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C H A P T E R
18MODULE 3
Linear programming
How do we graph linear inequations?
How do we graph systems of linear inequations?
How do we transfer from an optimisation problem to its mathematical formulation?
How do we use graphical methods to solve linear programming problems with two
decision variables?
Linear programming is used to solve problems in a wide variety of areas, such as planning,
design and finance. There are now many computer programs available for solving linear
programming problems, and these are used extensively in industry. Linear programming is
often used to determine a solution to a problem to maximise profit or minimise cost. Such
problems are called optimisation problems.
18.1 Regions defined by an inequalityIn Chapter 16, lines in the cartesian plane were discussed. The lines are described by
equations, for example:
0 x = 1x
y = 2
y
x0 10
10x + y = 10
y
Each line is an infinite set of points. The position of each point can be described by
coordinates. For example, points with the coordinates (0, 2) (1, 2) (1.5, 2) and (√
2, 2) all lie on
the line with the equation y = 2. Points with the coordinates (2, 8) (3, 7) (0, 10) (3.6, 6.4) all
lie on the line with the equation x + y = 10.
488
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Chapter 18 — Linear programming 489
InequalitiesEach point in the shaded region has an
x coordinate greater than 2. The shaded
region is described as the region for which
x ≥ 2 and formally {(x, y) : x ≥ 2}. The
boundary (the line with the equation x = 2)
is included. The convention is that if the
boundary is included it is represented with a
continuous line and if it is not included it is represented with a dotted line.
x0
(4, 5)
(3, 2)(3, 1)
(2, 0)
x = 2x > 2
required regionboundary included
y
Example 1 Sketching regions with horizontal or vertical lines as boundaries
Represent each of the following regions with a suitable graph.
a x < 2 b y ≤ −5 c y > 4
Solution
a
x0
x = 2
2
y
x < 2
b
x
0–5 y = –5
y
y < 5
c
x
4
0
y = 4
y
y > 4
The line x + y = 10 separates the plane into two regions.
Each point in the shaded region has coordinates whose sum
is greater than or equal to 10, for example:
(2, 11) 2 + 11 = 13 ≥ 10
(10, 2) 10 + 2 = 12 ≥ 10
(10, 0) 10 + 0 = 10 ≥ 10
x0 10
10
(10, 2)
(2, 11)
x + y > 10
y
In the diagram opposite, each point in the shaded
region has coordinates whose sum is less than 10,
for example:
x
(–1, 10)
(0, 0)
10(1, 8)
10x + y = 10x + y < 10
y
−1 + 10 < 10
1 + 8 < 10
0 + 0 < 10
The region to be shaded can always be established
by testing one point. The origin is often the easiest point
to check.
SAMPLE
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490 Essential Further Mathematics – Module 3 Graphs and relations
Example 2 Sketching regions defined by an inequality
Sketch the graph of the region 3x − 2y ≤ 6.
Solution
1 Sketch the graph of
3x − 2y = 6. The
straight line is drawn
as a continuous line
as the boundary
is included.
When x = 0, −2y = 6
∴ y = −3
When y = 0, 3x = 6
∴ x = 2 x(0, 0)
(2, 0)
(0, –3)
3x – 2y = 6
y
2 Check the origin (0, 0).
x0 2
–3
3x – 2y = 6
y
3 × 0 − 2 × 0 = 0 ≤ 6
(0, 0) is in the required region.
∴ the region above the line is
required.
3 Shade the required region.
Exercise 18A
1 Calculate the value of 5x + 6y corresponding to each of the given points and state whether
they are:
A in the region 5x + 6y > 22 B in the region 5x + 6y < 22
C on the line with equation 5x + 6y = 22
a (2, 2) b (3, 1) c (4, 2) d (0, 0)
2 Sketch the graph of each of the following regions:
a x ≥ 5 b x < 4 c x ≤ −2 d y > 4 e x > −5 f y ≤ 6
3 Sketch the graph of each of the following regions:
a y ≥ 2x + 5 b 3x − 2y < 6 c x + y ≤ 10 d x − y ≥ 10
e 2x − 6y ≤ 3 f y ≤ 2x g x − y > −3 h x − 2y > −3
i 2x + y ≥ 12 j 3x − 4y < 12
18.2 Regions defined by two inequalitiesExample 3 Sketching a region defined by two inequalities
Sketch the graph of the region that is defined by the inequalities x − y ≤ 10 and y ≤ x .
