24 Case Study Inequalities and Linear Programming 24.1Compound Linear Inequalities in One Unknown 24.2Quadratic Inequalities in One Unknown 24.3Linear

24

Case Study

Inequalities and Linear Programming

24.1 Compound Linear Inequalities in One Unknown

24.2 Quadratic Inequalities in One Unknown

24.3 Linear Inequalities in Two Unknowns

Chapter Summary

24.4 Linear Programming

24.5 Applications of Linear Programming

P. 2

In an art lesson, students are asked to make lantern frames using sticks and plasticine. Each group of students is given 100 sticks and 70 pieces of plasticine. There is also a requirement that the difference between the number of rectangular prisms and the number of pentagonal pyramids should not be more than 3.

To make the frame of a rectangular prism, we need 12 sticks and 8 pieces of plasticine, while 10 sticks and 6 pieces of plasticine are needed to make the frame of a pentagonal pyramid.

It is not difficult. Let me show you.

What is the maximum number of lantern frames we can make if we have 100 sticks and 70 pieces of plasticine?

Case StudyCase Study

P. 3

The following table shows some combinations of the lantern frames that can be made.

Case StudyCase Study

Number of rectangular prisms

Number of pentagonal pyramids

Number of sticks used

Number of pieces of plasticine used

3 6 96 60

5 3 90 58

5 4 100 64

It is not easy to list all possible combinations.

In fact, we can use linear programming to find the answer.

P. 4

For example: (a) x > 4 (b) x 6

24.1 24.1 Compound Linear Inequalities in Compound Linear Inequalities in One UnknownOne Unknown

When we solve these kinds of compound inequalities, we have to find the values of x that can satisfy all the given inequalities.

In junior forms, we learnt that the solutions of a linear inequality in one unknown can be represented on a number line.

Now, we are going to investigate the solution of compound inequalities connected by the word ‘and’.

A. A. Solving Compound Inequalities with ‘and’Solving Compound Inequalities with ‘and’

If we want to represent the solution of compound inequalities on a number line, we can shade the region represented by each inequality

Then the overlapping region is the solution: x > 4 and x 6, that is, 4 < x 6.

P. 5

Example 24.1T

Solution:

Solve 2x + 3 7x – 2 and 3x + 1 < 4x + 2, and represent the solution on the number line.

Solving 2x + 3 7x – 2, we have2x – 7x –2 – 3



–5x –5x 1………..(1)

Solving 3x + 1 < 4x + 2, we have3x – 4x < 2 – 1

–x x –1……..(2)

Combining (1) and (2), we have

Therefore, the required solution is –1 < x 1.

P. 6

Example 24.2TSolve the compound inequality and –2(x – 1) + 3 < x – 7,

and represent the solution on the number line.

Solution:



xx

2

33

Solving , we have

6 – x – 3 2x

–x – 2x < 3 – 6

x 1………..(1)

Solving –2(x – 1) + 3 < x – 7, we have

–2x + 2 + 3 < x – 7

–2x – x < –7 – 2 – 3

x 4………..(2)

Therefore, the required solution is x > 4.

xx

2

33

–3x < –3 –3x < –12

P. 7

Example 24.3TFanny has $200 and she wants to buy two types of gifts, which are pencil cases and mouse pads, for her students. The total number of gifts is 20 and the number of pencil cases should be at least three. If the prices of a pencil case and a mouse pad are $14 and $8 respectively, find the possible number of pencil cases that she can buy.

Solution:Let x be the number of pencil cases, then the number of mouse pads is (20 x). We have



14x + 8(20 x) 200 and x 314x + 160 8x 200 and x 3 160 + 6x 200 and x 3

6x 40 and x 3

and x 33

20x

Combining the above results, we have

The possible number of pencil cases are 3, 4, 5, 6.

P. 8

Now, we are going to study another kind of compound inequalities connected by the word ‘or’. For example, 3x – 4 > 5 or 2x + 5 < 2.

The solution is the whole shaded region on the number line.

Solving compound inequalities with ‘or’, we need to solve each inequality separately and shade the region represented by each inequality.


BB. . Solving Compound Inequalities with ‘or’Solving Compound Inequalities with ‘or’

P. 9

Example 24.4T

Solution:

Solve or , and represent the solution on the number line.



32

4 x8

3

2 x

Solving , we have

x + 4 6x 2………..(1)

Solving , we have

2x 24x 12………..(2)

Therefore, the required solution is x > 2.

32

4 x

83

2x

P. 10

Example 24.5T

Solution:

Solve or , and represent

the solution on the number line.



