LINEAR PROGRAMMING
UNIT 1
y
Couch bundles
20
15
10
5
0
0
5
10
15
20
10
15
20
Chair bundles
x
y
Couch bundles
20
15
10
5
0
0
5
Chair bundles
x
UNIT 1
EXPLORING AN OPTIMIZATION
PROBLEM
LESSON 1.1
LEARNING OUTCOMES
Upon completion of lesson 1.1 students will
be expected to:
11A2 relate sets of numbers to solutions of
inequalities
11B3 demonstrate an understanding of
the relationship between arithmetic
operations and operations on
equations and inequalities
11B4 use the calculator correctly and
efficiently
11C6 apply the linear programming process
to find optimal solutions
11C11 express and interpret constraints
11C18 interpolate and extrapolate to solve
problems
INTRODUCTION
Read Making Choices - Linear Programming,
page 1 in Constructing Mathematics, Book 2. The
numbered outcomes will provide direction for the
first chapter and define what you will learn in this
unit.
Before beginning this lesson, read the review
section in the text, pages 37-43. This section will
help you keep track of the important concepts and
terms in this unit.
SAMPLE PROBLEM 1
Procedure A
The table of values below shows the relationship
between number of couch bundles cut and money
earned.
# of couch
bundles cut
Amount
earned ($)
0
0
1
12
1.5
18
2
24
2.5
30
3
36
3.5
42
4
48
4.5
54
5
60
CORRESPONDENCE STUDIES
Graph the Values
Draw a graph of the values to visually display
the data. Should you join the dots? Yes, join the
dots, because Heather is paid for cutting partial
couch bundles. If Heather was not paid for partial
bundles, the data would be discrete and you would
not join the dots.
Heather's Earnings
60
54
48
42
earnings ($)
Read Exploring an Optimization Problem on text
page 2 and Investigation 1, Looking at Heather’s
Income. This sample problem works through
Procedures A, C and D ( for couch bundles only) of
the investigation.
36
30
24
18
12
6
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
# of couch bundles
PAGE 13
MATHEMATICS FOUNDATIONS 11
Domain:
Range:
{x ≥ 0, x ∈ R}
{y ≥ 0, y ∈ R}
The domain is read as “x is greater than or equal to
zero, and x is an element of the reals (real numbers)”.
This is true because you can not have negative
time. Time can only move forward, therefore
it is represented by any number form of real
numbers. The domain is usually written by using
restrictions, if there are any (in this case negative
numbers are not permitted), followed by the
number set to which the variable belongs.
The range is read as “y is greater than or equal to
zero, and y is an element of the reals”.
Heather's Earnings
60
54
48
42
earnings ($)
Domain and Range
Domain and range are written for a continuous
relationship. In this case, it is the relationship
between the number of bundles cut and the
money Heather earns.
36
30
24
18
12
6
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
# of couch bundles
To find the slope, m, count how far up is it from
one point to the next on the graph. This is the rise.
Going up from the first point at 0 to the second
point at 12 is “going up 12”. So, the rise is 12.
How far to the right? This is the run. The run goes
from 0 to 1, so the run is 1.
Procedure C
The relationship graphed in Procedure A is the
relationship between the number of bundles being
cut and the money that Heather can earn.
rise , so the slope for this line is
The slope = run
rise 12
run = 1 = 12. The slope is 12.
Note, from the second dot to the third, the rise is
6 and the run is 1 . The slope is 6 ÷ 1 which is 12.
2
2
Since the line of the graph is straight, the slope
between each pair, or any pair of points, must be
the same. On this graph, the slope is 12. From the
second point to the fifth point is a rise of 18 and a
run of 1.5, so the slope is 18 ÷ 1.5, which is 12.
This relationship can be expressed as an equation.
Since the data forms a straight line, the equation
will take the form of y = mx + b, or in this case
E = m(n) + b.
To find the y-intercept, b, where does the line
cross the vertical axis (y-axis)? The graph crosses
the vertical axis at (0, 0). So, the y-intercept, the
b-value in the equation, must be 0.