SAMPLE
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Chapter 18 — Linear programming 491
Solution
1 Sketch the graph of the region
x − y ≤ 10.
2 On the same set of axes sketch
the graph of y ≤ x .
x
x – y = 10
10
–10
y
0
The region required is the region
with overlap of the two shadings.
x
y
x – y = 10
10
–10
0
y = x
required region
(boundaries included)
Example 4
Sketch the graph of the region that is defined by the inequalities 3x + 2y ≤ 6 and y − x ≤ 1.
Solution
x
y
3x + 2y = 6
y = x + 1
3
required region
(boundaries included)
5
4
5
9,
0 2
1 Draw the two straight lines.
2 Find the point of intersection of the
two lines by solving the simultaneous
equations or using a graphics calculator.
3 Shade each of the required regions.
4 The defined region is shown by the overlap.
Exercise 18B1 Sketch the region defined by each of the following sets of inequalities and find the
coordinates of the point of intersection:
a y ≥ −2x + 10
y ≥ x − 2
b 3x − 2y < 6
x − 3y > 9
c 2x + 3y > −6
3x − y < 6
d 2x + 3y ≤ 6
x ≥ 0
e 2y − 3x < 6
x < 0
f y − 3x < 6
x − y > −3
g x − y < 1
y − x < 1
h 2x − 2 ≥ y
2x ≤ 3 − 2y
i 2y < 4x + 2
y < 2x + 1
SAMPLE
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492 Essential Further Mathematics – Module 3 Graphs and relations
18.3 Feasible regions
Example 5 Shading a feasible region
A hardware shop has room for no more than a total of 100 cans of paint of Brand X and
Brand Y. It is known that at least 30 cans of Brand X and no more than 50 cans of Brand Y will
be required.
By shading a suitable region of the cartesian plane, indicate the possible number of Brand X
and Brand Y that can be stored.
Solution
y
(60, 10)10
20
30
40
50
60
70
80
90
100
0 10 20 30 40 50 60 70 80 90 100
feasible region
x = 30
x + y = 100
(40, 50)
(50, 50)
(46, 20)
(80, 1)
y = 50
(30, 50)
x
Let x be the number of cans of brand X.
Let y be the number of cans of brand Y.
The constraints can be expressed by the
following inequalities:
x ≥ 30
0 ≤ y ≤ 50
x + y ≤ 100
The shaded region in Example 5 is called the feasible region. It is the graph of all possible
pairs (x, y) which satisfy the inequalities.
Points with coordinates such as (30, 50) (40, 50) (46, 20) (50, 50) (60, 10) and (80, 1) all lie
within the feasible region. Points on the boundary are included in the feasible region.
Example 6 Shading a feasible region
Indicate, by shading, the region that satisfies all the following constraints. Mark in the
coordinates of the ‘corner’ points of the feasible region:
3x + 8y ≥ 24
3x + 2y ≥ 12
x ≥ 0
y ≥ 0
Solution
x
y
feasible region
0 2
2
4
4
6
6
(8, 0)
(0, 6)
3x + 2y = 123x + 8y = 24
8
83
, 2
10
SAMPLE
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Chapter 18 — Linear programming 493
Exercise 18C
1 For each of the following indicate, by shading, the region that satisfies all of the constraints.
Mark in the coordinates of the ‘corner’ points of the feasible region.
a x + y ≤ 7
5x − 3y ≤ 15
x ≥ 0
y ≥ 0
b −x + 2y ≤ 12
y ≥ 2x − 9
x ≥ 2
y ≥ 0
c 3x + 4y ≤ 60
y ≥ 15 − 3x
x ≥ 0
y ≥ 3
d x + 2y ≥ 10
y ≤ 2x
0 ≤ x ≤ 6
0 ≤ y ≤ 14
e −4x + 5y ≤ 20
2x + 3y ≥ 6
x ≥ 0
y ≥ 0
f x ≤ 7
y ≥ 4
x + y ≤ 9
24x + 30y ≥ 360
g x + y ≥ 6
2x + 5y > 20
x ≤ 4
y < 5
h x ≥ 0
y ≥ 0
3x + y ≥ 15
2x + 3y ≥ 36
i x ≥ 0
y ≥ 0
x + y ≤ 4
x ≤ y
y ≤ 3
18.4 Objective functionsAn objective function is used to describe a quantity that you are trying to make as large as
possible (for example, profits) or as small as possible (for example, the amount of paint needed
to paint a house) subject to some constraints. Consider the following situation.