4

1

3

2

2

1

xxx)4(

3

21 xx

Solving , we have

Solving , we have4

1

3

2

2

1

xxx

4

1

6

)2(2)1(3 xxx

4

1

6

4233 xxx

)1(3)7(2 xx33142 xx

14332 xx17 x

)1.(..........17x

)4(3

21 xx

)4(2)1(3 xx8233 xx3823 xx

)2(..........11x

Therefore, the required solution is x –17 or x –11.

P. 11

Consider the graph of a quadratic function y ax2 + bx + c, where a > 0.

24.24.22 Quadratic Inequalities in One Quadratic Inequalities in One UnknownUnknown

If the graph meets the x-axis at x p and x q, the x-axis can be divided into three intervals:

AA. . Solving Quadratic Inequalities in One Solving Quadratic Inequalities in One Unknown by the Graphical MethodUnknown by the Graphical Method

Then we can solve the inequality by considering the value of y in each interval.

For example: (i) ax2 + bx + c > 0, where a > 0 : The solution is x < p or x > q.(ii) ax2 + bx + c < 0, where a > 0 : The solution is p < x < q.(iii) ax2 + bx + c 0, where a > 0 : The solution is x p or x q.(iv) ax2 + bx + c 0, where a > 0 : The solution is p x q.

P. 12

Example 24.6T

Solution:

Solve the inequality 6x – x2 0 graphically and represent the solution on the number line.

6x – x2 0

x2 – 6x 0

x(x – 6) 0


AA. . Solving Quadratic Inequalities in One Solving Quadratic Inequalities in One Unknown by the Graphical MethodUnknown by the Graphical Method

The solution of the inequality is 0 x 6.

P. 13

In addition to using the graphical method, a quadratic inequality can also be solved by the algebraic method. Before solving, it is necessary to factorize the quadratic equation to get the root(s) or the boundary point(s).

1. If ab > 0, then either a > 0 and b > 0 or a < 0 and b < 0.That is, the product of two numbers that have the same sign is positive.

(1) Sign Testing Method The basic multiplication sign rules state that:


BB. . Solving by the Algebraic MethodSolving by the Algebraic Method

2. If ab < 0, then either a > 0 and b < 0 or a < 0 and b > 0.That is, the product of two numbers that have the same sign is negative.

P. 14

Example 24.7T

Solution:

Solve the following quadratic inequalities algebraically.(a) x2 – 6x + 8 < 0 (b) x(3x – 8) –4

(a) x2 – 6x + 8 < 0 (x – 4)(x – 2) < 0



x – 4 < 0 and x – 2 > 0 or x – 4 > 0 and x – 2 < 0 x < 4 and x > 2 or x > 4 and x < 2

There is no solution for x > 4 and x < 2.

The solution is 2 < x < 4.

P. 15

Example 24.7T

Solution:(b) x(3x – 8) –4

3x2 – 8x + 4 0



x – 2 0 and 3x – 2 0 or x – 2 0 and 3x – 2 0

x2 and x or x 2 and x

The solution is x or x 2.

(x – 2)(3x – 2) 0

3

2

3

2

3

2

Solve the following quadratic inequalities algebraically.(a) x2 – 6x + 8 < 0 (b) x(3x – 8) –4

P. 16

(2) Test Value Method To solve a quadratic inequality by the test value method,

Remarks:We can choose any values of x in an interval.Remember that we are only interested in the sign (positive or negative) that we get after substituting the value x into the quadratic expression.



Boundary numbers are found by replacing the inequality symbol in the quadratic inequality with the sign ‘’, and then solving the equation.

first find the boundary numbers (the roots) and divide the number line into intervals by the boundary numbers,

then take a value of x from each interval and check whether it satisfies the inequality or not.

P. 17

Example 24.8T

Solution:

Solve the inequality by the test value method and

represent the solution on the number line.

When x 2,(2 4)[1(2) + 3] 6 > 0.



232

12xx

x

232

12xx

x

22612 xxx 01262 2 xxx01252 2 xx0)32)(4( xx

When x 0,(0 4)[1(0) + 3] 12 < 0.

When x 5,(5 4)[1(5) + 3] 13 > 0.

The solution is or x 4. 2

3x

–2 0 5

P. 18

Example 24.9T

Solution:

If the equation 2x2 + k(x – 1) + 4(1 – x) 0 has two distinct real roots, find the range of values of k.