E is the earning in dollars, and n is the number of
bundles cut.
Procedure D
State the equation for the line on the graph. This
equation also represents the values in the table.
The equation of the line is E = 12n + 0, or simply
E = 12n. Spoken, this is “earnings equals 12 times
the number of bundles.”
Independent and Dependent Variables
The domain (the number of couch bundles cut)
is the independent variable because the amount
Heather earns (dependent variable) depends on
how many bundles she cuts.
Remember that m is the slope of the line
rise
(m = slope = the ratio run
) , and b is the
y-intercept, or the point at which the line on the
graph passes through the vertical axis.
PAGE 14
CORRESPONDENCE STUDIES
UNIT 1
DO AND SEND 1
1. Do Procedures A, C and D using the data
for chair bundles from Investigation 1,
Looking At Heather’s Income on text page 3.
2. Do Investigation Questions 1, 2 and 3 on
text pages 3-4.
SAMPLE PROBLEM 2
Read Check Your Understanding question 5 on
text page 4, which examines the money Colin can
make for assembling bicycles.
Part A
What is the slope of the graph? The rise is 40, for
earnings of $40, and the run is 5, for 5 bicycles
rise = 40 = 8. Eight
assembled. The slope = run
5
represents $8 per bicycle assembled. So the slope
represents the earning ratio: how many dollars
earned per bicycle assembled.
Part B
Since the points on the graph are connected by
a line, the data is continuous. This means Colin
must be paid for partially assembling a bicycle.
Part C
Go to $68 on the vertical axis, slide over to the
graph of the line, then slide down to the value on
the x-axis, or horizontal axis. The value is about 7
or 8 bikes.
Part D
To write the equation, you need a slope and
y-intercept value. Since the slope is 8, and the
y-intercept value is 0, the equation must be
E = 8n + 0, or simply E = 8n.
Use this equation to check the answer in Part C.
The earnings are $68 and 68 = 8n; divide both
sides of the equation by 8.
68 = 8n , n = 8.5
8
8
Colin must have assembled 8.5 bikes.
DO AND SEND 2
3. Do Check Your Understanding questions 7
and 8 on text page 5.
SAMPLE PROBLEM 3
Read Investigation 2, More Thoughts on Income,
Procedures A and B on text page 6. This problem is
a continuation of Sample Problem 2 in this guide;
the money Colin earns for assembling bicycles.
Procedure A
The line y = 8x has many points. Four points are:
(0, 0), (5, 40), (10, 80), and (15, 120).
The points represent different amounts of earnings
for different numbers of bikes assembled. For
example, (15, 120) indicates that 15 bikes were
assembled, for which Colin earned $120.
Procedure B
For the points mentioned in part i) in the text,
(10, 90) represents earning $90 for assembling
10 bikes. This is $10 more than Colin earns to
assemble the same number of bikes. The second
point mentioned, (12, 100), represents earning
$100 for assembling 12 bikes. This is $4 more
than what Colin earns for assembling the same
number of bikes.
For part ii), four other points that lie above the
line are:
(5, 60), (5, 100), (10, 120), and (15, 140).
CORRESPONDENCE STUDIES
PAGE 15
MATHEMATICS FOUNDATIONS 11
These all represent amounts of earnings greater
than the amount Colin earns for assembling that
number of bikes.
DO AND SEND 3
4. Do Procedure C on text page 6 and write
the inequality to represent the points
given in Procedure C.
Hint: if the points lie above the line
y = 8x, then the y-values are greater than
(>) $8 per bike.
SAMPLE PROBLEM 4
Read Investigation Question 9 on text page 7.
The problem deals with the relationship between
chocolate bars sold and money earned.
Part A
To write the equation, y = mx + b, first find the
rise = 10 = 2
slope of the graph: run
5
Then find the y-intercept, 0.
The equation would be y = 2x + 0, or y = 2x.
Part B
Read the definition of an inequality on text page 6.
Since our equation is y = 2x, the points above the
line would be greater than 2x, so the inequality for
points above the line would be written as y > 2x.
