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TABLE OF CONTENTS
Introduction to the Program Flowchart ... 0.4 Program Flowchart ... 0.5 How to Use This Guide ... 0.6 General Introduction ... 0.9 Intermediate and Terminal Objectives of the Module ... 0.11 Diagnostic Test on the Prerequisites ... 0.17 Answer Key for the Diagnostic Test on the Prerequisites ... 0.21 Analysis of the Diagnostic Test Results ... 0.25 Information for Distance Education Students ... 0.27
1. Determining the Maximum ... 1.1 2. Equations Involving a Maximum ... 2.1 3. Graphing an Equation of the Form y = ax2... 3.1 4. Graphing an Equation of the Form y = ax2 + c... 4.1 5. Solving a Second-Degree Equation by Factoring ... 5.1 6. Solving a Second-Degree Equation Using the Quadratic Formula ... 6.1 7. Graphing a Second-Degree Equation ... 7.1 8. Determining the Maximum or Minimum, Given a
Second-Degree Equation ... 8.1 9. Solving a Problem That Can Be Written as a
Second-Degree Equation ... 9.1 Final Review ... 10.1 Answer Key for the Final Review Exercises ... 10.4 Terminal Objectives ... 10.5 Self-Evaluation Test... 10.7 Correction Key for the Self-Evaluation Test ... 10.13 Analysis of the Self-Evaluation Test Results ... 10.17 Final Evaluation... 10.18 Answer Key for the Exercises ... 10.19 Glossary ... 10.93 List of Symbols ... 10.99 Bibliography ... 10.100 Review Activities ... 11.1
INTRODUCTION TO THE PROGRAM FLOWCHART
Welcome to the World of Mathematics!
This mathematics program has been developed for the adult students of the Adult Education Services of school boards and distance education. The learning activities have been designed for individualized learning. If you encounter difficulties, do not hesitate to consult your teacher or to telephone the resource person assigned to you. The following flowchart shows where this module fits into the overall program. It allows you to see how far you have progressed and how much you still have to do to achieve your vocational goal. There are several possible paths you can take, depending on your chosen goal.
The first path consists of modules MTH-3003-2 (MTH-314) and MTH-4104-2 (MTH-416), and leads to a Diploma of Vocational Studies (DVS).
The second path consists of modules MTH-4109-1 (MTH-426), MTH-4111-2 (MTH-436) and MTH-5104-1 (MTH-514), and leads to a Secondary School Diploma (SSD), which allows you to enroll in certain Gegep-level programs that do not call for a knowledge of advanced mathematics.
The third path consists of modules MTH-5109-1 (MTH-526) and MTH-5111-2 (MTH-536), and leads to Cegep programs that call for a solid knowledge of mathematics in addition to other abiliies.
MTH-5110-1 Introduction to Vectors
MTH-5109-1 Geometry IV
MTH-5108-1 Trigonometric Functions and Equations
MTH-5107-1 Exponential and Logarithmic Functions
MTH-5106-1 Real Functions and Equations
MTH-5104-1 Optimization II
MTH-5103-1 Probability II
MTH-5102-1 Statistics III
MTH-5101-1 Optimization I
MTH-4110-1 The Four Operations on
MTH-4109-1 Sets, Relations and Functions
MTH-4108-1 Quadratic Functions
MTH-4107-1 Straight Lines II
MTH-4106-1 Factoring and Algebraic Functions
MTH-4105-1 Exponents and Radicals
MTH-4103-1 Trigonometry I MTH-4102-1 Geometry III MTH-536 MTH-526 MTH-514 MTH-436 MTH-426 MTH-416 MTH-314 MTH-216 MTH-116 MTH-3002-2 Geometry II
MTH-3001-2 The Four Operations on Polynomials
MAT-2008-2 Statistics and Probabilities I
MTH-2007-2 Geometry I
MTH-2006-2 Equations and Inequalities I
MTH-1007-2 Decimals and Percent
MTH-1006-2 The Four Operations on Fractions
MTH-1005-2 The Four Operations on Integers
MTH-5111-2 Complement and Synthesis II
MTH-4111-2 Complement and Synthesis I
MTH-4101-2 Equations and Inequalities II
MTH-3003-2 Straight Lines I
DVSMTH-5112-1 Logic 25 hours = 1 credit 50 hours = 2 credits MTH-4104-2 Statistics II
THE PROGRAM FLOWCHART
Hi! My name is Monica and I have been asked to tell you about this math module. What’s your name?