2x2 + k(x – 1) + 4(1 – x) 0



2x2 + kx – k + 4 – 4x 02x2 + (k – 4)x + (4 – k) 0

The discriminant of the equation (k – 4)2 – 4(2)(4 – k) k2 – 8k + 16 – 32 + 8k k2 – 16

Since the equation has two distinct real roots, we have > 0,

k2 – 16 > 0(k + 4)(k – 4) > 0

k < – 4 or k > 4

P. 19

24.24.33 Linear Inequalities in Two Linear Inequalities in Two UnknownsUnknowns

AA. . Linear Inequalities in Two UnknownsLinear Inequalities in Two Unknowns

Linear inequalities in two unknowns are inequalities which can be represented in the form Ax + By + C > 0 or Ax + By + C 0, where A, B and C are real numbers, with A and B not both equal to zero.

Consider the graph of the straight line x + y 1.

The straight line divides the coordinate plane into two parts, one above the line and the other below the line. Each region is called a half-plane.

The region above the line is called the upper half-plane, and that below the line is called the lower half-plane.

The straight line itself is called the boundary line of each half-plane.

P. 20


A dashed line is used to indicate that all points on the line do not satisfy the inequality x + y > 1. It means that all the points on the line are not included in the solution of the inequality x + y > 1.

If we want to represent the solution of the inequality x + y 1, we have to draw a solid line. In this case, the points on the line x + y 1 are also included in the solution of x + y 1.


Every point in the upper half-plane satisfies the inequality x + y > 1. Conversely, every point in the lower half-plane satisfies the inequality x + y < 1.

P. 21


We can summarize the steps of finding the solution of a linear inequality in two unknowns as follows:

1. Draw the line for the corresponding equality first. Use a solid line if the straight line is a part of the solution and a dashed line if the line is not a part of the solution.

2. Choose a point not on the line and substitute its coordinates into the inequality. Check whether the point satisfies the inequality.

3. The solution of the inequality is (a) the half-plane containing the test point if the inequality is

satisfied by the coordinates of the point; or (b) the half-plane not containing the test point if the inequality

is not satisfied by the coordinates of the point.


P. 22

Example 24.10T

Solution:Solve 2x + y < 10 graphically.



x 0 1 2 3 4y 10 8 6 4 2

Step 1: Plot the graph of 2x + y 10 with a dashed line.

Step 2: Choose (0, 0) as a test point and substitute its coordinates into 2x + y (the L.H.S. of the inequality):

2(0) + 0 0< R.H.S.

which satisfies the given inequality.Step 3: Since the test point satisfies the inequality, the graph of

2x + y < 10 is the half-plane containing (0, 0).

P. 23

BB. . System of Linear Inequalities in Two System of Linear Inequalities in Two UnknownsUnknowns

We have just studied the method to represent the solution of a linear inequality in two unknowns graphically.


We can use the same technique to solve two or more linear inequalities in two unknowns graphically.

P. 24


Combining the two graphs of the inequalities, the overlapping region is the solution of the system of inequalities.Remarks: Alternatively, we can use arrows to indicate the solution of each inequality.


For example, to solve the system of inequalities :

Step 1:Consider the graph of x + y 8, and choose (0, 0) as the test point.

68

yxyx

∵ 0 + 0 0 < 8 ∴The lower half-plane represents the

solution of x + y 8.Step 2:Consider the graph of x – y 6 and (0, 0).∵ 0 + 0 0 < 6 ∴The upper half-plane represents the

solution of x – y 6.

P. 25

Example 24.11T

Solution:

(a) Plot the straight lines y – x 0 and x + y + 2 0 on the same coordinate plane.

(b) Hence solve the system of inequalities .

(a) For y x 0, we have

For x + y + 2 0, we have

(b) The shaded region represents the solution of the system of inequalities.

x 3 2 0 2 3

y 3 2 0 2 3

x 3 2 0 2 3

y 1 0 2 4 5


2yxxy


P. 26

Example 24.12T

Solution:

(a) Solve the system of inequalities graphically.

(b) How many integral pairs of (x, y) satisfy the above system of inequalities?

(a) For 2x + y 12,


410

122

yyx

yx

x 3 4 5 6

y 6 4 2 0

For x y 10,

x 4 6 8 10

y 6 4 2 0


The shaded region represents the solution of the system of inequalities.

(b) 44 pairs

P. 27

A linear function in two variables x and y is a function in the form C ax + by, where a and b are constants.

24.24.44 Linear ProgrammingLinear Programming

Usually, we want to find the optimum values (maximum or minimum) of a linear function C, where x and y are under some restrictions.

