Linear Programming Class 12th Notes - Free NCERT Class 12 Maths Chapter 12 Notes - Download PDF

Linear Programming Class 12th Notes - Free NCERT Class 12 Maths Chapter 12 Notes - Download PDF

Edited By Ramraj Saini | Updated on Apr 23, 2022 01:25 PM IST

Linear Programming belongs to the 12 chapter of NCERT. The NCERT Class 12 Maths chapter 12 notes cover up the main portions of the chapter. Class 12 Math chapter 12 notes are mainly focused on the important formulas, the way to approach the problem. A Class 12 Maths chapter 12 note has the objective to either maximize or minimize the numerical values.

Notes for Class 12 Maths chapter 12 is made in such a way that it contains linear function. These functions are based on the factor in the form of linear equations or inequalities. CBSE class 12 maths chapter 12 notes not only covers the NCERT notes but also covers NCERT notes for Class 12 Maths chapter 12.

After going through Class 12 Linear Programming notes

Also, students can refer,

NCERT Class 12 Maths Chapter 12 Notes

Linear programming is generally defined as the technique for maximizing or minimizing a linear function of several variables like input or output cost.

Linear Programming Problem And Its Mathematical Formula

A Mountaineering equipment dealer deals in only two items–tent and rucksacks. He has Rs 50,000 for investment and has storage of compiling a maximum of 60 pieces. A tent is priced at Rs 2500 and a rucksack Rs 500. He estimates that by selling one tent he can get a profit of Rs 250 and that from the. sale of one rucksack he earns a profit of Rs 75. So how many tents and rucksacks should he buy from the money to maximize his total profit, assuming: he can sell all the products he buys.

Here we can observe

  • The dealer can invest his money in buying a tent or rucksacks or a combination thereof. Later he would earn different ways of profits by following different investment strategies.
  • There are a lot of conditions i.e.His investment is of the maximum of Rs 50,000 and he has storage of a maximum of 60 pieces.

Let us assume that he decides to buy a tent only and no rucksacks, so he can buy 50000 ÷ 2500, i.e., 20 tents. Thus his profit can be written as: Rs (250 × 20), i.e., Rs 5000.

Suppose he chooses to buy rucksacks only and no tent. With the capital of Rs 50,000, he can buy 50000 ÷ 500, i.e. 100 rucksacks. But he can only store 60 pieces. Therefore, he can only buy 60 rucksacks which will give a profit of Rs (60 × 75), i.e., Rs 4500.

There are other possibilities too; like, he may choose to buy 10 tents and 50 rucksacks, as his storage is limited to only 60 pieces. The total profit would be calculated as Rs (10 × 250 + 50 × 75), i.e., Rs 6250 and so on.

Thus, we find that the dealer can invest his money in many different ways so that he would earn different profits by the given different investment strategies.

Now the question arises how should he invest to get the maximum profit. For this let us see the process mathematically.

Mathematical Formulation Of The Problem

Let x be the number of tents and y be the number of rucksacks that the dealer buys. Here, x and y must be non-negative, i.e.,

\\ x \geqslant 0..............1 \\ y \geqslant 0 ..............2 \text{(Non-negative constraints)}

Mathematically,

2500x + 500y ≤ 50000 (investment)

or 5x + y ≤ 100 ... (3)

and x + y ≤ 60 (storage) ... (4)

Z = 250x + 75y ( objective function) ... (5)

Now, the given problems now reduces to:

Maximize Z = 250x + 75y

5x + y ≤ 100

x + y ≤ 60

x ≥ 0, y ≥ 0

Such a problem is called linear programming.

Some Basic Terminology

Objective function :

Let Z = ax + by being a linear function, where a, b are the constants, Linear objective function goes by the computation of the maximum or the minimum of X

Let, Z = ax + by be a linear objective function. Then the variables x and y are called decision variables.

Constraints :

Constraints are the limitations or the restrictions imposed on the decision variables.

Optimization problem:

A problem that asks to maximize or minimize a linear function limited to certain constraints.

Graphical Method Of Solving Linear Programming Problems

Let us see a graph

5x + y ≤ 100 ... (1)

x + y ≤ 60 ... (2)

x ≥ 0 ... (3)

y ≥ 0 ... (4)

1646395703418

Each point in the shaded region represents a feasible choice, therefore, is called the feasible region.

Any points outside feasible points are called infeasible points.

Optimal (feasible) solution:

Any point in the feasible region that gives the maximum or minimum value of the objective function is called an optimal solution.

Methods Of Solving A Linear Problem

  1. Find the feasible region of the problem and find the vertices.

  2. Find the objective function Z = ax + by. Let M and m be the largest and the smallest point of the problem

  3. i) When the area is bounded. "M" and "m" are maximum and minimum values If a feasible area is unbounded then

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  • ax +by > M no common points common with the feasible region.

  • ax +by < m no common points common with the feasible region.

Different Types Of Linear Programming Problems

  • Manufacturing Problem: - Here we describe the different types of unit that should be produced in a firm

  • Diet Problems: - Here we discuss the requirement of essential nutrients required for our body.

  • Transportation Problem: - Here we discuss the cheapest way of transporting materials from factories to different locations

This brings us to the end of the chapter.

Significance Of NCERT Class 12 Math Chapter 12 Notes

Class 12 linear programming notes will be helping the students to revise the chapter and get a gist of the important topics required. Here all the important parts are nicely covered starting from basic to advance. Also, Class 12 Math chapter 12 notes are useful for completing the Class 12 CBSE Syllabus and for competitive exams like BITSAT, and JEE MAINS. Class 12 Maths chapter 12 notes pdf download can be used to prepare in offline mode.

Linear programming Class 12 notes pdf download: link

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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