this smallest value of $\text{Z}$ is the minimum value of $\text{Z}$  (Theorem $\left( 2 \right)$).

But here, we have seen that the feasible part is unbounded.

Therefore, $-300$ may or may not be the minimum value of $\text{Z}$.

We use a graph to decide on this topic.

$\text{-50x + 20y -300}$ that is,

$\text{-5x + 2y -30}$  (By dividing the above Equation by $\text{10}$)

And need to confirm whether the resulting open half-plane has points in common with the feasible region or not.

If it has common attributes, then $-300$ will not be the minimum value of $\text{Z}$ . Otherwise, $-300$ will be the minimum value of $\text{Z}$ .

As shown in the above graph, it has common points, and hence,

$\text{Z = -50x + 20y}$ has no minimum value subject to the given constraints.

General Features of Linear Programming Problems

1. A convex region is always the viable zone.

2. The vertex (corner) of the feasible region is where the objective function's maximum (or most minor) solution occurs.

3. If two corner points have the same maximum (or lowest) objective function value, then every point on the line segment connecting them has the same top (or minimum) value.

Different Types of Linear Programming Problems:

The following are some of the most crucial linear programming problems:

1. Manufacturing Problems:

When each product requires a fixed quantity of workforce, machine hours, labour hour per unit of product, warehouse space per unit of output, and so on, determine the number of units of various things that a company should manufacture and sell to optimise profit.

1. Diet Problems:

Determine the number of constituents/nutrients that should be included in a diet to keep the expense of the intended diet as low as possible while ensuring that each constituent/nutrient is present in a minimum amount.

1. Transportation problems:

Determine a transportation timetable to determine the most cost-effective method of moving a product from various plants/factories to multiple to keep the intended diet’s expense markets.

Revision Notes CBSE Class 12 Maths Notes Chapter 12 Linear Programming

What is Linear Programming

Linear programming can be defined as the method that can be used for optimizing various operations. This is done with the help of some constraints. Linear programming also consists of linear functions. Linear functions, in turn, are subjected to various constraints in the form of linear equations. These constraints can also be in the form of inequalities.

An important point to stress in these revision notes of Class 12 Maths Chapter 12 is that linear programming can also be used for finding the optimum resource utilization. Also, if we look at the term ‘linear programming,’ then you will realize that it is made from two words. These words are linear and programming.

The word ‘linear’ refers to the relationship that exists between various variables with degree one. On the other hand, the word programming refers to the process of selecting the best solution out of all alternatives.

Students who are going through these Class 12 Notes on Linear Programming should note that apart from mathematics, the concept of linear programming is also used in other fields. Some of those fields are manufacturing, business, economics, and telecommunication.

It would be best if you also remembered that linear programming is also known as linear optimization. It is also known by its abbreviated form, which is LP. Linear programming can also be used for calculating loss and profit.

According to various NCERT Class 12 Revision Notes Maths Chapter 12, there are different assumptions that are made while working with linear programming. And these assumptions are mentioned below.

• The linear function or the objective function has to be optimized

• The total number of constraints should be expressed in quantitative terms

• The relationship that exists between the objective function and the constraints should also be linear

It is vital for all students to be familiar with these assumptions before they practice questions related to the Maths Class 12 Chapter 12 Revision Notes.

Apart from the assumptions, there are also some essential components that are related to linear programming. Do you want to know what those components are? If yes, then go through the list that is mentioned below.

• Data

• Constraints

• Decision variables

• Objective functions

The topic of Class 12 Revision Notes Chapter 12 cannot be complete without talking about the vital characteristics of linear programming. So, this is precisely what we will discuss now.

There are many NCERT solutions Chapter 12 Class 12 Maths Revision Notes that state that there are five characteristics of a linear programming problem. These characteristics are:

Constraints

Constraints can be explained as the limitations that are imposed in a mathematical form. This is usually regarding the resource.

Objective Function

The objective function should always be specified in a quantitative way when it comes to a linear programming problem.

