Class 9 Chapter 3 (Coordinate Geometry) Class Notes

 

Coordinate Geometry

Cartesian System

If we take two number lines, one horizontal and one vertical, and then combine them in such a way that they intersect each other at their zeroes, and then they form a Cartesian Plane.

• The horizontal line is known as the x-axis and the vertical line is known as the y-axis.

• The point where these two lines intersects each other is called the origin. It is represented as ‘O’.

• OX and OY are the positive directions as the positive numbers lie in the right and upward direction.

• Similarly, the left and the downward directions are the negative directions as all the negative numbers lie there.

Quadrants of the Cartesian Plane

The Cartesian plane is dividing into four quadrants named as Quadrant I, II, III, and IV anticlockwise from OX.

 

Coordinates of a Point

To write the coordinates of a point we need to follow these rules-

• The x - coordinate of a point is marked by drawing perpendicular from the y-axis measured a length of the x-axis .It is also called the Abscissa.

• The y - coordinate of a point is marked by drawing a perpendicular from the x-axis measured a length of the y-axis .It is also called the Ordinate.

• While writing the coordinates of a point in the coordinate plane, the x - coordinate comes first, and then the y - coordinate. We write the coordinates in brackets.

In the above figure, OB = CA = x coordinate (Abscissa), and CO = AB = y coordinate (Ordinate).

We write the coordinate as (x, y).

Remark: As the origin O has zero distance from the x-axis and the y-axis so its abscissa and ordinate are zero. Hence the coordinate of the origin is (0, 0).

The relationship between the signs of the coordinates of a point and the quadrant of a point in which it lies.

Quadrant	Coordinate	Sign in the quadrant
I	(+, +)	1st quadrant is enclosed by the positive x-axis and the positive y-axis.
II	(-, +)	2nd quadrant is enclosed by the negative x-axis and the positive y-axis.
III	(-, -)	3rd quadrant is enclosed by the negative x-axis and the negative y-axis.
IV	(+, -)	4th quadrant is enclosed by the positive x-axis and the negative y-axis

Plotting a Point in the Plane if its Coordinates are Given

Steps to plot the point (2, 3) on the Cartesian plane -

• First of all, we need to draw the Cartesian plane by drawing the coordinate axes with 1 unit = 1 cm.

• To mark the x coordinates, starting from 0 moves towards the positive x-axis and count to 2.

• To mark the y coordinate, starting from 2 moves upwards in the positive direction and count to 3.

• Now this point is the coordinate (2, 3)

Likewise, we can plot all the other points, like (-3, 1) and (-1.5,-2.5) in the right site figure.

Is the coordinates (x, y) = (y, x)?

Let x = (-4) and y = (-2)

So (x, y) = (- 4, – 2)

(y, x) = (- 2, - 4)

Let’s mark these coordinates on the Cartesian plane.

You can see that the positions of both the points are different in the Cartesian plane. So,

If x ≠ y, then (x, y) ≠ (y, x), and (x, y) = (y, x), if x = y.

 

 

Example: Write the coordinates of the points marked on the axes.From the figure,
(i) The coordinates of point A are (4, 0).
(ii) The coordinates of point B are (0, 3).
(iii) The coordinates of point C are (-5, 0).
(iv) The coordinates of point D are (0, -4).
(v) The coordinates of point E are (2/3, 0).

 

 

 

 

Example:

Plot the points (6, 4), (- 6,- 4), (- 6, 4) and (6,- 4) on the Cartesian plane.

Solution:

As you can see in (6, 4) both the numbers are positive so it will come in the first quadrant.

For x coordinate, we will move towards the right and count to 6.

Then from that point go upward and count to 4.

Mark that point as the coordinate (6, 4).

 Similarly, we can plot all the other three points.

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Class 9 Chapter 2 (Polynomials) Class Notes

 

Notes of Polynomials

Topics in the Chapter

• Polynomials
• Classification of polynomials on the basis of number of terms
• Degree of a Polynomial
• Values of polynomials at different points
• Zeroes of a polynomial
• Identity: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
• Remainder Theorem
• Factor Theorem
• Identity: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

Polynomials

→ An algebraic expression in which the exponents of the variables are non-negative integers
are called polynomials.
For example: 3x⁴ + 2x³ + x + 9, 3x² etc.

Constant polynomial
→ A constant polynomial is of the form p(x) = k, where k is a real number.
For example, –9, 10, 0 are constant polynomials.

Zero polynomial
A constant polynomial ‘0’ is called zero polynomial.

General form of a polynomial
A polynomial of the form where are p(x) = a_nxⁿ + a_n-1x^n-1 + .... + a₁x + a₀x where a₀, a₁... a_r, are constant and a_n≠ 0.
Here, a₀, a₁... a_n are the respective coefficients of x⁰,x¹, x² ... xⁿ and n is the power of
the variable x.
a_nxⁿ + a_n-1x^n-1 - a₀ and are called the terms of p(x).

Classification of polynomials on the basis of number of terms

• A polynomial having one term is called a monomial. e.g. 3x, 25t³ etc.

• A polynomial having one term is called a monomial. e.g. 2t-6, 3x⁴ +2x etc.

• A polynomial having one term is called a monomial. e.g. 3x⁴ + 8x + 7 etc.

Degree of a Polynomial

• The degree of a polynomial is the highest exponent of the variable of the polynomial.
For example, the degree of polynomial 3x⁴ + 2x³ + x + 9 is 4.

• The degree of a term of a polynomial is the value of the exponent of the term.

Classification of polynomial according to their degrees 

• A polynomial of degree one is called a linear polynomial e.g. 3x+2, 4x, x+9.

• A polynomial of degree is called a quadratic polynomial. e.g. x²+ 9, 3x² + 4x + 6

• A polynomial of degree three is called a cubic polynomial e.g. 10x³+ 3, 9x³

Note: The degree of a non-zero constant polynomial is zero and the degree of a zero polynomial is not defined.

Values of polynomials at different points

• A polynomial is made up of constants and variables. Hence, the value of the polynomial changes
as the value of the variable in the polynomial changes.
• Thus, for the different values of the variable x, we get different values of the polynomial.