Rhonda and her friends are earning money at a
rate of $2 per bar, the points above the line y > 2x
represent earnings greater than the $2 per bar.
Part C
From part B, you know that the points above
the line are greater than 2x, so points below the
line must be less than 2x, therefore the inequality
for points below the line is y < 2x; these points
represent earnings less than $2 per bar.
PAGE 16
EVALUATING AN INEQUALITY
Read Focus A, A Further Look at Inequalities on
text page 8. This focus shows how to evaluate an
inequality. When a coordinate is substituted into
an inequality and evaluated, and the evalutation
proves that the inequality is true, then the
inequality is said to be satisfied. The following
problem will show how to evaluate an inequality.
SAMPLE PROBLEM 5
Read Focus Question 11 on text page 8. This
question asks if there are any other solutions to the
inequality from Focus A.
Part A
To graph the line y = 10x + 2 first find the
y-intercept: +2 is the y-intercept.
Put a dot on the y-axis at 2.
From there rise 10 and run 1 to place a second
dot. Join the dots.
y
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
1
2
3
4
5
x
Part B
The table above right shows four points that
satisfy the inequality and two that do not.
CORRESPONDENCE STUDIES
UNIT 1
x
y
y < 10x + 2
Yes or No
1.5
20
yes
2.5
30
yes
0
4
yes
-0.5
-2
yes
-1.5
-15
no
-2
-20
no
DESIGNATING POINTS IN INEQUALITIES
There are many, many points that will satisfy
an inequality. Read Focus B, Shading on how to
designate (shade) all these points.
WHEN THE SLOPE AND Y-INTERCEPT ARE NOT VISIBLE
The same points are shown plotted on a graph,
note the position of the two points that did not
satisfy the inequality y ≤ 10x + 2.
30
20
15
USING A GRAPHING CALCULATOR
10
5
-22
-11
Sometimes equations and inequalities are given
in forms where the slope and y-intercept are not
visible, like 3x + 5y ≤ 110. Read Focus C, Think
About Inequalities on text page 12 to learn how to
re-arrange inequalities so they can be graphed.
The example in the text, y ≤ 22 - 0.6x shows a
y-intercept of 22, and a slope of -0.6, or - 6 .
10
25
-33
The point (2, 5) is lower than the point (2, 7),
which is on the line, so an inequality that (2, 5)
would satisfy is: y ≤ 3x + 1 or y < 3x + 1.
To graph using the TI-83 graphing calculator:
1
2
3
-55
Press
-10
10
-15
15
ON
to turn on the calculator, then
Y=
to get the function screen.
Your function screen should look
like the one below.
-20
20
Part C
All the “yes” points from the table are on the line
y = 10x + 2 or above it.
SAMPLE PROBLEM 6
Read Check Your Understanding question 20, part
A on text page 10. In this problem you are given a
line and a point on the graph. You are to write an
inequality that will be satisfied by that point.
To clear any equations that might
be on the screen:
The line has a y-intercept of 1.
rise = 3 = 3
The slope of the line is run
1
The equation of the line is therefore y = 3x + 1.
CORRESPONDENCE STUDIES
Press
CLEAR
which is on the right under the
blue cursor keys, then
ENTER
until all the equations are gone.
PAGE 17
MATHEMATICS FOUNDATIONS 11
Return the cursor, using the blue cursor keys, to
the first equation entry place, Y1 =.
To graph y = 22 - 0.6x
22
then
−
0.6
the blue subtract key, then
then x, which is the button
X,T,θ,n
to show 22 - 0.6x
Press
GRAPH
At this point you may or may
not see the graph depending
on the WINDOW settings.
Press
ZOOM
Press
6
This selects the standard
window with x-values ranging
from -10 to 10, and the same
for y-values. However, this
graph will not show on those
window settings.
Next we want to graph the inequality
y ≤ 22 - 0.6x. In the first window of the
function screen (Y = button), see the diagonal
line to the left of the Y= ? Move the cursor key
to the left of the Y= to sit on the diagonal line,
then press ENTER, ENTER, ENTER
Your screen should look like the one below,
only your cursor (black square) will be
flashing.