A hardware shop has room for no more than a total of 100 cans of paint of Brand X and
Brand Y. It is known that at least 30 cans of Brand X and no more than 50 cans of Brand Y will
be required. The profit on each can of Brand X is $2.00 and $2.50 for each can of Brand Y. If
the entire stock can be sold, how many of each brand should there be to yield maximum profit?
Let x be the number of cans of brand X.
Let y be the number of cans of brand Y.
The constraints for this situation can be expressed by the following inequalities:
x ≥ 30 (number of cans of Brand X required)
0 ≤ y ≤ 50 (number of cans of Brand Y required)
x + y ≤ 100 (total number of Paint required)
The profit function is given by:
P = 2x + 2.5y
where the value of P yields the number of dollars profit.
To be a solution, (x, y) must be in the feasible region.
Consider the point (60, 10).
For x = 60, y = 10 P = 2 × 60 + 2.5 × 10 = 145
SAMPLE
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494 Essential Further Mathematics – Module 3 Graphs and relations
There are other values of x and y that yield a profit of $145
e.g. x = 55, y = 14; x = 50, y = 18; x = 45, y = 22; x = 30, y = 34.
All these points lie on the straight line 2x + 2.5y = 145 (a profit of $145).
For a profit of $185 the straight line is 2x + 2.5y = 185.
For a profit of $225 the straight line is 2x + 2.5y = 225.
All these lines are parallel. They have been added to the graph of the feasible region.
y
10
20
30
40
50
60
70
80
90
100
0 10 20 30 40 50 60 70 80 90 100
required region
x
2x + 2.5y = 2502x + 2.5y = 185
2x + 2.5y = 2252x + 2.5y = 145
(50, 50)
(30, 50)
It can be seen that the maximum profit possible is $225 and this is only achievable at one
point in the feasible region, i.e. (50, 50).
It can be seen that a larger value of P (e.g. P = 250) will not yield points in the feasible
region.
The function with the rule P = 2x + 2.5y is an example of an objective function. In linear
programming problems, the aim is to find the maximum or minimum value of an objective
function for a given feasible region. To help us do this, we can make use of the corner point
principle.
The corner point principleIn linear programming problems, the maximum or minimum value of a linear objective
function will occur at one of the corners of the feasible region.Note: If two corners give the same maximum or minimum value, then all points along a line joining the twopoints will also the same maximum or minimum values.
This means that we only need to evaluate the objective function at each of the corner points to
find the maximum or minimum value of an objective function.SAMPLE
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Chapter 18 — Linear programming 495
Example 7 Finding the minimum value of an objective function
Find the minimum value of the objective function
Z = 3x + 2y subject to the constraints:
x ≥ 0, y ≥ 0
4x + y ≥ 12
3x + 15y ≥ 44
3x + 4y ≥ 22
as displayed in the feasible region opposite.
y
x0
feasibleregion
(0, 12)
(2, 4)143
, 2443
, 0
A
B
C
D
Solution
1 Write down the objective function.
2 Evaluate the objective function at each
of the corners.
3 Identify the corner point giving the minimum
value and write down your answer.
Z = 3x + 2y
A (0, 12) : Z = 3 × 0 + 2 × 12 = 24
B (2, 4) : Z = 3 × 2 + 2 × 4 = 14
C
(14
3, 2
): Z = 3 × 14
3+ 2 × 2 = 18
D
(44
3, 0
): Z = 3 × 44
3+ 2 × 0 = 44
The minimum value is Z = 14 when
x = 2 and y = 4.