Zeroes of a polynomial

• A real number a is said to be the zero of polynomial p(x) if p(a) = 0, In this case, a is also called the root of the equation p(x) = 0.
Note:
• The maximum number of roots of a polynomial is less than or equal to the degree of the polynomial.
• A non-zero constant polynomial has no zeroes.
• A polynomial can have more than one zero.

Division of a polynomial by a monomial using long division method

Example: Divide x⁴ -2x³- 2x²+7x-15 by x-2

Steps of the Division of a Polynomial with a Non –Zero Polynomial

Divide x² - 3x -10 by 2 + x

Step 1:  Write the dividend and divisor in the descending order i.e. in the standard form. x² - 3x -10 and x + 2

Divide the first term of the dividend with the first term of the divisor.

x²/x = x this will be the first term of the quotient.

Step 2: Now multiply the divisor with this term of the quotient and subtract it from the dividend.

Step 3: Now the remainder is our new dividend so we will repeat the process again by dividing the dividend with the divisor.

Step 4: – (5x/x) = – 5 

Step 5: 

The remainder is zero.

Hence x² - 3x – 10 = (x + 2)(x - 5) + 0

Dividend = (Divisor × Quotient) + Remainder

Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).

As we know that

P(x) = g(x) q(x) + r(x)

If p(x) is divided by (x-t) then

If x = t

P (t) = (t - t).q (t) + r = 0

To find the remainder or to check the multiple of the polynomial we can use the remainder theorem.

Example:

What is the remainder if a⁴ + a³ – 2a² + a + 1 is divided by a – 1.

Solution:

P(x) = a⁴ + a³ – 2a² + a + 1

To find the zero of the (a – 1) we need to equate it to zero.

a -1 = 0

a = 1

p (1) = (1)⁴ + (1)³ – 2(1)² + (1) + 1

= 1 + 1 – 2 + 1 + 1

= 2

So by using the remainder theorem, we can easily find the remainder after the division of polynomial.

Factor Theorem

Factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then

1. (y - t) is a factor of p(y), if p(t) = 0, and

2. P (t) = 0 if (y – t) is a factor of p(y).

Example: 1

Check whether g(x) = x – 3 is the factor of p(x) = x³ - 4x² + x + 6 using factor theorem.

Solution:

According to the factor theorem if x - 3 is the factor of p(x) then p(3) = 0, as the root of x – 3 is 3.

P (3) = (3)³ - 4(3)² + (3) + 6

= 27 – 36 + 3 + 6 = 0

Hence, g(x) is the factor of p(x).

Example: 2

Find the value of k, if x – 1 is a factor of p(x) = kx² – √2x + 1

Solution:

As x -1 is the factor so p(1) = 0

Factorization of Polynomials

Factorization can be done by three methods

1. By taking out the common factor

If we have to factorize x² –x then we can do it by taking x common.

x(x – 1) so that x and x-1 are the factors of x² – x.

2. By grouping

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

3. By splitting the middle term

x² + bx + c = x² + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

Example: 1

Factorize 6x² + 17x + 5 by splitting the middle term.

Solution:

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x² + 17x + 5 =6 x² + (2 + 15) x + 5

= 6 x² + 2x + 15x + 5

= 2 x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Algebraic Identities

1. (x + y)2 = x2 + 2xy + y2

2. (x - y)2 = x2 - 2xy + y2

3. (x + y) (x - y) = x2 - y2

4. (x + a) (x + b) = x2 + (a + b)x + ab

5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2

7. (x - y)3 = x3- y3 - 3xy(x - y) = x3 - y3 - 3x2y + 3xy2

8. x3 + y3 = (x + y)(x2 – xy + y2)

9. x3 - y3 = (x - y)(x2 + xy + y2)

10. x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz - zx)       x3 + y3 + z3 = 3xyz if x + y + z = 0

Example: 2

Factorize 8x³ + 27y³ + 36x²y + 54xy²

Solution:

The given expression can be written as

= (2x)³ + (3y)³ + 3(4x²) (3y) + 3(2x) (9y²)

= (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)²

= (2x + 3y)³ (Using Identity VI)

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.

Example: 3

Factorize 4x² + y² + z² – 4xy – 2yz + 4xz.

Solution:

4x² + y² + z² – 4xy – 2yz + 4xz = (2x)² + (–y)² + (z)² + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)

= [2x + (- y) + z]² (Using Identity V)

= (2x – y + z)² = (2x – y + z) (2x – y + z)

Identity: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

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Class 9 Chapter 1 (Number System) Class Notes

 

Notes on Number Systems

 

Introduction to Natural Numbers

Non-negative counting numbers excluding zero are called Natural Numbers.

N = 1, 2, 3, 4, 5, ……….

Whole Numbers

All natural numbers including zero are called Whole Numbers.

W = 0, 1, 2, 3, 4, 5, …………….

Integers

All natural numbers, negative numbers and 0, together are called Integers.

Z = – 3, – 2, – 1, 0, 1, 2, 3, 4, …………..

Rational Numbers

The number ‘a’ is called Rational if it can be written in the form of r/s where ‘r’ and ‘s’ are integers and s ≠ 0,

Q = 2/3, 3/5, etc. all are rational numbers.

 

How to find a rational number between two given numbers?

To find the rational number between two given numbers ‘a’ and ‘b’ is \(\dfrac{{a + b}}{2}\)

Example: Find 2 rational numbers between 4 and 5.

Solution: To find the rational number between 4 and 5

\(\dfrac{{a + b}}{2} = \dfrac{{4 + 5}}{2} = \dfrac{9}{2}\)

To find another number we will follow the same process again.

\(\dfrac{1}{2}\left( {4 + \dfrac{9}{2}} \right) = \dfrac{1}{2} \times \dfrac{{17}}{2} = \dfrac{{17}}{4}\)

Hence the two rational numbers between 4 and 5 are \(\dfrac{9}{2}\) and \(\dfrac{{17}}{4}\).