Press
GRAPH to see:
Press WINDOW to change to these settings to
the ones shown below:
Press GRAPH to see the graph of the line.
30
-30
-5
5
40
-30
(your graph will not show numbers)
PAGE 18
CORRESPONDENCE STUDIES
UNIT 1
DO AND SEND 4
5. Do Investigation Question 10 on text
pages 7-8.
6. Do Focus Questions 12 to 15 on text
page 9.
7. Do Check Your Understanding questions
19 and 20, parts B, C, and D on text
page 10.
8. Do Check Your Understanding questions
21-27 on text pages 11-13.
9. Do Check Your Understanding questions
28-30 on text page 13 using your
graphing calculator. Copy what you see
on the screen on your work pages. Mark
the extreme x- and y-values onto your
sketch of the graph. For example, for
the graph of y ≤ 22 - 0.6x, your sketch
should look like this:
30
-30
40
-30
CORRESPONDENCE STUDIES
PAGE 19
MATHEMATICS FOUNDATIONS 11
EXPLORING POSSIBLE SOLUTIONS
LESSON 1.2
LEARNING OUTCOMES
Upon completion of lesson 1.2 students will
be expected to:
11A2 relate sets of numbers to solutions of
inequalities
11C6 apply the linear programming process
to find optimal solutions
11C8 demonstrate an understanding of
real-world relationships by translating
between graphs, tables, and written
descriptions
11C11 express and interpret constraints
11C18 interpolate and extrapolate to solve
problems
11C20 solve systems of equations and
inequalities both with and without
technology
11E3 represent systems of inequalities as
feasible regions
11E4 represent linear programming
problems using the Cartesian
coordinate system
INTRODUCTION
Lesson 2 begins with a continuation of Heather’s
problem from Lesson 1. This lesson examines
other constraints on Heather’s situation that affect
her earnings. Constraints are conditions that must
be met.
SAMPLE PROBLEM 1
Read Investigation 3, Part 1: Look at the Constraints
on Chairs and the information in the margin on
text page 14.
Procedure A
Since Heather has to cut at least 8 chair bundles
every two weeks, this means that she will cut
more than 8 bundles or only 8 bundles. Using the
language of inequalities this would be written as:
•
The number of chair bundles Heather cuts every
two weeks must be greater than or equal to 8.
•
The inequality for this is: x ≥ 8.
Ten points that satisfy the inequality organized
in a table are:
# of chair
bundles
8
9
# of couch
bundles
0
10 20
10 12 15 8.5 9.75 10.25 11 13
3
11
13
14
12
4
The number of couch bundles represent the
y-values. There are no constraints on couch
bundles, so digits were randomly chosen from 0
to 20, since twenty was on the coordinate grid.
These points are shown on the following grid.
PAGE 20
7
Couch bundles
•
y
20
15
10
5
0
•
0
5
10
Chair bundles
15
20
x
The points are found in the region to the right
of the line x = 8 (blue dotted line), which is a
vertical line rising up from x = 8.
CORRESPONDENCE STUDIES
UNIT 1
DO AND SEND 1
DO AND SEND 2
1. Read and do Procedures A and B of
Investigation 3, Part 2 on text page 15.
2. Complete Check Your Understanding
question 4 on text page 17, then do
questions 1-3 on text page 16.
SAMPLE PROBLEM 2
Read Procedure C of Investigation 3, Part 2: Look at
the Constraints on Couches. Refer to the grids you
have completed for Do and Send 1.
PART A
The grid below shows the points from Procedure
A, Part 1 as blue circles, and the points from Part
2 as black crosses. The common region is shaded
in blue.
3. Read Focus D and explain in your own
words what you learned from reading this
focus.
4. Do Check Your Understanding question 7,
parts A, C, and F on text page 19.