Example 8 Finding integer solutions
A courier has to transport 900 parcels using large vans, which can take 150 parcels, and
smaller vans, which can take 60 parcels. The costs of each journey are $5.00 by large van and
$4.00 by small van. The total cost must be less than $44.
a Represent the feasible region for these constraints on a graph.
b Indicate the possible values of x and y on the diagram by using a ×.
c What is:
i the largest number of vehicles that could be used?
ii the smallest number of vehicles that could be used?
Solution
y
x0
5
10
15
2 4 6 8 10
150x + 60y = 900
5x + 4y = 44
boundary
excluded
(3.2, 7)
a Let x be the number of large vans and
y the number of small vans.
constraints:
x ≥ 0
y ≥ 0
150x + 60y ≥ 900 (5x + 2y ≥ 30)
5x + 4y < 44
SAMPLE
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496 Essential Further Mathematics – Module 3 Graphs and relations
b The crosses in the feasible region indicate the possible numbers of small vans and large
vans.
c i The largest number of vehicles that could be used is 9: either 5 large and 4 small or 4 large
and 5 small or 6 large and 3 small or 7 large and 2 small.
ii The smallest number of vehicles that could be used is 6 large vans.
Example 9 Setting up and solving a linear programming problem
A business produces imitation antique vases in two styles: Ming Dynasty vases and Geometric
Period Greek vases.
Each vase requires:
potters to make the vase
artists to decorate the vase
During one week the business employs 10 potters and 4 artists. Each employee works for a
total of 40 hours from Monday to Friday.
The time spent making each vase is shown in the table below.
Employee Ming Geometric
Potter 8 hours 4 hours
Artist 2 hours 2 hours
Let x represent the number of Ming Dynasty vases made in a week and y represent the number
of Geometric Period vases made in a week.
a The business has a permanent weekly order of 10 Ming vases and 20 Geometric vases.
These give the inequalities
x ≥ 10 and y ≥ 20
The inequality formed from the total time available to the potters is
8x + 4y ≤ 400
Write down the corresponding inequality formed from the total time available to the artists.
b Draw the graphs of the four inequalities in part a.
c Clearly show the feasible region on the graph in part b.
The profit is $50 on each Ming vase is $50 and $30 on each Geometric vase.
d All vases produced in a week are sold. Write down an expression in terms of x and y for the
total profit $A that the business will receive.
e Determine the number of each type of vase that should be produced in a week to result in
the maximum profit.SAMPLE
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Chapter 18 — Linear programming 497
Solution
y
10
20
30
40
50
60
70
80
90
100
0 10 20 30 40 50 60 70 80x
x + y < 80
(20, 60)
(10, 20)
(10, 70)
x = 10
y = 20
8x + 4y = 400
(40, 20)
a 2x + 2y ≤ 160 or, equivalently, x + y ≤ 80 b, cd P = 50x + 30y
e Evaluate at each of the vertices of the feasible region:
(10, 70) P = 50 × 10 + 30 × 70 = 2600
(40, 20) P = 50 × 40 + 30 × 20 = 2600
(10, 20) P = 50 × 10 + 30 × 20 = 1100
(20, 60) P = 50 × 20 + 30 × 60 = 2800
To maximise the profit, 20 Ming vases and 60 Geometric
vases should be produced.
Exercise 18D1 The region that satisfies all of the following constraints: 5x – 2y = 20
–x + 2y = 8
x
y
AD
B C0
2468
10(7, 7.5)
2 4 6 8–2–2–4–6–8
–10
–4–6–8
5x − 2y ≤ 20
−x + 2y ≤ 8
x ≥ 0
y ≥ 0
is as shown.
a Write down the values of the coordinates of A, B, C and D.
b Find the maximum value of z = x + 2y subject
to the set of constraints above.
2 The region that satisfies all the following constraints
is shown:
x
y
103 C
–6
–4
–2
2
4
0B
6
8A E
D
y = 2x – 6
y = 6
4x + 5y = 40
4x + 5y ≤ 40
y ≥ 2x − 6
x ≥ 0
0 ≤ y ≤ 6
a Find the coordinates of A, B, C, D and E.
b Find the maximum value of z = 2x + y subject
to the set of constraints above.