Remark: There could be unlimited rational numbers between any two rational numbers.

 

Irrational Numbers

The number ‘a’ which cannot be written in the form of p/q is called irrational, where p and q are integers and q ≠ 0 or you can say that the numbers which are not rational are called Irrational Numbers.

Example - \(\sqrt 7 \), \(\sqrt {11} \), π  etc.

 

Real Numbers

All numbers including both rational and irrational numbers are called Real Numbers.

R = – 2, \( - \dfrac{2}{3}\), 0, 3 and \(\sqrt {2} \)

Real Numbers and their Decimal Expansions

1. Rational Numbers

If the rational number is in the form of \(\dfrac{a}{b}\) then by dividing a by b we can get two situations.

a). If the remainder becomes zero

While dividing if we get zero as the remainder after some steps then the decimal expansion of such number is called terminating.

Example: \(\dfrac{7}{8} = 0.875\)

b). If the remainder does not become zero

While dividing if the decimal expansion continues and not becomes zero then it is called non-terminating or repeating expansion.

Example: \(\dfrac{1}{3} = 0.3333.... \). It can be written as  \(0.\overline 3 \).

Hence, the decimal expansion of rational numbers could be terminating or non-terminating recurring and vice-versa.

2. Irrational Numbers

If we do the decimal expansion of an irrational number then it would be non –terminating non-recurring and vice-versa. i. e. the remainder does not become zero and also not repeated.

Example:  π = 3.141592653589793238……

 

Representing Real Numbers on the Number Line

To represent the real numbers on the number line we use the process of successive magnification in which we visualize the numbers through a magnifying glass on the number line.

Example: Mark \(4.\overline {26} \) on the number line up to 4 decimal places.

Step 1: The number lies between 4 and 5, so we divide it into 10 equal parts. Now for the first decimal place, we will mark the number between 4.2 and 4.3.
Step 2: Now we will divide it into 10 equal parts again. The second decimal place will be between 4.26 and 4.27.
Step 3: Now we will again divide it into 10 equal parts. The third decimal place will be between 4.262 and 4.263.
Step 4: By doing the same process again we will mark the point at 4.2626.

Operations on Real Numbers

1. The sum, difference, product and quotient of two rational numbers will be rational.

Example:

\( \Rightarrow \dfrac{3}{4} + \dfrac{7}{4} = \dfrac{{10}}{4} = \dfrac{5}{2}\)

 \( \Rightarrow \dfrac{7}{4} - \dfrac{3}{4} = \dfrac{4}{4} = 1\)

 \( \Rightarrow \dfrac{7}{4} \times \dfrac{3}{4} = \dfrac{{21}}{{16}}\)

 \( \Rightarrow \dfrac{7}{4} \div \dfrac{3}{4} = \dfrac{7}{3}\)

2. If we add or subtract a rational number with an irrational number then the outcome will be irrational.

Example: If 5 is a rational number and \(\sqrt 7 \) is an irrational number then 5 + \(\sqrt 7 \) and 5 - \(\sqrt 7 \) are irrational numbers.

3. If we multiply or divide a non-zero rational number with an irrational number then also the outcome will be irrational.

Example: If 7 is a rational number and \(\sqrt 5 \) is an irrational number then 7\(\sqrt 7 \) and \(\dfrac{7}{{\sqrt 5 }}\) are irrational numbers.

4. The sum, difference, product and quotient of two irrational numbers could be rational or irrational.

Example:

\(\sqrt 3  + \sqrt 3  = 2\sqrt 3 \;\;\;\;\;\;(irrational + irrational = irrational)\)

\(\sqrt 2  - \sqrt 2  = 0\;\;\;\;\;\;(irrational - irrational = rational)\)

\(\sqrt 6  \times \sqrt 6  = 6\;\;\;\;\;\;(irrational \times irrational = rational)\)

\(\dfrac{{\sqrt {13} }}{{\sqrt {13} }} = 1\;\;\;\;\;\;(irrational \div irrational = rational)\)

 

Finding Roots of a Positive Real Number ‘x’ geometrically and mark it on the Number Line

To find \(\sqrt x \) geometrically

1. First of all, mark the distance x unit from point A on the line so that AB = x unit.

2. From B mark a point C with the distance of 1 unit, so that BC = 1 unit.

3. Take the midpoint of AC and mark it as O. Then take OC as the radius and draw a semicircle.

4. From the point B draw a perpendicular BD which intersects the semicircle at point D.

The length of BD = \(\sqrt x \).

To mark the position of \(\sqrt x \) on the number line, we will take AC as the number line, with B as zero. So C is point 1 on the number line.

Now we will take B as the centre and BD as the radius, and draw the arc on the number line at point E.

Now E is \(\sqrt x \) on the number line.

 

Identities Related to Square Roots

If p and q are two positive real numbers, then

1. \(\sqrt {pq}  = \sqrt p  \times \sqrt q \) 
2. \(\sqrt {\dfrac{p}{q}}  = \dfrac{{\sqrt p }}{{\sqrt q }}\)
3. \(\left( {\sqrt p  + \sqrt q } \right)\left( {\sqrt p  - \sqrt q } \right) = p - q\)
4. \(\left( {p + \sqrt q } \right)\left( {p - \sqrt q } \right) = {p^2} - q\)
5. \(\left( {\sqrt p  + \sqrt q } \right)\left( {\sqrt r  + \sqrt s } \right) = \sqrt {pr}  + \sqrt {ps}  + \sqrt {qr}  + \sqrt {qs} \)
6. \({\left( {\sqrt p  + \sqrt q } \right)^2} = p + 2\sqrt {pq}  + q\)
   

Example 1: Simplify: \(\left( {3 + \sqrt 7 } \right)\left( {5 - \sqrt {11} } \right)\)

We will use the identity

\(\left( {\sqrt p  + \sqrt q } \right)\left( {\sqrt r  + \sqrt s } \right) = \sqrt {pr}  + \sqrt {ps}  + \sqrt {qr}  + \sqrt {qs} \)

\(\left( {3 + \sqrt 7 } \right)\left( {5 - \sqrt {11} } \right) = 15 + 5\sqrt 7  + 3\sqrt {11}  + \sqrt {77} \)

Eample 2: Simplify:\(\left( {\sqrt 5  + \sqrt {11} } \right)\left( {\sqrt 5  - \sqrt {11} } \right)\)

We will use the identity

\(\left( {p + \sqrt q } \right)\left( {p - \sqrt q } \right) = {p^2} - q\)
\(\left( {\sqrt 5  + \sqrt {11} } \right)\left( {\sqrt 5  - \sqrt {11} } \right) = 5 - 11 =  - 6\)

Rationalizing the Denominator

Rationalize the denominator means to convert the denominator containing square root term into a rational number by finding the equivalent fraction of the given fraction.