Whether you are registered at an adult education center or at Formation à distance, ...
You’ll see that with this method, math is a real breeze!
... you have probably taken a placement test which tells you exactly which module you should start with.
My results on the test indicate that I should begin with this module.
Now, the module you have in hand is divided into three sections. The first section is...
... the entry activity, which contains the test on the prerequisites.
By carefully correcting this test using the corresponding answer key, and record-ing your results on the analysis sheet ...
The memo pad signals a brief reminder of concepts which you have already studied.
The calculator symbol reminds you that you will need to use your calculator.
The sheaf of wheat indicates a review designed to reinforce what you have just learned. A row of sheaves near the end of the module indicates the final review, which helps you to interrelate all the learning activities in the module.
The starting line shows where the learning activities
The little white question mark indicates the questions
for which answers are given in the text.
... you can tell if you’re well enough prepared to do all the activities in the module.
The boldface question mark indicates practice exercices
which allow you to try out what you have just learned.
And if I’m not, if I need a little review before moving on, what happens then?
In that case, before you start the activities in the module, the results analysis chart refers you to a review activity near the end of the module.
In this way, I can be sure I have all the prerequisites for starting.
Exactly! The second section contains the learning activities. It’s the main part of the module.
Look closely at the box to the right. It explains the symbols used to identify the various activities.
The target precedes the
objective to be met.
Lastly, the finish line indicates
that it is time to go on the self-evaluation test to verify how well you have understood the learning activities.
A “Did you know that...”?
For example. words in bold-face italics appear in the glossary at the end of the module...
... statements in boxes are important points to remember, like definitions, for-mulas and rules. I’m telling you the format makes everything much easier.
The third section contains the final re-view, which interrelates the different parts of the module.
Yes, for example, short tidbits on the history of mathematics and fun puzzles. They are in-teresting and relieve tension at the same time.
No, it’s not part of the learn-ing activity. It’s just there to give you a breather. There are also many fun things
in this module. For example, when you see the drawing of a sage, it introduces a “Did you know that...”
Must I memorize what the sage says?
It’s the same for the “mathwhiz” pages, which are designed
espe-cially for those who love math. They are so stimulating thateven if you don’t have to do them, you’ll still want to.
And the whole module has been arranged to make learning easier.
There is also a self-evaluation test and answer key. They tell you if you’re ready for the final evaluation.
Thanks, Monica, you’ve been a big help.
I’m glad! Now, I’ve got to run. See you!
This is great! I never thought that I would like mathematics as much as this!
In this module you will be learning about parabolas. A parabola is the graphic representation of a situation where a maximum or minimum quantity, or vertex, is reached. For example, a ball thrown up into the air rises to a maximum height before it begins to fall again. The graph representing the ball as it rises to its maximum height and falls to the ground takes the shape of a parabola.
Mathematically, the parabola is represented by an equation of the form y = ax2 + bx + c. This type of equation is called a second-degree equation because
the variable “x” has the exponent “2.” Second-degree equations differ from first-degree equations in that they involve a maximum or minimum value that is not found in a first-degree equation in two variables.
All cases similar to that of a ball thrown up into the air can be expressed as second-degree equations and are represented graphically by a parabola. In business and the physical sciences, the parabola is often used to find the maximum or minimum quantity in different situations. You will be examining several such examples in this module.
You will also learn to graph a second-degree equation of the form y = ax2 + bx + c, where
a is a rational number other than 0 and b = c = 0, where a and c are rational numbers other than 0 and b = 0, and where a, b and c are rational numbers other than 0. Among other things, this involves drawing up a table of values and determining the axis of symmetry.
You will then use factoring to solve second-degree equations of the form ax2 + bx + c = 0. All five factoring methods will be used. You should be familiar
with these five methods as they were covered in a previous module entitled “Factoring.”
You will also learn a new method for solving equations algebraically. This method involves using the quadratic formula, which is essential for solving
polynomials that cannot be factored.
Whatever method you choose, you will have to find the value or values of the variable that make the equation equal to zero. These values are called the roots
of the equation.
You will also learn to use algebra to find the coordinates of the maximum point in a situation that can be expressed as a quadratic function. Lastly, given an everyday situation or a problem related to numbers or applied geometry, you will learn how to express that situation mathematically and solve the resulting second-degree equation.