5. Do Check Your Understanding questions
10-12 on text pages 20-21.
y
Couch bundles
20
15
10
5
0
0
5
10
Chair bundles
15
20
x
The new graph of the common region is shown
below.
y
Couch bundles
20
15
10
5
0
0
5
10
Chair bundles
CORRESPONDENCE STUDIES
15
20
x
PAGE 21
MATHEMATICS FOUNDATIONS 11
CONECTING THE REGION AND
THE SOLUTION
LESSON 1.3
INTRODUCTION
In Lesson 3, you will investigate where optimal
feasible points can be found in the feasible region.
This means examining points in the feasible region
to determine which points determine maximum
or minimum earnings, income, or profit.
SAMPLE PROBLEM 1
Read Investigation 4, Finding the Best Solution on
text pages 22-23. Note the definition of optimal
solution in the margin on text page 22. In this
problem, you will find the optimal solution for
maximum profit.
Procedure A
Time
cutting machine
Time
sewing machine
Profit
Hats (h)
4 min
3 min
$1.10
Visors (v)
3 min
1 min
$0.60
LEARNING OUTCOMES
Upon completion of lesson 1.3 students will
be expected to:
11A2 relate sets of numbers to solutions of
inequalities
11B3 demonstrate an understanding of
the relationship between arithmetic
operations and operations on
equations and inequalities
11C6 apply the linear programming process
to find optimal solutions
11C8 demonstrate an understanding of
real-world relationships by translating
between graphs, tables, and written
descriptions
11C11 express and interpret constraints
11C20 solve systems of equations and
inequalities both with and without
technology
11E3 represent systems of inequalities as
feasible regions
11E4 represent linear programming
problems using the Cartesian
coordinate system
Procedure B
The cutting machine can only be used for a
maximum of 120 minutes a day, so
v
60
55
4h + 3v ≥ 120 is one inequality.
3h + 1v ≥ 60 is the other inequality.
Procedure C
Remember that h ≥ 0, and v ≥ 0, and v is on the
horizontal axis. Notice the white space above the
horizontal axis and to the right of the vertical axis.
This is the feasible region.
PAGE 22
50
45
40
Visors
The sewing machine can only be used for a
maximum of 60 minutes each day, so
Spinney Manufacturing
65
35
30
25
20
15
10
5
0
0
5
10
15
20
25
Hats
30
35
40
45
h
CORRESPONDENCE STUDIES
UNIT 1
Procedure D
P for Profit = 1.1h, or $1.10 per hat made + 0.6v,
or 60 cents per visor made: P = 1.1h + 0.6v
40
30
$23.50
10
20
$23.00
10
26
$26.60
10
10
$17.00
15
10
$22.50
16
10
$23.60
try 5 hats and 33 visors = $25.30. The point
is lower but the profit is higher. The top of the
feasible region does not give the greatest profit.
Let’s examine why ... If you graph the profit line
P = 1.1h + 0.6v for P= $20 (see the dotted line in
the first diagram above right) then, with P=$30
25
20
15
10
5
0
0
5
10
15
20
25
Hats
30
35
40
45
h
(see the dotted line in the diagram below), the last
point touched in the feasible region as the profit
line moved from its position in the first diagram to
the second diagram was the intersection point at
(12, 24).
v
Spinney Manufacturing
65
60
55
50
45
40
35
30
25
e
lin
By trying other points, you will discover the point
that gives the most profit is the intersection point.
The maximum or minimum profits, or incomes,
etc. will occur (most of the time) at intersection
lines.
30
it
of
pr
try the intersection point (12, 24) = $27.60.
This is greater than $26.60.
35
e
lin
5
CORRESPONDENCE STUDIES
45
it
of
pr
•
50
profit
Procedures F and G
Since the most profit ($26.60) occurs when 10
hats and 30 visors are made - it seems like the
greatest profit will occur near the top of the feasible
region.
• try zero hats and 40 visors - no, only $24.
•
55
Visors
number of hats number of visors
Spinney Manufacturing
60
Visors
Procedure E
Six points in the feasible region recorded in a
table:
v
65
20
15
10
5
0
0
5
10
15
20
25
Hats
30
35
40
45
h
Note: if the profit line is parallel to one of the
constraint lines, then the point that gives the
maximum profit would be any point on the
constraint line that is in the feasible region.