For which we can use the identities of the real numbers.

Example: Rationalize the denominator of \(\dfrac{7}{{7 - \sqrt 3 }}\).

Solution:

We will use the identity \(\left( {p + \sqrt q } \right)\left( {p - \sqrt q } \right) = {p^2} - q\) here.

\( \Rightarrow \frac{7}{{7 - \sqrt 3 }} \times \frac{{7 + \sqrt 3 }}{{7 + \sqrt 3 }} = \frac{{7(7 + \sqrt 3 )}}{{49 - 3}} = \frac{{49 + 7\sqrt 3 }}{{46}}\)

 

Laws of Exponents for Real Numbers

If we have a and b as the base and m and n as the exponents, then

1. a^m × aⁿ =a^m+n
2. (a^m)ⁿ = a^mn
3. \(\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}};\,m > n\)
4. a^m b^m = (ab)^m
5. a⁰ = 1
6. a¹ = a
7. \(\dfrac{1}{{{a^n}}} = {a^{ - n}}\)

• Let a > 0 be a real number and n is a positive integer,  \(\sqrt[n]{a} = b\), if \({b^n} = a\) and b > 0, then \(\sqrt[n]{a} = {a^{\frac{1}{n}}}\)

• Let a > 0 be a real number m and n be integers such that m and n have no common factors other than 1, and n > 0. Then \({a^{\frac{m}{n}}} = {\left( {\sqrt[n]{a}} \right)^m}\)
  

 

 

Example: Simplify the expression (2x³y⁴) (3xy⁵)².

Solution: Here we will use the law of exponents

a^m × aⁿ =a^m+n and (a^m)ⁿ = a^mn

(2x³y⁴)(3xy⁵)²

(2x³y⁴)(3 ² x ² y¹⁰)

18. x³. x². y⁴. y¹⁰

18. x³⁺². y⁴⁺¹⁰

18x⁵y¹⁴

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Class 12 Chapter 13 (Probability) Class Notes)

Probability

Introduction:

Probability: probability measures how much chance is there for my expected
result to happen.

\[{\rm{Probability  =  }}\dfrac{{{\rm{Number \,\, of \,\, Favourable \,\, Outcomes}}}}{{{\rm{Total \,\, Number \,\, of \,\, Outcomes}}}}\]

Example: Let there be a basket with 3 balls: Red, Green, Blue. If I want to pick a Red ball,
we can calculate the probability of picking a red ball.

Total number of possibilities when we pick a ball from the basket = 3 ( The ball we pick can be Red, Green or blue)

Total number of favorable possibility i.e. possibility to get a red ball = 1

Therefore, Probability of getting a red ball, \(P(red) = \dfrac{1}{3}\)

Sample space: set of all possible outcomes of a random experiment is called Sample space. It is denoted by S.

Sample point: each element of the sample space is called the sample point.
Example: for Tossing a coin, the sample space is S = {HEAD, TAIL} and and the element in sample space namely head, or tail, each of them can be called as a sample point

Event: it is the set of favorable outcome.
Example: if you throw a coin and if you are looking for head, then head is the favorable outcome.

Mutually Exclusive events: two events are said to be mutually exclusive if
there is no common element between them.

Exhaustive events: the given events are Exhaustive if when I take the
elements in those events forms the given sample space.

 

Conditional Probability:

Given a condition and you have to find the probability

Example: when I throw a dice, what is the probability that the outcome is 3 given that the outcome is odd.

The answer is 1/3 as the odd numbers are 1, 3 and 5. So we need the probability of getting the number 3 out of the numbers 1,3,5. The chance is one out of three so The probability is 1/3.

∴ if number is odd,then the probability that it is 3 is = P(3|number is odd) = 1/3

Note:But P(3) = 1/6 If there is no condition. As the probability of getting a single number named 3, out of total 6 numbers namely 1,2,3,4,5,6 is 1/6

 

Conditional Probability Definition:

Probability of event E is called the conditional probability of F given that E has already occurred, and it's defined by P(F|E).

Formula for conditional probability
P(F|E) = P(F∩E) / P(E)

 

Conditional Probability Problems:

Example: In a survey in a class it was found that the probability of a student watching ABC videos is 0.8. and the probability that a student is both topper and also watches ABC videos is 0.792. what is the probability that a student is a topper if he watches ABC videos?

Solution:
Let E denotes the event that a person watches ABC videos 
Let F denotes the event that a person is topper .

Then, P(E) = The probability that a person watches ABC videos = 0.8
P(E ∩ F ) = probability that one is both topper and also watches ABC videos = 0.792
and P(F|E) = the probability that a person is a topper if he watches ABC videos = ?
then according to Formula for conditional probability: P(F|E) = P(F∩E)/ P(E)
= 0.792/0.8
= 0.99
∴ The probability that a person is a topper if he watches ABC videos is 0.99

  

Properties of conditional probability:

• P(S|F) = 1
• P(F|F) = 1
• P((A∪B)|F) = P(A|F) + P(B|F) – P((A∩B)|F)
• P(E'|F) = 1- P(E|F)

Example: Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) =2/5
Solution: ∵ 2P(A) = P(B) = 5/13
=> P(B) = 5/13 and P(A) = 1/2 x P(B) = 1/2  × 5/13 = 5/26

Now the formula of conditional probability is P(A|B) = P( A∩B)/ P(B)
∴ 2/ 5 = P( A∩B) /P(B)  [Since it was given in question that P(A|B)=2/5 ]

=> P(A∩B) = 2/5 X P(B)
= 2/5 × 5/13
= 2/13

So, P(A∩B)= 2/13 , but we need to find P(A ∪ B)
Now is there any formula relating P(A ∪ B) and P(A∩B) ?
Yes , the formula is P(A ∪ B)=P(A)+P(B)−P(A∩B)
Now you know values of P(A)= 5/ 26 , P(B)= 5/13 and P(A∩B)= 2/13.