3 a Illustrate the region that satisfies all the following constraints:
x + 3y ≤ 17
5x + 3y ≥ 25
0 ≤ x ≤ 8
0 ≤ y ≤ 6
b Find the maximum value of z = x + 3y subject to the set of constraints in a.
SAMPLE
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498 Essential Further Mathematics – Module 3 Graphs and relations
4 The region that satisfies the following constraints
is shown:
x
y
0 3
4x + y = 12
3x + 4y = 22
3x + 15y = 44C
B
A12
443
4415
112
223
D
4x + y ≥ 12
3x + 4y ≥ 22
3x + 15y ≥ 44
x ≥ 0
y ≥ 0
a Find the coordinates of points A, B, C and D.
b Find the minimum value of z = 3x + 2y
subject to the set of constraints above.
5 a Illustrate the region that satisfies all the following constraints:
4x + 5y ≥ 52
y ≥ 0.5x
y ≤ 1.8x
x ≥ 4
y ≥ 0
b Find the minimum value of z = 4x + 10y subject to the set of constraints in a.
6 The region that satisfies all the following constraints
is as shown:
04x + 3y = 39
CB
D
x + 7y = 16
3x + y = 8
y = x + 6
x
y
A
3x + y ≥ 8
y ≤ x + 6
4x + 3y ≤ 39
x + 7y ≥ 16
x ≥ 0
y ≥ 0
a Find the coordinates of points A, B, C and D.
b Find the minimum value of z = 7.5x + 2.5y subject to the set of constraints above.
Questions 7−12 provide practice in setting up constraints and forming the objective function
7 A manufacturer makes two types of skateboard, A and B. Each skateboard is cut out in the
pattern shop, constructed in the assembly shop and then sanded and painted in the finishing
shop. Each skateboard of type A requires 1.2 units of labour in the pattern shop, 2.0 units
of labour in the assembly shop and 1.3 units of labour in the finishing shop. Each
skateboard of type B requires 1.0 units, 1.4 units and 1.5 units in the pattern, assembly and
finishing shops respectively. The numbers of units of labour available per week in the
pattern, assembly and finishing shops are 180, 240 and 200 respectively.
The profit from the sale of each skateboard of type A is $35, and from each skateboard
of type B is $30. The manufacturer can sell all the skateboards he can make. The number of
skateboards of type A made per week is a, and the number of skateboards of type B made
per week is b.
SAMPLE
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Chapter 18 — Linear programming 499
The constraints are:
Pattern Shop: 1.2a + b ≤ 180
Assembly shop: 2a + 1.4b ≤ 240
a Give the inequation which describes the constraints given for the finishing shop.
b Give the rule for the profit per week, $P in terms of a and b.
8 A pheasant farmer wishes to blend food for her birds from two commercially available
products, Wheato and Oatex. Each kilogram of Wheato costs $1.80 and contains 200 units
of vitamins and 150 units of protein. Each kilogram of Oatex costs $2.10 and contains
180 units of vitamins and 260 units of protein. Each pheasant requires 12 units of vitamins
and 16 units of protein per day and the farmer has 290 pheasants.
The farmer wishes to satisfy her birds’ dietary requirements for the least cost. If x is the
number of kilograms of Wheato that she buys per day and y is the number of kilograms of
Oatex that she buys per day, write down the constraints on x and y.
Minimum units of vitamins required = 12 × 290
= 3480
∴ 200x + 180y ≥ 3480
Minimum units of protein required = 16 × 290
= 4640
a Give the inequality derived from the information about minimum protein.
b Give the rule for $C, the cost of the food per day.
9 A self-employed computer serviceman has more work than he can manage to do in a single
8-hour day. He can work for up to 5 hours on tasks of type A, which pay him $60 per hour,
and he can work for up to 6 hours on tasks of type B, which pay him $70 per hour. He has
at hand 41 new components and knows that he will use them at a rate of four per hour on
tasks of type A and six per hour on tasks of type B.
The serviceman wishes to decide the best mix of tasks for the day. If he spends x hours
on tasks of type A and y hours on tasks of type B, write down the constraints on x and y.