PAGE 23
MATHEMATICS FOUNDATIONS 11
DO AND SEND 1
1. Do Investigation Questions 1, 3 and 4,
parts a) and b) on text pages 23-24.
I
s= I - 8
or
s=
- 0.5t
16
16 16t
Let I = $320, so the equation becomes
s = 25 - 0.5t
30
SAMPLE PROBLEM 2
Part C
i) Heather is paid $12 for each couch bundle
and $4.25 for each chair bundle (text page
14). Since “chair bundles” are on the x-axis
and “couch bundles” are on the y-axis the
income equation is: I = 12y + 4.25x.
ii) Predicted maximum income is at (8, 15);
I = 12(15) + 4.25(8) = $214.
Predicted minimum income is at (8, 10);
I = 12 (10) + 4.25(8) = $154.
iii) Since these are the maximum and minimum
points, they are the only points in the feasible
region to give these incomes.
25
# of sweatshirts
Read Investigation Question 4, part C on text page
24. This is a continuation of Heather’s problem.
SAMPLE PROBLEM 3
Read Check Your Understanding question 6, part A,
on text page 25. In this problem, you are to write
an income equation, then graph it to decide where
the maximum income will occur.
The income equation is: I = 8t + 16s
To graph this, you need to rearrange the equation
(solve for s), then pick a value for an income, I.
PAGE 24
L1
15
10
5
0
0
5
10
15
20
25
30
35
40
# of T-shirts
45
50
55
60
65
According to the graph of the income line, L1,
the last point in the feasible region (shaded blue)
touched by L1 as it moves through the region is
the point (45, 10). So, Sheila must sell 45
T-shirts, and 10 sweat shirts to maximize her
profits.
DO AND SEND 3
DO AND SEND 2
2. Do Investigation Question 5 on text
page 25.
20
3. Do Investigation Question 6, part b on
text page 26.
SAMPLE PROBLEM 4
Read Focus E, Solving a System of Equations by
Estimating the Vertices on text pages 26-27. The
text shows three ways to solve a system of equations.
Solving a system means finding the coordinates for
the intersection point of the two lines or curves
when they are graphed.
•
In Method 1, you graph the two lines using
the y-intercept, slope method. The intersection
point actually looks like (16, 12.5), not the
(18, 12) mentioned in the text.
CORRESPONDENCE STUDIES
UNIT 1
•
Method 2 involves graphing, but uses
graphing technology to produce the same
graphs. Sometimes when you use TRACE to
find the intersection point, you miss the actual
point, when this occurs try using the CALC
feature found above the TRACE button. With
your calculator showing the graph screen,
press:
From there, rise 4 and run 1, then join the points.
The intersection seems to be (1, 1).
5
4
3
2
1
2nd
TRACE
cursor down the menu to 5:
intersect, then press
ENTER
At the bottom of the screen the
calculator asks: “is this the first
curve (or line)”, press
ENTER
again to say “yes”. Then it asks: “is
this the second curve (or line)”,
press
ENTER
again to indicate “yes”. Now it asks
you to guess. To do this, move the
cursor close to the intersection
point and press
ENTER
one more time. At the bottom
of the screen it will tell you the
intersection point. It gives the
x-value on the left and the y-value
on the right.
SAMPLE PROBLEM 5
Read Check Your Understanding question 9,
part A, on text page 28. You are to solve a system
of equations by graphing.
To graph y = 2x - 1, put a dot on -1 on the y-axis.
From there, rise two and run one, then join the
points.
To graph y = 4x - 3, put a dot on -3 on the y-axis.
CORRESPONDENCE STUDIES
-33
-22
-11
1
0
2
3
4
5
-11
-22
-3
SOLVING ALGEBRAICALLY
Sometimes, it is easier to solve a system using
algebraic methods. This gives a more accurate
answer, provided you do not make an algebraic
error. Read Focus F, Solving Systems of Equations
Algebraically on text pages 28 -29.