So substitute these values in the formula, you will get value of P(A∪B).
That is P(A∪B)=P(A)+P(B)−P( A∩B)
=> P(A∪B)= 5/26 + 5/13 - 2/13
=> P(A∪B)= (5+10−4 )/26
=> P(A∪B)=11/26

∴ The value of P(A∪B) is 11/26

 

Example: Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle , what is the probability that the son is on one end given that the father is in middle?

Solution:

let Mother be denoted with M, Father be denoted with F and Son be denoted with S. If the sample space for all the possible ways of arranging them is denoted by S.
Then the possible elements in S are, S = {MFS, MSF, FMS, FSM, SMF, SFM}
Now E denotes son on one end as given in question so A={MFS, FMS, SMF, SFM}
and F denoted Father in middle as given in question , so B={MFS, SFM}
Since the common elements between A and B are MFS and SFM ∴A∩B ={MFS, SFM}
Now, P(E∩F) = probability of getting two elements of A∩B I.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM
= 2/6 = 1/3
and P(F)= probability of getting two elements of B i.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM.
= 2/6 = 1/3
∴ P(E|F) = P(E∩F) /P(F)
= (1/3)  / (1/3)
= 1
∴The probability that the son is on one end given that the father is in middle is 1.

Example: If A and B are events such that P(A|B) = P(B|A), then
(A)A ⊂ B but A ≠ B

(B)A = B

(C)A ∩ B = Φ

(D)P(A) = P(B)

Solution:
Given in question that P(A|B) = P(B|A)
But we know P(A|B) = P( A∩B) /P(B)
and P(B|A) = P( A∩B)/ P(A)
since P(A|B) = P(B|A)
=> P( A∩B)/ P(B) = P( A∩B)/ P(A)
=> P(A) = P(B)
Thus the correct option is (D).

 

Multiplication Theorem on Probability:

Let E and F be two events associated with a sample space of an experiment.
Then 

P(E ∩ F) = P(E) P(F|E), P(E) ≠ 0
= P(F) P(E|F), P(F) ≠ 0

If E, F and G are three events associated with a sample space, then

P(E∩F∩G) = P(E) P(F|E) P(G|E∩F)

Example: In a survey in a class, the probability for a person to watch ABC videos is 0.8 and the probability for a person to be a topper, if given that he watches ABC videos is 0.99. find the probability for a person to be both topper and watches ABC videos.

Solution:
let Event E denotes the event that a person watches ABC videos
let Event F denotes the event that a person is topper
then P(E) = the probability that a person watches ABC videos =0.8
and P(F|E)= the probability that a person is a topper if he watches ABC videos=0.99
then P (E ∩ F)= the probability that a person is both topper and also watches ABC videos
then according to Multiplication Theorem on probability
P (E ∩ F) = P (E) P (F | E)
= 0.8 x 0.99
= 0.792
∴ The probability that a person is both topper and also watches ABC videos = 0.792

 

Independent Events:
Let E and F be two events associated with a sample space S. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we say that the two events are independent. Thus, two events E and F will be independent, if

(a) P(F | E) = P(F), provided P (E) ≠ 0
(b) P(E | F) = P(E), provided P (F) ≠ 0

Using the multiplication theorem on probability, we have

(c) P(E ∩ F) = P(E) P(F)

Three events A, B and C are said to be mutually independent if all the
following conditions hold:

P(A ∩ B) = P(A) P(B)
P(A ∩ C) = P(A) P(C)
P(B ∩ C) = P(B) P(C)
and P(A ∩ B ∩ C) = P(A) P(B) P(C)

Example: If you throw two coins, then probability of getting head or tail in second coin is independent of probability of getting head or tail in first coin.

Example: Let A and B be independent events with P(A)=0.3 and P(B) = 0.4.
Find(i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A|B) (iv) P(B|A)
Solution:
(i) we have to find the value of P(A ∩ B) and they gave P(A) =0.3 and P(B) = 0.4.
given A and B are independent events.
We know that if A and B are independent events then P(A ∩ B) = P(A) P(B)
=>P(A ∩ B) = 0.3×0.4 = 0.12

(ii)we have to find the value of P(A ∪ B)
the formula for P(A ∪ B) is P (A ∪B)=P(A)+P(B)−P( A∩B)
=> P(A ∪ B)=0.3+0.4-0.12 = 0.58

(iii) P(A|B)=?
the formula for P(A|B) is P(A|B)= P( A∩B)/ P(B)
=> P(A|B)= 0.12/ 0.4 = 0.3

(iv) P(B|A)=? 
the formula for P(B|A) is P(B|A)= P( A∩B)/ P(A)
=> P(B|A)= 0.12/0.3 = 0.4

Example:If A and B are two events such that P(A)=1/4 ,P(B)=1/2 , ,P(A∩B)=1/8
find P(not A and not B).
Solution:
we know that P(not A and not B)=P(A'∩B')
But we also know that A'∩B'=(A∪B)' => P(A'∩B')=P((A∪B))'
∴P(not A and not B)=P(A'∩B')
=P((A∪B))'
=1-P(A∪B) ( ∵ P(A)' = 1-P(A) )
=1- (P(A)+P(B)−P( A∩B)) (∵ P(A∪B)=P(A)+P(B)−P( A∩B) )
=1- ( 1/4 + 1/2 - 1/8 )
=1- 5/8
=3/8

 

Bayes' Theorem:

If E1, E2,..., En are mutually exclusive and exhaustive events associated
with a sample space, and A is any event of non zero probability, then

 

Example: In a survey in a classroom it was found that chance of a person becoming a topper is 10% . if a person is a topper, there is a 99% chance that he watches ABC videos. One who is not a topper there is 1% chance of watching ABC videos. Then, what is the Probability for one to be a Topper , given if he watches ABC videos?