INTERMEDIATE AND TERMINAL OBJECTIVES OF
Module MTH-4108-1 consists of nine units and requires 25 hours of study, distributed as follows. The terminal objectives appear in boldface.
Objectives Number of Hours* % (evaluation)
1 to 6 4 20%
7 10 40%
8 5 20%
9 5 20%
* One hour are allotted for the final evaluation.
1. Determining the Maximum
Given a word problem dealing with an everyday situation that can be expressed as an equation of the form y = ax2 + bx + c, find the value of the
variables x and y that corresponds to the required maximum quantity (maximum output, maximum profit, maximum height) by using one of the following methods:
• filling in a partially completed table of values;
• substituting different values for x in a second-degree
equation. The values of a, b and c are rational numbers and a≠ 0.
The equation and the values of x are given. The given values of x are usually natural numbers. The answer must be written as an ordered pair (x, y) and the steps in the solution must be shown.
2. Equations Involving a Maximum
Given a word problem dealing with an everyday situation and using a partially completed table of values, formulate a second- degree equation of the form y = ax2 + bx + c, where a, b and c are rational numbers and a≠ 0.
3. Graphing an Equation of the Form y = ax2
Graph a second-degree equation of the form y = ax2, where a is a rational
number between – 5 and 5 (a≠ 0). The result should be the graph of a parabola with the vertex, the axis of symmetry and the equation of the axis of symmetry clearly indicated. The scale for each of the two axes must also be indicated.
4. Graphing an Equation of the Form y = ax2 + c
(1) Graph a second-degree equation of the form y = ax2+ c, where a is a rational
number between – 5 and 5 (a≠ 0) and where c is a rational number. The result should be the graph of a parabola with the vertex, the axis of symmetry and the equation of the axis of symmetry clearly indicated. The scale for each of the two axes must be shown as well. (2) Indicate whether the vertex of the parabola is a maximum or a minimum point.
5. Solving a Second-Degree Equation by Factoring
(1) Solve a second-degree equation of the form ax2 + bx + c = 0, where a, b and
c are rational numbers and a ≠ 0, using the appropriate factoring method (removing the common factor, factoring by grouping, factoring a trinomial of the form ax2 + bx + c or factoring the difference of squares) as well as the
multiplication property of zero. The steps in the solution must be shown. (2) Given the roots of a quadratic equation, determine the coordinates of the points on the graph of the equation ax2 + bx + c = y that correspond to these
6. Solving a Second-Degree Equation Using the Quadratic Formula
(1) Find the value of the discriminant ∆ = b2 – 4ac in order to
determine the number of roots (0, 1 or 2) of a second-degree equation of the form ax2 + bx + c = 0, where a, b and c are rational numbers and
a ≠ 0. (2) If necessary, solve this equation using the quadratic
formula: x = –b±± b2 – 4ac
2a . The resulting values are real numbers
and the steps in the solution must be shown. (3) Given the graph of
a second-degree equation of the form y = ax2 + bx + c, find the number
of zeros for this equation and indicate the coordinates of the points corresponding to these zeros.
7. Graphing a Second-Degree Equation
Graph a second-degree equation of the form y = ax2 + bx + c, where a,
b and c are rational numbers and a≠ 0. The result should be the graph
of a parabola with the following information clearly indicated: the vertex, the axis of symmetry and its equation, the y-intercept, the
point symmetric to the y-intercept and, if necessary, the coordinates
corresponding to the zeros of this equation. The scale for each axis and the calculations involved in finding each of these points must also be indicated.
8. Determining the Maximum or Minimum, Given a Second-Degree Equation
Determine the abscissa and the ordinate of the maximum or minimum point of a parabola, given a second-degree equation of the form y = ax2 + bx + c, where a, b and c are rational numbers and a≠ 0.
The problems deal with situations related to science or business. The steps in the solution must be shown.
9. Solving a Problem That Can Be Written as a Second-Degree Equation
Using factoring or the quadratic formula, solve a word problem that can be written as a second-degree equation of the form
ax2 + bx + c = 0, where a, b and c are rational numbers and a≠ 0. Any
irrelevant solutions should be disregarded. Each problem involves finding a maximum of two solutions and deals with computation, geometry or everyday situations. The steps in the solution must be shown.