Linearity

Linearity can be described as the relationship that exists between two or more variables in a function. It must always be linear. This means that the degree of the variable should always be equal to one.

Finiteness

According to the Class 12 Maths Revision Notes Chapter 12, there should always be finite and infinite numbers of output and input. What we mean by this is that, if the function has infinite factors, then the optimal solution should not be feasible.

Non-Negativity

Non-negativity refers to the fact that the value of a variable should either be zero or position. A variable should not have a negative value.

(The image has been taken from Google. Apart from this image, there was no specific image that could be added in this article to enhance the understanding of the readers.)

The Linear Programming Simplex Method

The linear programming simplex method is an important topic that students should learn about. This is why students should focus on this topic in these Linear Programming Class 12 Maths Revision Notes.

Simply put, the simplex method is used to solve linear programming models. This method can help an individual find the perfect solution to any problem. It consists of a tableau, slack variables, and pivot variables for the optimization of a problem. The algorithm can also be used for:

• To normalize restrictions

• Change various variables and normalize the sign of independent terms

• Helps in matching the objective functions to zero

• Write the first tableau of the simplex method

• Stopping condition

• Output and input variable choices

• Again updating tableau

• Continuing the iteration until the user gets the optimal solution

The Applications of Linear Programming

There are many applications of linear programming. If we were to look at a real-time example, then it could be considering the limitations of materials and labour and finding the best production levels for the highest profit in various circumstances. This example is an integral part of optimization techniques, which is an important area of study in mathematics.

The applications of linear programming in other fields include solving problems related to design and manufacturing for shape optimization in the field of engineering. In the field of efficient manufacturing, linear programming helps in maximizing profits.

For the energy industry, linear programming plays a vital role by providing various methods for the optimization of the electric power system. Also, in transportation optimization, it can help with time efficiency and cost optimization.

Did you know that apart from optimizing various solutions, linear programming also solves many functional problems in operations analysis? These problems are represented as linear programming problems.

There are also some particular linear programming problems like multi-commodity flow queries and network flow queries. These problems are said to be important because these problems produce various research solutions on functional algorithms.

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• Boosts student confidence for exams.

Conclusion

At Vedantu, our Class 12 Maths Revision notes are your go-to buddies for mastering Linear Programming. We break down the important stuff, like understanding problems and finding solutions. Our organized format makes studying a breeze, helping you remember things better. We keep it real with practical examples, showing how it all connects to everyday life. These notes are the secret sauce for a strong grip on Linear Programming and acing your exams. Students love them because they make learning easy, giving you the skills to shine in math. Trust Vedantu's notes – we're here to make your math journey awesome!

FAQs on Linear Programming Class 12 Notes CBSE Maths Chapter 12 [Free PDF Download]

1. Define linear programming.

Linear programming can be defined as the process that can be used for optimizing various problems that are subjected to some constraints. In other words, linear programming can also be explained as the process of minimizing and maximizing different linear functions under the linear inequality constraints. Many experts consider the task of solving problems related to linear programming as relatively easy.

2. What are the different types of linear programming?

The different types of linear programming are:

• Using the simplex method to solve linear programming problems

• Using R to solve linear programming questions

• Solving linear programming questions by using the graphical method

• Using the open solver technique to solve linear programming problems

3. Mention the requirements of linear programming.

There are five basic requirements for linear programming. These requirements are:

• Constraints

• Linearity

• Finiteness

• Non-negativity

• Objective function

4. What are the advantages of using linear programming?

Some advantages of using linear programming are:

• It allows one to solve multi-dimensional issues

• Linear programming provides insights into various business problems

• Linear programming also helps in making adjustments. This is according to the condition change theory

• One can find the best solution to a problem by using linear programming. This is done by calculating the profit and cost of various things

5. What do you understand about linear programming problems?

The linear programming problems or LPP can help an individual to find the best solution to a linear function, which is also known as the objective function. This is done when the linear function is placed under various constraints or a set of linear inequality constraints.