Solution:
Let “A person to be a Topper” is Event A and “To watch ABC videos” is Event B

Given P(A) = 0.1 = probability for a person to be a Topper.

So P(A')= 1-P(A)=1 - 0.1= 0.9 = probability for a person to be not a Topper

Given P(B|A) = 0.99 =  probability that a person watches ABC videos, if he is a topper, 

so P(B'|A) = 1 - P(B|A) = 1 - 0.99 = 0.01 = probability that a person not watches ABC videos, if he is a topper

Given P(B|A')=0.01=probability that a person watches ABC videos , if he is not a topper 
so P(B'|A')=1-0.01=0.99=probability that a person not watches ABC videos , if he is not a topper

Therefore according to Bayes' Theorem,Probability for one to be a Topper , if he watches ABC videos is

= (0.1x0.99)/ {( (0.1x0.99)+(0.01x0.9))}
= 0.9909

∴ if one watches ABC videos , for him there is 99.09% chance to become topper.

Example: A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution:
Let E1= the event of selecting first bag.
E2=the event of selecting second bag.
A = The event of getting red ball.
Since there is equal chance of selecting first bag or selecting second bag,
P(E1)=P(E2)= 1/2

now P(A|E1)=P(Drawing a red ball from first bag )= 4/8
and P(A|E2)=P(Drawing a red ball from second bag )= 2/8 =1/4

probability that ball is drawn from the first bag given that the ball drawn is red is P(E1|A)

 

Example:Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Solution:
Let E1= event that the first group wins the competition
E2 = event that the second group wins the competition
A = event of introducing a new product.
Then,
P(E1) = Probability that the first group wins the competition = 0.6
P(E2) = Probability that the second group wins the competition = 0.4
P(A|E1)=Probability of introducing new product if the first group wins =0.7
P(A|E2)=Probability of introducing new product if the second group wins =0.3
The probability that the new product is introduced by the second group is given by

 

 

Random Variables and its probability distributions:

A random variable is a real valued function whose domain is the sample space of a random experiment.
The probability distribution of a random variable X is the system of numbers

where P_i > 0, i = 1 to n and P₁ + P₂ + P₃ + …..... + P_n =1

Example: Suppose that a coin is tossed twice so that the sample space is S = {HH,HT,TH,TT}. Let X represent the number of heads that can come up. So with each sample point we can associate a number for X so for HH the value of X is 2 as in HH there are 2 heads, for HT the value of X is 1 as there is only one head in HT. similarly for TH, X = 1 and for TT value of X = 0 as there are no heads in TT.

So in the sample space S = {HH,HT,TH,TT} assuming that the coin is fair,
P(HH)= ¼ , P(HT)= ¼ , P(TH)= ¼ , P(TT)= ¼
therefore P(X=0) = P(TT) = ¼
P(X=1) = P(HT ∪ TH) = P(HT) + P(TH) = ¼ + ¼ = ½
P(X=2) = P(HH) = ¼
Thus the table formed is shown below:

Example: A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Solution:
Let the probability of getting a tail in the biased coin be x.
∴ P (T) = x
⇒ P (H) = 3x
For a biased coin, P (T) + P (H) = 1
⇒ x + 3x = 1
⇒ 4x = 1
⇒ x = \(\frac{1}{4}\)
∴ P(T) =  \(\frac{1}{4}\)   and   P(H) =  \(\frac{3}{4}\)

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.
Let X be the random variable representing the number of tails.
∴ P (X = 0) = P (no tail) = P (H) × P (H)= \(\frac{3}{4}\)  × \(\frac{3}{4}\)
=  \(\frac{9}{16}\)

P (X = 1) = P (one tail) = P (HT) + P (TH) = \(\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}\)
= \(\frac{3}{{16}} + \frac{3}{{16}}\)
= \(\frac{3}{8}\)

P (X = 2) = P (two tails) = P (TT)= \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{{16}}\)

Therefore, the required probability distribution is as follows.

 

 

Mean and Variance of a Random variable:

Let X be a random variable assuming values x₁, x₂,...., x_n with probabilities p₁, p₂, ..., p_n, respectively such that p_i ≥ 0,  

Mean of X, denoted by μ [or expected value of X denoted by E (X)] is defined as:

and variance denoted by σ² , is defined as

Standard deviation of the random variable X is defined as

Let us take the above table

Now mean is

 
= X₁P₁ + X₂P₂ + X₃P₃
= (0X¼ ) + (1 X ½)+ (2X ¼)
= 0 + ½ + ½
=1

Now Variance = σ_x² = E(X²) − (E(X))²

so to find Variance let us find E(X²) and (E(X))²

= 0 + ½ + 1
= 3/2
= 1.5

and (E(X))² = 1²=1

∴ Var(X)= E(X²) − (E(X))² = 1.5 − 1 = 0.5

now standard deviation of X = √(variance (X ))
= √0.5
≈ 0.707

Example: In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed,
and X = 1 if he is in favour. Find E(X) and Var(X).
Solution:
It is given that P(X = 0) = 30% = 30/100 = 0.3
P(X=1)=70%= 70/100 = 0.7

Therefore, the probability distribution is as follows.

x	0	1
P(x)	0.3	0.7

Then, Mean= E(X) =

 
= 0×0.3+1×0.7
=0.7

= 02×0.3+12×0.7 =0.7

Now,
∴ Var(X)= E(X²)-(E(X))²
= 0.7−(0.7)²
= 0.7 − 0.49
= 0.21

Example:The mean of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face is (A) 1 (B) 2 (C) 5 (D) 8/3

Solution:
Let X be the random variable representing a number on the dice.

The total number of observations is six.
∴P(X=1)= 3/6 = 1/2

P(X=2)= 2/6 = 1/3

P(X=5)= 1/6

∴ The probability distribution is as follows.