AA. . Optimum Values of a Linear Function in Optimum Values of a Linear Function in Two VariablesTwo Variables

For each pair of x and y, there is a corresponding value of the function.

Consider a linear function C x + 2y.

For different values of C, the graphs of the linear function C x + 2y are a group of parallel lines and the value of C increases as the line shifts to the right.

x + 2y = 2

x + 2y = 4

x + 2y = 6

P. 28


AA. . Optimum Values of a Linear Function in Optimum Values of a Linear Function in Two VariablesTwo Variables

The above results demonstrate a fundamental concept of a new branch of mathematics, which is called linear programming.

Linear programming is the method of finding the optimum values of a linear function under some restrictions, which are called constraints. The linear functions we want to optimize are called the objective functions.

P. 29

There are two methods to find the optimum values of a linear function under given constraints.


BB. . Methods of Finding the Optimum Values of Methods of Finding the Optimum Values of a Linear Functiona Linear Function

Method I: Method of Sliding Line

First draw the line x + y 0 which is parallel to the graph of the objective function C x + y. Then move the line in a parallel direction to obtain the optimum values of C.

The value of C increases as the line shifts to the right, so the maximum value of C will be obtained at the extreme right vertex of the solution region, while the minimum value of C will be obtained at the extreme left vertex of the solution region.

From the graph, C attains its maximum and minimum values at R(8, –1) and Q(0, –1) respectively.

Maximum value of C 8 + (–1) 7 Minimum value of C 0 + (–1) –1

P. 30



Method II: Method of Testing Vertices

When solving problems of linear programming, the objective function always attains its maximum and minimum values at the vertices of the shaded region. We can find the optimum values of C by substituting the coordinates of all the vertices into the objective function.

At P(2, 3), C 2 + 3 5

At Q(0, –1), C 0 + (–1) –1

Maximum value of C 7minimum value of C –1

At R(8, –1), C 8 + (–1) 7

Sometimes the objective function attains its optimum value at more than one point.

P. 31

Example 24.13T

Solution:

Find the maximum and minimum values of the linear function C x – 2y, subject to the constraints:

At A(0, 15), C 0 2(15) 30


015

33

xyxyx

At B(0, 1), C 0 2(1) 2 At C(12, 3), C 12 2(3) 6 Maximum

value6

minimum value 30


Solve the system of inequalities graphically.By the method of testing vertices,

P. 32

Example 24.14T

Solution:

Find the maximum and minimum values of the linear function C x – 2y, where x and y are integers, subject to the constraints x + y 2, 3y – 2x 10 and y 4x – 19.

Using the method of sliding line,P attains its maximum value at (4, 2) and minimum value at (4, 6) and (6, 7).

Maximum value of C 4 – 2(–2)


8

Minimum value of C 6 – 2(7)8


Solve the system of inequalities graphically.

P. 33

24.24.55 Applications of Linear ProgrammingApplications of Linear Programming

There are many applications of linear programming in daily life.

To solve a real-life problem of linear programming, we can follow the steps below.

Step 1: Identify the variables x and y. Step 2: Set up the system of inequalities for the constraints. Step 3: Solve the system of inequalities graphically and shade the r

egion of feasible solution. Step 4: Write down the objective function C. Step 5: Find the optimal solution of the objective function by the

methods we learnt in the previous section.

People can use it to allocate resources (such as capital, materials, labour force, land, etc.) in order to achieve the best performance.

P. 34

Example 24.15TThe table below shows the ingredients per unit of product A and product B.

Suppose there are 60 units of chemical R and 25 units of chemical S. Let x units and y units be the amounts of product A and product B that are produced respectively. (a) Set up the system of constraints in terms of x and y. (b) Draw and shade the region of feasible solution that represents the

constraints in (a).(c) If the profit on selling product A is $100 and the profit on selling

product B is $120, find the maximum profit.

Chemical R Chemical S

Product A 2 units 1 unitProduct B 3 units 1 unit


(a) The constraints are:

integers negative-non are ,25

6032

yxyx

yxSolution:

P. 35

Example 24.15T

Solution:

Let x units and y units be the amounts of product A and product B that are produced respectively. (a) Set up the system of constraints in terms of x and y. (b) Draw and shade the region of feasible solution that represents the constraints in (a). (c) If the profit on selling product A is $100 and the profit on selling product B is $120, find the maximum profit.

(b) Refer to the figure.

(c) The objective function is P 100x + 120y. Draw a straight line 100x + 120y 0.By the method of sliding line, we observe P attains its maximum value at (15, 10).