DIAGNOSTIC TEST ON THE PREREQUISITES
1. Answer as many questions as you can. 2. You may use a calculator.
3. Write your answers on the test paper.
4. Don’t waste any time. If you cannot answer a question, go on to the next one immediately.
5. When you have answered as many questions as you can, correct your answers using the answer key which follows the diagnostic test.
6. To be considered correct, your answers must be identical to those in the key. In addition, the various steps in your solution should be equivalent to those shown in the answer key.
7. Transcribe your results onto the chart which follows the answer key. This chart gives an analysis of the diagnostic test results. 8. Do only the review activities that apply to your incorrect
9. If all your answers are correct, you may begin working on this module.
1. Find the value of y in the equation y = –x2 + 5x – 7 if:
a) x = 3 b) x = –2
2. Find the value of y in the equation y = 0.25x2 – 0.3x + 5 if x = 4.
3. a) Find the value of b2 – 4ac b) Find the value of b2 – 4ac
if b = –5, a = –3 and c = 0. if a = 1
2, b = 0.3 and c = 1 2.
4. Carry out the following operations on the polynomials below.
a) (5x2 – x + 3) + (x2 + 2x – 5) =
b) (x2 + 7x – 1) – (3x2 + 4 – 2x) =
y x 1 1 y x 1 1 x y x y
5. a) Graph the equation y = 5x – 3 and make a table of at least three related values.
Table of values
b) Graph the equation x = –2 and make a table of values for this equation.
Table of values
6. Solve the following equations by showing all the steps and checking the results.
b) – 5x= 0
c) – 8x+ 6 = 0
7. Factor the following polynomials.
a) 6x2 + 3x b) x2 + 5x – 2x – 10
c) x2 + 7x + 6 d) 3x2 + 11x – 20
y x 1 1
•x y – 1 – 8 0 – 3 1 2 2 7
ANSWER KEY FOR THE DIAGNOSTIC TEST
ON THE PREREQUISITES1. a) y = –x2 + 5x – 7 b) y =–x2 + 5x – 7 y = – (3)2 + 5(3) – 7 y = – (– 2)2 + 5(–2) – 7 y = – 9 + 15 – 7 y = – 4 – 10 – 7 y = – 1 y = – 21 2. y = 0.25x2 – 0.3x + 5 y = 0.25(4)2 – 0.3(4) + 5 y = 0.25(16) – 1.2 + 5 y = 4 – 1.2 + 5 y = 7.8 3. a) b2 – 4ac = (– 5)2 – 4(– 3)(0) b) b2 – 4ac = (0.3)2 – 4 1 2 1 2 b2 – 4ac = 25 – 0 b2 – 4ac = 0.09 – 1 b2 – 4ac = 25 b2 – 4ac = – 0.91 4. a) 6x2 + x – 2 b) – 2x2 + 9x – 5 c) 2x2– 8x – 24 d) –x2 + 4x – 1 2 5. a)
x y – 2 3 – 2 1 – 2 0 – 2 – 2 y x 1 1
•b) 6. a) Solution: Check: 2x + 7 = 0 2x + 7 = 0 2x= – 7 2(– 3.5) + 7 = 0 x= –7 2 or – 3.5 – 7 + 7 = 0 0 = 0 b) Solution: Check: – 5x= 0 – 5x = 0 x= 0 – 5 – 5(0) = 0 x= 0 0 = 0 c) Solution: Check: – 8x + 6 = 0 – 8x + 6 = 0 – 8x = – 6 – 8(0.75) + 6 = 0 x = – 6 – 8 –6 + 6 = 0 x = 3 4 or 0.75 0 = 0 7. a) 6x2 + 3x b) x2 + 5x – 2x – 10 3x(2x + 1) x(x + 5) – 2(x + 5) (x + 5)(x – 2)
c) x2 + 7x + 6 d) 3x2 + 11x – 20
x2 + 6x + x + 6 3x2 + 15x – 4x – 20
x(x + 6) + 1(x + 6) 3x(x + 5) – 4(x + 5) (x + 6)(x + 1) (x + 5)(3x – 4)
• Trinomial of the form • Trinomial of the form x2 + bx + c ax2 + bx + c
e) 4x2 – 25
(2x)2 – (5)2
(2x + 5)(2x – 5)
ANALYSIS OF THE DIAGNOSTIC TEST RESULTS
Question Answer Review Before Going
Correct Incorrect Section Page to Unit(s)
1. a) 11.1 11.4 Unit 1 b) 11.1 11.4 Unit 1 2. 11.1 11.4 Unit 1 3. a) 11.1 11.4 Unit 1 b) 11.1 11.4 Unit 1 4. a) 11.2 11.11 Unit 2 b) 11.2 11.11 Unit 2 c) 11.2 11.11 Unit 2 d) 11.2 11.11 Unit 2 5. a) 11.3 11.23 Unit 3 b) 11.3 11.23 Unit 3 6. a) 11.4 11.31 Unit 5 b) 11.4 11.31 Unit 5 c) 11.4 11.31 Unit 5 7. a) 11.5 11.40 Unit 5 b) 11.5 11.40 Unit 5 c) 11.5 11.40 Unit 5 d) 11.5 11.40 Unit 5 e) 11.5 11.40 Unit 5
• If all your answers are correct, you may begin working on this module.