X	1	2	5
P(X)	1/2	1/3	1/6

Now, Mean=E(X)

 
=1×1/2 +2×1/3 +5×1/6
= 1/2 +2/3 +5/6
= 12/6
= 2

∴ The correct option is B.

 

Bernoulli Trials and Binomial Distribution:

Bernoulli Trials:
Trials of a random experiment are called Bernoulli trials, if they satisfy the following
conditions:
(i) There should be a finite number of trials
(ii) The trials should be independent
(iii) Each trial has exactly two outcomes: success or failure
(iv) The probability of success(or failure) remains the same in each trial.

Example: A tossed coin shows a ‘head’ or ‘tail’. If the occurrence of head is considered as success, then the occurrence of tail is a failure .so, if we consider tossing a coin two times as Bernoulli trials, because 

(i)There are finite number of trials that is 2
(ii)tossing a coin second time is completely independent of the result of first time tossing
(iii)each trial has only two outcomes namely either head or tail and
(iv)The probability of getting head or tail remains the same in each trial.

 

 

Binomial Distribution:

A random variable X taking values 0, 1, 2, ..., n is said to have a binomial distribution
with parameters n and p, if its probability distribution is given by 

P (X = r) = ⁿc_r p^r q^n–r,
where q = 1 – p and r = 0, 1, 2, ..., n.

Example: If a coin is tossed 10 times, find the probability of exactly 5 heads.

Solution: let X denote the number of heads.
So X has the binomial distribution with n = 10 an p = ½
∴ P(X = x) = ⁿC_x q^n-x p^x , x = 1,2,3....,n
∴ P(X = x) = ¹⁰c_x(1/2)^10-x(1/2)^x = ¹⁰c_x(1/2)¹⁰.
so P(X = 6) = ¹⁰c₅(1/2)¹⁰= (10 !)/ (5 !×5!) ×( 1/2¹⁰) = 63/ 256

Example:A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0? 

Solution:
Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.
Since the balls are drawn with replacement, the trials are Bernoulli trials.
X has a binomial distribution with n = 4 and P= 1/10

∴ Q = 1 − P =1 − 1/10 = 9/10

∴ P(X = x) = ⁿC_x q^n-x p^x , x = 1,2,3....,n
= ⁴C_x ( 9/10 )^n−x( 1/10 )^x

P (none marked with 0) = P (X = 0)
= ⁴C₀ ( 9/10 )⁴ ( 1/10 )⁰
= 1×( 9/10)⁴ 

= ( 9/10 )⁴

 

Example: Find the probability of getting 5 exactly twice in 7 throws of a dice.

Solution:
The repeated tossing of a dice are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the dice.
Probability of getting 5 in a single throw of the dice is P=16
∴ Q = 1−P = 1 − 1/6 = 5/6

Clearly, X has the probability distribution with n = 7 and P=16
∴P(X = x) = ⁿC_x q^n-x p^x
= ⁷C_x (5/6)^7−x (1/6)^x

P (getting 5 exactly twice) = P(X = 2)
=⁷C₂ (5/6)⁵ (1/6)²
= 21×(5/6)⁵× 1/36
= 7 /12 ×(5/6)⁵

 

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Class 12 Chapter 12 (Linear Programming) Class Notes

 Linear Programming

Introduction:

Optimization is the way of life. We all have finite resources and time and we want to make the most of them. Different type of problems which seek to maximize or minimize profit or cost, form a general class of problems called optimization problems. So, an optimization problem may involve finding maximum profit, minimum cost, or minimum use of resources etc.

Linear programming problems are one of the important class of optimization problem. Linear Programming is a mathematical modeling technique in which a linear function is maximized or minimized when subjected to various constraints. A typical example would be taking the limitations of materials and labor, and then determining the best production levels for maximal profits under those conditions.

In general, a linear programming problem is specified as follows:

Given:

(i) n variables x₁, x₂, x₃,... ,x_n.

(ii) m linear inequalities in these variables.

E.g., 2x₁ + 5x₂ ≤ 10, 0 ≤ x₁ ≤ 2, etc.

(iii) We may also have a linear objective function.

E.g. Maximize/Minimize 3x₁ + 5x₂ + 7x₃

Goal: Find values for the x_i’s that satisfy the constraints and maximize/minimize the objective function.

 

Linear Programming Problem and its Mathematical Formulation:

Formulation of an LPP refers to translating the real-world problem into the form of mathematical equations which could be solved. It usually requires a thorough understanding of the problem.

Let us take an example to illustrate the formulation of linear programming problem in various different situations.

Example: Two food stuffs F₁ and F₂ contain vitamins A, B, C. The minimum daily requirements of these vitamins for a certain diet are 3 mg of A, 50 mg of B and 40 mg of C. One unit of the food -stuff F₁ contain 1 mg of A, 25 mg of B and 10 mg of C whereas one unit of the food-stuff F₂ contains 1 mg of A,

10 mg of B and 20 mg of C. The cost of one unit food-stuff F₁ is Rs 1 and that of F₂ is Rs 2. Formulate the problem as a linear programming problem.

Solution:

Let x units of F₁ and y units of F₂ are used in the diet.

Since 1 mg of F₁ contains 1 mg of A and one unit of F₂ contains 1 mg of A and the minimum requirement of A is 3 mg.

Therefore, x + y ≥ 3

Similarly for vitamins B and C, we have

25x + 10y ≥ 50 and 10x + 20y ≥ 40

Also, x ≥ 0, y ≥ 0

The cost of x units of F₁ and y units of F₂ is x + 2y

Since the cost is to be minimized, therefore the LPP is

Minimize Z = x + 2y 

Subjected to

x + y ≥ 3

25x + 10y ≥ 50

10x + 20y ≥ 40

x ≥ 0, y ≥ 0

This is the required LPP.

•  Linear function Z = x + 2y, which has to be minimized is called a linear objective function.
•  Variables x and y are called decision variables.
•  The linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. So, x + y ≥ 3, 25x + 10y ≥ 50, 10x + 20y ≥ 40 are constraints.
•  The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.