The maximum profit $[100(15) + 120(10)] 2700$


P. 36

Example 24.16T

Solution:

In a candy shop, there are two packets of candies. Packet A contains 5 pieces of milk candies and 20 pieces of fruit candies. Packet B contains 25 pieces of milk candies and 25 pieces of fruit candies. Peter wants to buy at least 140 pieces of milk candies and 375 pieces of fruit candies. If the prices of packets A and B are $15 and $20 respectively, find the minimum cost that Peter needs to pay.

Let x be the number of packets A and y be the number of packets B.The constraints are:

The objective function is C 15x + 20y.


140255

yxyx

yx


P. 37

Example 24.16T

Solution:

In a candy shop, there are two packets of candies. Packet A contains 5 pieces of milk candies and 20 pieces of fruit candies. Packet B contains 25 pieces of milk candies and 25 pieces of fruit candies. Peter wants to buy at least 140 pieces of milk candies and 375 pieces of fruit candies. If the prices of packets A and B are $15 and $20 respectively, find the minimum cost that Peter needs to pay.

As x, y are non-negative integers,C attains its minimum value at (15, 3).

The minimum cost $[15(15) + 20(3)]

285$


In the figure, we draw a line 15x + 20y 0 and move it to the right in a parallel direction.

P. 38

Example 24.17T

Solution:

David buys x toy cars and y toy robots from a factory. The cost of a car and a robot is $35 and $15 respectively and David’s budget is $900. He uses a plastic bag to carry the toys. The capacity of the bag is 50 items. The weight of a toy car and a toy robot are 300 g and 550 g respectively and David can carry a load of 20 kg. If the profit from selling a car and a robot is $15 and $10 respectively, find the maximum profit he can make and the number of toy cars and robots he should buy.

The constraints are:

The objective function is P 15x + 10y.


509001535

yxyx

yxyx


P. 39

Example 24.17T

Solution:

David buys x toy cars and y toy robots from a factory. The cost of a car and a robot is $35 and $15 respectively and David’s budget is $900. He uses a plastic bag to carry the toys. The capacity of the bag is 50 items. The weight of a toy car and a toy robot are 300 g and 550 g respectively and David can carry a load of 20 kg. If the profit from selling a car and a robot is $15 and $10 respectively, find the maximum profit he can make and the number of toy cars and robots he should buy.

Draw a line 15x + 10y 0 and move it to the right in a parallel direction.We observe that P attains its maximum value at (13, 29). The maximum profit )]29(10)13(15$[

Number of toy cars 13; number of robots 29.


485$

P. 40

24.1 Compound Linear Inequalities in One Unknown

Compound inequalities that contain the word ‘and’ are the set of values which satisfy both inequalities.

Chapter Chapter SummarySummary

Compound inequalities that contain the word ‘or’ are the set of values which satisfy one or both of the inequalities.

P. 41

A quadratic inequality can be solved by the:

Chapter Chapter SummarySummary24.2 Quadratic Inequalities in One Unknown

1. graphical method

2. algebraic method(a) Sign testing method(b) Test value method

P. 42

Chapter Chapter SummarySummary24.3 Linear Inequalities in Two Unknowns

1. Linear inequalities in two unknowns can be represented on the coordinate plane.

2. The line y mx + c divides the plane into two parts, the upper half-plane and the lower half-plane. (a) All the points in the upper half-plane satisfy the inequality

y > mx + c. (b) All the points in the lower half-plane satisfy the inequality y < mx + c.

3. The plane which satisfies the inequality is called the solution region.

4. The common region of all the inequalities represents the feasible solution of the system of inequalities.

Notes: The straight line y mx + c is included as part of the solution for inequalities with signs ‘’ or ‘’.

P. 43

Chapter Chapter SummarySummary24.4 Linear Programming

Linear programming is a method for finding the optimum values of a linear function under given constraints.

P. 44

Chapter Chapter SummarySummary24.5 Applications of Linear Programming

Steps in solving problems of linear programming:

(i) Identify the decision variables x and y.

(ii) Set up the system of inequalities for the constraints.

(iii) Set up the system of inequalities for the constraints.

(iv) Solve the system of inequalities graphically and shade the region of feasible solution.

(v) Write down the objective function C.

(vi) Find the optimum solution of the objective function.

Documents

24 Case Study Inequalities and Linear Programming 24.1Compound Linear Inequalities in One Unknown 24.2Quadratic Inequalities in One Unknown 24.3Linear