• For each incorrect answer, find the related section in the Review column and do the review exercises for that section before starting the unit(s) listed in the right-hand column under the heading Before Going on to.
INFORMATION FOR DISTANCE
You now have the learning material for MTH-4108-1 together with the home-work assignments. Enclosed with this material is a letter of introduction from your tutor indicating the various ways in which you can communicate with him or her (e.g. by letter, telephone) as well as the times when he or she is available. Your tutor will correct your work and help you with your studies. Do not hesitate to make use of his or her services if you have any questions.
DEVELOPING EFFECTIVE STUDY HABITS
Distance education is a process which offers considerable flexibility, but which also requires active involvement on your part. It demands regular study and sustained effort. Efficient study habits will simplify your task. To ensure effective and continuous progress in your studies, it is strongly recommended that you:
• draw up a study timetable that takes your working habits into account and is compatible with your leisure time and other activities;
The following guidelines concerning the theory, examples, exercises and assign-ments are designed to help you succeed in this mathematics course.
To make sure you thoroughly grasp the theoretical concepts:
1. Read the lesson carefully and underline the important points.
2. Memorize the definitions, formulas and procedures used to solve a given problem, since this will make the lesson much easier to understand.
3. At the end of an assignment, make a note of any points that you do not understand. Your tutor will then be able to give you pertinent explanations.
4. Try to continue studying even if you run into a particular problem. However, if a major difficulty hinders your learning, ask for explanations before sending in your assignment. Contact your tutor, using the procedure outlined in his or her letter of introduction.
The examples given throughout the course are an application of the theory you are studying. They illustrate the steps involved in doing the exercises. Carefully study the solutions given in the examples and redo them yourself before starting the exercises.
The exercises in each unit are generally modelled on the examples provided. Here are a few suggestions to help you complete these exercises.
1. Write up your solutions, using the examples in the unit as models. It is important not to refer to the answer key found on the coloured pages at the end of the module until you have completed the exercises.
2. Compare your solutions with those in the answer key only after having done all the exercises. Careful! Examine the steps in your solution carefully even if your answers are correct.
3. If you find a mistake in your answer or your solution, review the concepts that you did not understand, as well as the pertinent examples. Then, redo the exercise.
4. Make sure you have successfully completed all the exercises in a unit before moving on to the next one.
Module MTH-4108-1 contains three assignments. The first page of each assignment indicates the units to which the questions refer. The assignments are designed to evaluate how well you have understood the material studied. They also provide a means of communicating with your tutor.
When you have understood the material and have successfully done the perti-nent exercises, do the corresponding assignment immediately. Here are a few suggestions.
1. Do a rough draft first and then, if necessary, revise your solutions before submitting a clean copy of your answer.
2. Copy out your final answers or solutions in the blank spaces of the document to be sent to your tutor. It is preferable to use a pencil.
3. Include a clear and detailed solution with the answer if the problem involves several steps.
4. Mail only one homework assignment at a time. After correcting the assign-ment, your tutor will return it to you.
In the section “Student’s Questions,” write any questions which you may wish to have answered by your tutor. He or she will give you advice and guide you in your studies, if necessary.
In this course
Homework Assignment 1 is based on units 1 to 7. Homework Assignment 2 is based on units 8 and 9. Homework Assignment 3 is based on units 1 to 9.
When you have completed all the work, and provided you have maintained an average of at least 60%, you will be eligible to write the examination for this course.