 Some important Definition:

•  A set of values of the decision variables which satisfy the constraints of a LPP is called a solution  of the LPP.
•  A solution of a LPP which also satisfy the non-negativity restrictions of the problem is called its feasible solution. The set of all feasible solutions of a LPP is called the feasible region.
•  Any point outside the feasible region is an infeasible solution.
•  A feasible solution which optimize (maximize or minimize) the objective function of a LPP is  called an optimal solution of the LPP.

 Graphical method of solving linear programming problems:

 The graphical method for solving linear programming problems is applicable to those problems which involve only two variables. This method is based upon a theorem called extreme point theorem which is stated as follows:

Corner Point Method:

 This method comprises of the following steps:

 

1. Find the feasible region of the linear programming problem and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point.
2. Evaluate the objective function Z = ax + by at each corner point. Let M and m respectively denote the largest and smallest values of these points.
3. (i) When the feasible region is bounded, M and m are the maximum and minimum values of Z.

    (ii) In case, the feasible region is unbounded, we have:

4. (a) M is the maximum value of Z, if the open half plane determined by ax + by > M has no point in common with the feasible region. Otherwise, Z has no maximum value.

(b) Similarly, m is the minimum value of Z, if the open half plane determined by ax + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value.

Problem:

Maximize Z = 3x + 4y

Subject to the constraints:

x + y ≤ 4, x ≥ 0, y ≥ 0

Solution:

The feasible region determined by the constraints x + y ≤ 4, x ≥ 0, y ≥ 0 is as follows:

                                                          

The corner points of the feasible region are O (0, 0), A (4, 0), and B (0, 4).

The values of Z at these points are as follows:

                                             

Therefore, the maximum value of Z is 16 at the point B (0, 4).

Problem:

Minimize Z = x + 2y

subject to 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0.

Solution:

The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, and y ≥ 0 is as follows:

                                                        

The corner points of the feasible region are A (6, 0) and B (0, 3).

The values of Z at these corner points are as follows:

                     

It can be seen that the value of Z at points A and B is same. If we take any other point such as (2, 2) on

line x + 2y = 6, then Z = 6

Thus, the minimum value of Z occurs for more than 2 points.

Therefore, the value of Z is minimum at every point on the line x + 2y = 6

Problem:

Maximize Z = x + y

subject to x – y ≤ -1, -x + y ≤ 0, x, y ≥ 0.

Solution:

The region determined by the constraints

x – y ≤ -1, -x + y ≤ 0, x, y ≥ 0 is as follows:

From the figure, we can see that there is no point satisfying all the constraints simultaneously. Thus, the problem is having no feasible region and hence no feasible solution.                                                                         

Thus, Z has no maximum value.

 

Different Types of Linear Programming Problems:

 

1. Manufacturing problems:

In these problems, we determine the number of units of different products which should be produced and sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output etc., in order to make maximum profit.

 

Problem: A manufacturer produces nuts ad bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit, of Rs 17.50 per package on nuts and Rs. 7.00 per package on bolts. How many packages of each should be produced each day so as to maximize his profit, if he operates his machines for at the most 12 hours a day?

 

Solution:

Let the manufacturer produce x packages of nuts and y packages of bolts. Therefore,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as follows:

                               

The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7.

Therefore, the constraints are

x + 3y ≤ 12

3x + y ≤ 12

Total profit, Z = 17.5x + 7y

The mathematical formulation of the given problem is

Maximize Z = 17.5x + 7y    …….. (1)

subject to the constraints,

x + 3y ≤ 12    ………… (2)

3x + y ≤ 12    ………… (3)

x, y ≥ 0    …………….. (4)

The feasible region determined by the system of constraints is as follows:

                                                          

The corner points are A (4, 0), B (3, 3), and C (0, 4).

The values of Z at these corner points are as follows:

                                             

The maximum value of Z is Rs 73.50 at (3, 3).

Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the

maximum profit of Rs 73.50.

 

2. Diet problems:

In these problems, we determine the amount of different kinds of constituents/nutrients which should be included in a diet so as to minimize the cost of the desired diet such that it contains a certain minimum amount of each constituent/nutrients.

 

Problem:

Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B. Determine the minimum cost of the mixture?

Solution:

Let the mixture contain x kg of food P and y kg of food Q. Therefore,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as follows:

                                    

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B.

Therefore, the constraints are

3x + 4y ≥ 8

5x + 2y ≥ 11

Total cost, Z, of purchasing food is, Z = 60x + 80y

The mathematical formulation of the given problem is

Minimize Z = 60x + 80y   …….. (1)

subject to the constraints,

3x + 4y ≥ 8     …….. (2)

5x + 2y ≥ 11   ……. (3)

x, y ≥ 0            …… (4)

The feasible region determined by the system of constraints is as follows:

                                                       

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(8/3, 0), B(2, 1/2) and C(0, 11/2).

The values of Z at these corner points are as follows:

                                                    

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of

1. For this, we graph the inequality, 60x + 80y < 160 or 3x + 4y < 8, and check whether the

resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with 3x + 4y < 8

Therefore, the minimum cost of the mixture will be Rs 160 at the line segment joining the

points (8/3, 0) and (2, 1/2).

 

3. Transportation problems:

 In these problems, we determine a transportation schedule in order to find the cheapest way of transporting a product from plants/factories situated at different locations to different markets.

 

Problem:

An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?

Solution:

Let the airline sell x tickets of executive class and y tickets of economy class.

The mathematical formulation of the given problem is as follows:

Maximize z = 1000x + 600y    …..... (1)

subject to the constraints,

x + y ≤ 200   ………….(2)

x ≥ 20            …………(3)

y – 4x ≥ 0     ………….(4)

x, y ≥ 0         ………….(5)

The feasible region determined by the constraints is as follows:

                                                    

The corner points of the feasible region are A (20, 80), B (40, 160) and C (20, 180).

The values of z at these corner points are as follows:

                   

 

The maximum value of z is 136000 at (40, 160).

Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs 136000.

 

 

 

 

 

 

 

 

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