SAMPLE PROBLEM 6
Look at Check Your Understanding question 15,
part A on text page 29, which asks you to solve a
system of equations algebraically. Use Method 1,
shown in Focus F to solve this problem.
y = x - 4 and y = 4x + 5
Since both right sides in the equations = y,
rearrange the equation:
x - 4 = 4x + 5
subtract x from both sides
x - x - 4 = 4x - x + 5
-4 = 3x + 5
which gives
subtract 5 from both sides
-4 - 5 = 3x + 5 - 5 which gives
-9 = 3x
-9 = 3x
3 3
-3 = x
divide both sides by 3
which gives
PAGE 25
MATHEMATICS FOUNDATIONS 11
Substitute x = -3 into y = x - 4
SAMPLE PROBLEM 8
y = -3 - 4 or y = -7
Read Focus Question 21, part A, on text page 31,
which asks you to solve a system of equations with
fractions algebraically.
The intersection point is (-3, -7)
Note: this point also satisfies the other equation.
SAMPLE PROBLEM 7
Read Check Your Understanding question 17,
part A, which asks you to solve a system of
equations algebraically. This problem will be
solved using a method slightly different than the
method shown in Focus F.
2x + y = 5 and 4x - y = 9
First, rearrange one of the equations to equal y.
4x - y + y = 9 + y
add y to both sides
4x - 9 = 9 - 9 + y
subtract 9 from both sides
4x - 9 = y
substitute this equation for y
in the other equation.
2x + 4x - 9 = 5
simplify and add 9 to both
sides
6x - 9 + 9 = 5 + 9 which gives
6x = 14
divide both sides by 6
6x = 14 , x = 7
6
6
3
1
7
Substitute x = into 2x + y = 5, y = .
3
3
SOLVING SYSTEMS OF EQUATIONS WITH FRACTIONS
Read Focus G, Solving Systems of Equations
With Fractions on text pages 30-31. Sometimes
equations have fractions. Focus G reminds you to
eliminate the fractions before solving. To eliminate
the fractions, multiply all terms in the equation by
the lowest common denominator (LCD).
PAGE 26
1
2
x - y = -2 and x - y = 1
3
3
First, multiply both equations by the LCD.
1
x - (3)y = (3)-2
3
x - 3y = -6
(3)
(3)x - (3) 2 y = (3)1
3
3x - 2y = 3
Now, solve the first equation for x by adding 3y to
both sides
x - 3y + 3y = 3y - 6
x = 3y - 6
Substitute this x-value into the second equation for x
3(3y - 6) - 2y = 3
Remove the brackets by multiplying
9y - 18 - 2y = 3
Combine the y-terms
7y - 18 = 3
Add 18 to both sides
7y = 21
Divide both sides by 7
y=3
If y = 3, then x = 3(3) - 6 = 9 - 6 = 3
So the point of intersection is (3, 3).
SAMPLE PROBLEM 9
Look at Focus Question 23, part A, i), on text page
32, which asks you to draw a graph of the region
defined by a pair of inequalities and find the
maximum profit.
CORRESPONDENCE STUDIES
UNIT 1
To graph the line y = 4 - 2x put a dot at 4 on the
y-axis, rise negative 2, and run positive 1. Shade
below the line (shaded blue).
Graph the line y = 1, which is a horizontal line
passing through the y-axis at y =1.
Shade above this line (shaded grey).
y
5. Do Focus Question 12 on text page 29.
4
6. Do Check Your Understanding questions
15-17, 19, and 20 on text pages 29-30
using the algebraic approach. Question
15, parts A and b, have been done for you.
3
2
1
-11
4. Do Focus Questions 7, 9, 10 and 11 on
text pages 27-28. If the text does not
specify which method to use, you may
choose any method to find the point of
intersection. Question 9, part A, is done
for you.
5
L1
-22
DO AND SEND 4
0
1
2
3
4
5
x
-1
L1 is the profit line (black line). When P = 6, then
y = - 3 x + 3. It shows the maximum point will
2
be at the top vertex of the triangle (0, 4). The
maximum profit will be P = 3 (0) + 2 (4) = $8.