DETERMINING THE MAXIMUM
SETTING THE CONTEXT
Mr. Pitt owns a peach orchard with 30 trees yielding an average of 400 peaches each. He wants to plant more peach trees in order to get the maximum yield from his orchard; however, for each additional tree planted, the average yield per tree will drop by 10 peaches. How many peach trees should Mr. Pitt plant to ensure a maximum harvest?
To achieve the objective of this unit, you should be able to solve
problems involving maximum yield, profit or height. The problems are
based on everyday situations. You will be required to fill in partially
completed tables of values and find the value of x or y that corresponds
to the required maximum. You should also be able to find the maximum value for situations whose equation is given.
The table below should help Mr. Pitt solve his problem.
Table 1.1 Projected yield of Mr. Pitt’s orchard
Number of Total Average Total yield peach trees number of yield per of orchard
to be added peach trees tree y x 0 30 400 30 × 400 = 12 000 1 30 + 1 400 – 10 × 1 (30 + 1) × (400 – 10 × 1) = 12 090 2 30 + 2 400 – 10 × 2 (30 + 2) × (400 – 10 × 2) = 12 160 3 30 + 3 400 – 10 × 3 (30 + 3) × (400 – 10 × 3) = 12 210 4 30 + 4 400 – 10 × 4 (30 + 4) × (400 – 10 × 4) = 12 240 5 30 + 5 400 – 10 × 5 (30 + 5) × (400 – 10 × 5) = 12 250 6 30 + 6 400 – 10 × 6 (30 + 6) × (400 – 10 × 6) = 12 240 7 30 + 7 400 – 10 × 7 (30 + 7) × (400 – 10 × 7) = 12 210 8 30 + 8 400 – 10 × 8 (30 + 8) × (400 – 10 × 8) = 12 160 9 30 + 9 400 – 10 × 9 (30 + 9) × (400 – 10 × 9) = 12 090 10 30 + 10 400 – 10 × 10 (30 + 10) × (400 – 10 × 10) = 12 000
The first column shows the number of trees Mr. Pitt will have to add to his orchard. The numbers range from 0 to 10 inclusive and are identified by x.
The second column shows the total number of peach trees obtained by adding the current number of peach trees and the additional number of trees to be planted with a view to maximizing the orchard’s yield.
The third column shows the average yield per tree. This yield is equal to the average number of peaches produced by each tree (400) less the production loss for each additional tree (10/additional tree).
The fourth column shows the orchard’s total yield after additional trees have been planted. This yield is equal to the product of the total number of trees (column 2) and the average yield per tree (column 3). This product is identified as y.
For example, the third line of Table 1.1 tells you that if two trees were added, the total number of peach trees would be 30 + 2 = 32, that the average yield per tree would be 400 – 10 x 2 and that the orchard’s total yield would be
(30 + 2) x (400 – 10 x 2) = 12 160 peaches. If you have understood this,
you should be able to answer the following questions without any difficulty.
?What is the orchard’s greatest total yield? ...
?How many trees should Mr. Pitt add to his orchard in order to obtain the greatest total yield? ...
?What is the total number of peach trees required for the orchard to yield a maximum amount of fruit? ...
?What will be each tree’s average yield when the total yield will have reached its maximum? ...
?If Mr. Pitt adds between 0 and 5 trees, will his orchard’s total yield increase or decrease? ...
?What would happen to the orchard’s total yield if Mr. Pitt were to add between 6 and 10 trees? ...
The greatest total yield for Mr. Pitt’s orchard is 12 250 peaches, which corresponds to the addition of 5 peach trees. Consequently, he will need a total of 35 trees for his orchard to yield the maximum amount of fruit. The average yield per peach tree when the total yield has reached its maximum is equal to 400 – 10 × 5, or 350 peaches.
You will no doubt have noticed from Table 1.1 that with the addition of between 0 and 5 trees, the total yield of Mr. Pitt’s orchard increases, whereas with the addition of between 6 and 10 peach trees, the total yield decreases.
Lastly, the ordered pair corresponding to the orchard’s maximum yield is 5 peach trees and 12 250 peaches, that is, (5, 12 250). Problem solved!
Many other problems involve finding a maximum value. Below is an example taken from the printing industry.
N.B. You may use a calculator for all calculations in the module, unless otherwise indicated.
The XYZ publishing company prints 36 books a day, which are sold for $40 each. The company can print 50 books a day. However, for each additional book printed, the selling price per unit decreases by $1. How many additional books must the company print in order to maximize its daily sales? What is this maximum sales figure?
?To solve this problem, complete the table below by following the example of the lines that are already completed.
Table 1.2 Number of books sold by the XYZ publishing company
Additional Total number Selling price Daily sales books of books per book
x y 0 36 $40 36 × $40 = $1 440 1 36 + 1 = 37 $40 – $1 = $39 37 × $39 = ... 2 36 + 2 = 38 $40 – $2 = ... 38 × .... = ... 3 36 + .... = ... $40 – ... = ... .... × .... = ... 4 .... + .... = ... ... – ... = ... .... × .... = ...
?What ordered pair corresponds to the maximum sales figure? ...
?How many additional books must the company print in order to maximize its daily sales? ...
?What is the maximum sales figure? The completed table looks like this:
Table 1.2 Number of books sold by the XYZ publishing company
Additional Total number Selling price Daily sales books of books per book
x y 0 36 $40 36 × $40 = $1 440 1 36 + 1 = 37 $40 – $1 = $39 37 × $39 = $1 443 2 36 + 2 = 38 $40 – $2 = $38 38 ×$38 = $1 444 3 36 + 3 = 39 $40 – $3 = $37 39×$37 = $1 443 4 36 + 4 = 40 $40 – $4 = $36 40×$36 = $1 440
The ordered pair corresponding to the maximum sales figure is 2 additional books and $1 444 per day, or (2, 1 444). To reach the maximum sales figure of
$1 444, the company must print 2 additional books each day.
Procedure for finding the value of x or y that
corresponds to a maximum, using a table of values:
1. Read the problem carefully.
2. Complete the table of values that represents the situation. 3. Find the ordered pair that corresponds to the maximum. 4. Depending on the question asked, find the value of x or y in the ordered pair that corresponds to the required maximum.
N.B. The ordered pair corresponding to the maximum is the vertex of a
parabola, which we will study in a later unit. Now it’s time for you to put what you have learned into practice.
1. Mickey Mouse Electronics can sell 300 VCRs at a profit of $60 per unit. Seeking to increase its profits, the company decides to take its VCRs off the market to increase the demand. It calculates that its sales will increase by 50 units for each week that the VCRs are off the market, but that its profit will drop by $5 per unit owing to storage costs. Complete the following table based on this information.
Table 1.3 Profits on VCR sales
Number of weeks Number Profit on Total profit off the market of units each unit
x sold y 0 300 $60 $60 × 300 = $18 000 1 300 + (1 × 50) = 350 $60 – ($5 × 1) = $55 $55 × 350 = $9 250 2 3 4 5 6
a) What ordered pair corresponds to the maximum profit generated by VCR sales?...
b) How many weeks should the company keep its product off the market in order to earn a maximum profit? ...
Did you know that...
…the idea of withholding an item in order to raise its market value is not new? Indeed, one of the basic principles of economics is that “scarcity creates demand.” This means that a certain clientele is ready to pay dearly and sometimes even very dearly for a product that is not easily found. The purchase of a luxury car is a good example. Some people wait for over two years for their Ferrari Testarossa, which sells for approximately $250 000, service and taxes not included!
The height reached by a ball t seconds after it is thrown up into the air is represented by the functionh = – 5t2 + 30t. How many seconds will it take
for the ball to reach its maximum height?
Throughout this module, we will use the term function to designate
equations relating to problems involving a maximum or minimum value. We use the term function because, in this type of situation, we find the values of a variable as a function of the values given to the other variable. You will have the opportunity to study this concept in more depth in a subsequent module.
?To solve this problem, complete the following table of values, where t represents the time and h the height.
Equation: h = – 5t2 + 30t
Table 1.4 Calculating the height reached by a ball
t h Breakdown of calculations for the table of values 0 0 If t = 0, then h = – 5(0)2 + 30(0) = 0.
1 25 If t = 1, then h = – 5(1)2 + 30(1) = 25.
2 40 If t = 2, then h = – 5(2)2 + 30(2) = 40.
3 45 If t = 3, then h = – 5(3)2 + 30(3) = 45.