CORRESPONDENCE STUDIES
Note: question 15, parts A and B were
solved algebraically two different ways.
Question A was solved like Focus F;
question B was solved a little differently.
Explain how B was solved.
7. Do Focus Questions 21 and 23 on text
pages 31- 32. Part A of both questions
has been solved for you.
PAGE 27
MATHEMATICS FOUNDATIONS 11
LINEAR PROGRAMMING
PUTTING IT ALL TOGETHER
LESSON 1.4
INTRODUCTION
In this lesson you will re-examine Heather’s
problem and finally solve it. Heather’s problem is a
Linear Programming Problem. It is important to be
able to describe all the steps (read the Note in the
margin, text page 32) in the linear programming
method. After you have solved Heather’s problem
there are a several similar problems to solve. As
you solve the problems be careful to note all the
steps involved.
Read Focus H, Heather’s Maximum Income on text
page 32 to review all the information in Heather’s
problem. Through the sample problems and
your Do and Send assignments, you have already
completed many of the steps for this problem. At
this point, you will collect all the information and
present the solution.
DO AND SEND 1
1. a) Do Procedures A and B on text page 32.
In Procedure A, list the constraints on
Heather’s income and describe how you
wrote them algebraically. Then, draw
the graph to indicate the feasible region
and explain the steps involved. From
the graph, indicate how to determine
Heather’s maximum income.
Procedure B asks you to determine the
important intersection points using an
algebraic approach.
PAGE 28
LEARNING OUTCOMES
Upon completion of lesson 1.4 students will
be expected to:
11A2 relate sets of numbers to solutions of
inequalities
11B3 demonstrate an understanding of
the relationship between arithmetic
operations and operations on
equations and inequalities
11C6 apply the linear programming process
to find optimal solutions
11C8 demonstrate an understanding of
real-world relationships by translating
between graphs, tables, and written
descriptions
11C11 express and interpret constraints
11C20 solve systems of equations and
inequalities both with and without
technology
11E3 represent systems of inequalities as
feasible regions
11E4 represent linear programming
problems using the Cartesian
coordinate system
b) When you are finished Heather’s
Problem, make sure you stated her
maximum income and what she must
do to achieve that income.
2. At this point you will practice the steps
for linear programming problems. Do
Check Your Understanding question 24 on
text page 33. The diagram for question
24 is shown on the next page, as well as a
few additional questions. Copy and label
the graph before answering the questions.
CORRESPONDENCE STUDIES
UNIT 1
y
y = 2000 - 50x is the same as ...
30
Maximum Profit
100
2000 - 50x
30 30
The y-intercept is 200 ÷ 3 = 66.6,
and the slope is a rise of -5 and a run
of 3.
y=
tricycles
80
60
40
x+
2x
80
+y
20
y=
00
=1
0
0
20
40
60
bicycles
80
100
x
a) The diagram below shows the profit
equation, L1, plotted on the graph.
y
c) Create a new set of profits so the
intersection point (0, 80) results in
the maximum profit. Explain your
thought process for creating the new
profit set (how you came up with the
new profit set).
Maximum Profit
100
80
tricycles
b) Change the profit statement to give
a profit of $30 on each bicycle and a
profit of $50 on each tricycle. Draw
another graph to show the new profit
line and use it to find the maximum
profit.
60
3. Complete Check Your Understanding
questions 26 and 28 on text pages 33-34.
L1
40
x+
2x
+y
20
y=
4. Select any problem on text page 44 and
submit a complete solution.
80
=1
00
0
0
20
40
60
bicycles
80
100
x
To graph the profit line from the
equation P = 50x + 30y, let the profit,
P, be $2000, then solve the equation
y = 2000 - 50x
30
Explain how the profit line can
be used to determine the point of
maximum profit. If graphing by
hand:
for y.
CORRESPONDENCE STUDIES
THIS IS THE END OF UNIT 1
Make sure you have completed all the Do
and Send assignments in this unit.
SEND ALL ASSIGNMENTS TO
YOUR MARKER NOW
PAGE 29