To complete Table 1.4, replace t by its given value on each line. The result is the following table.
t h Breakdown of calculations for the table of values 0 0 If t = 0, then h = – 5(0)2 + 30(0) = 0. 1 25 If t = 1, then h = – 5(1)2 + 30(1) = 25. 2 40 If t = 2, then h = – 5(2)2 + 30(2) = 40. 3 45 If t = 3, then h = – 5(3)2 + 30(3) = 45. 4 40 If t = 4, then h = – 5(4)2 + 30(4) = 40. 5 25 If t = 5, then h = – 5(5)2 + 30(5) = 25. 6 0 If t 6, then h = – 5(6)2 + 30(6) = 0.
The ordered pair (3, 45) corresponds to the maximum point of the equation h = – 5t2 + 30t; thus, it will take 3 seconds for the ball to reach its maximum
Procedure for finding the value of x or y that
corresponds to a maximum, given an equation:
1. Determine what x and y represent.
2. Complete a table of values for the given values of x. 3. Find the ordered pair that corresponds to the maximum. 4. Depending on the question asked, find the value of x or y in the ordered pair that corresponds to the required maximum.
The following exercise will test your ability to find the value of x or y that corresponds to the required maximum.
1. Consider the function h = – 5t2 + 20t + 30 representing the height in metres
attained by an object t seconds after it is thrown up into the air. Find the time required for this object to reach its maximum height. What is its maximum height?
Let t = 0, 1, 2, 3 and 4 and complete the four steps for solving a problem that involves finding a maximum value.
1. ... .
2. Table of values:
Table 1.5 Height reached by an object after it is thrown into the air
t h Breakdown of calculations for the table of values for the equation h = – 5t2 + 20t + 30
0 1 2 3 4 3. ... 4. ... ...
The theory presented in this unit can be summarized by the two procedures shown in the boxes on pages 1.6 and 1.9. Review them before moving on to the
1. A survey conducted during an amateur theatre festival in Sherbrooke indicated that 100 people would attend the festival if the tickets were sold for $4.50 apiece. For each $0.30-decrease in the price of a ticket, 20 more people would attend the festival.
a) What ticket price would bring in the most revenue? b) What would be the maximum revenue?
Complete the following table in order to find the answers to these questions.
Table 1.6 Revenue for the Sherbrooke theatre festival
Number of Total number Ticket price Revenue extra people of people y
x× 20 0 100 $4.50 $4.50 × 100 = $450 1 × 20 100 + 20 = 120 $4.50 – ($0.30 × 1) $4.20 × 120 = $504 2 × 20 3 × 20 4 × 20 5 × 20 6 × 20 7 × 20 a) ... . b) ... .
2. The function h = 40t – 5t2 represents the height of an object t seconds after it
is thrown up into the air. Draw up a table of values for t = 0, 1, 2, 3, 4, 5, 6, 7 and 8. How many seconds will it take for the object to reach its maximum height? To answer this question, complete the four steps for finding the maximum value. Write your answers in the spaces provided below.
1. ... .
Table 1.7 Height of an object t seconds after being thrown
3. ... .
1. Complete the following sentences.
a) Using a table of values representing a situation or using an ... , we can find the value of x or y that corresponds to a ... .
b) Using a table of values, we can find the value of x or y that corresponds to a maximum by using the following procedure:
1. Read the ... carefully.
2. ... the table of values that represents the situation. 3. Find the ordered pair that corresponds to the ... . 4. Depending on the question asked, find the value of ... or y in the ordered pair that corresponds to the required ... .
2. Given an equation, we can use a four-step procedure to find the value of x or y corresponding to a maximum. Describe the procedure.
1. ... ... . 2. ... ... . 3. ... ... . 4. ... ... .
THE MATH WHIZ PAGE
A Well-Hidden Maximum!
A little challenge for the more daring among you!
At present, Chip Electronics could sell 100 calculators at a profit of $5.00 per unit. The company decides to withdraw its calculators from the market in order to boost demand. The company estimates that for each week that the calculators are off the market, it can sell 20 more calculators. Furthermore, for each week that the calculators are off the market, the profit per calculator goes down by $0.25 owing to the suspension of sales and storage costs. How long should Chip Electronics wait if it wishes to earn maximum profits?
To solve this problem, complete the table of values on the following page.
Table 1.8 Profit on calculator sales
Number of Number of Profit on Total weeks off calculators each profit the market sold calculator
x y 0 100 $5 100 × $5 = $500 1 100 + 1 × 20 = 120 $5 – 1 × $0.25 = $4.75 120 × $4.75 = $570 2 3 4 5 6 7 8 9