Also write down the formula for $P, the amount of pay received, in terms of x and y.
10 An ice-cream manufacturer makes 2000 litres per day of just three flavours: strawberry,
chocolate and vanilla. The profits on one litre of strawberry, chocolate and vanilla
ice-cream are, respectively, 50 cents, 40 cents and 30 cents. Demand is such that the
manufacturer should never make more than 600 litres of vanilla ice-cream in a day and
should never make more strawberry ice-cream than chocolate ice-cream.
On a particular day, the supply of raw materials is such that a maximum of 800 litres of
chocolate ice-cream can be made.
a If x and y are the numbers of litres of strawberry and chocolate ice-cream made that day,
respectively, how many litres of vanilla ice-cream are made?
b Write down, in simplest form, the constraints on x and y, and also the rule for the profit,
$P, assuming that all ice-cream made that day is sold.
SAMPLE
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500 Essential Further Mathematics – Module 3 Graphs and relations
11 A manufacturer makes two articles, X and Y, which require three different component
parts, A, B and C. Each X requires 12A, 8B and 10C components, and each Y requires 9A,
11B and 13C components. The profit to the manufacturer on each X is $84 and on each Y
is $72.
In the period just before Christmas the manufacturer wishes to decide how many of X
and how many of Y to make, in order to maximise profit. There are 280A, 260B and 320C
components available at that time. If x is the number of X produced and y is the number of
Y produced, write down the constraints on x and y. Also write down the rule for the profit,
$P, assuming that all articles are sold.
12 A small service station sells petrol and diesel. The costs to the proprietor are 35 cents for
each litre of petrol and 42 cents for each litre of diesel. She can store at most 20 000 litres
of petrol and 15 000 litres of diesel at any given time. She has found from experience that
the demand for diesel is never more than half as great as the demand for petrol.
The proprietor wishes to arrange for deliveries of fuel on the day before a holiday
weekend. At that time her storage tanks are each at 50% capacity and she can afford to
spend $5000 buying fuel. If x is the number of litres of petrol she buys and y is the number
of litres of diesel she buys, write down the constraints on x and y.
13 For a journey across Ellesmene
Island in the Arctic Circle, an
explorer wishes to travel as lightly
as possible. He can obtain supplies
of two types of lightweight food,
type X and type Y, in order to supply him with energy, protein and carbohydrate. The
details about them are shown in the table.
Energy per Protein per CarbohydrateFood serve (units) serve (units) per serve (units)
Type X 600 3.0 20
Type Y 400 3.5 35
The explorer estimates that his minimum daily requirements of energy, protein and
carbohydrate will be 2600 units, 19 units and 150 units respectively. Each serve of type X
food weighs 36 g and each serve of type Y food weighs 56 g.
a If x and y are the number of serves per day of type X and type Y foods respectively that
the explorer takes, four of the constraints on x and y are:
x ≥ 0
y ≥ 0
600x + 400y ≥ 2600
3x + 3.5y ≥ 19
Give the constraint determined by the amount of carbohydrate required.
b If W grams is the weight per day of food that the explorer takes, write down the rule for
W in terms of x and y.
c Use a graphical method to determine how many serves per day of each food type the
explorer should take in order to minimise weight while still satisfying his daily dietary
requirements.
SAMPLE
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Chapter 18 — Linear programming 501
14 Sam wants to buy some CDs, which cost $30 each, and some books, which cost $20 each.
She can spend up to $300 and wants to spend at least twice as much on CDs as she spends
on books. In addition, she wants to buy at least two CDs and no more than three books.
a If x denotes the number of CDs and y denotes the number of books that Sam buys, two
of the constraints that x and y must satisfy are:
3x + 2y ≤ 30
3x ≥ 4y
Write down two further constraints.
b Indicate on a graph the region for which x and y satisfy the constraints of a.
c Find the maximum number of items that Sam can buy.
15 A computer importer has two depots, D1 and D2,
and two retail outlets, R1 and R2. At the beginning
of October the importer receives orders for Versanote
180 computers, which can be summarised as shown:
Number of computersOutlet ordered
R1 123
R2 86
The numbers of computers available can be
summarised as shown:Number of computers
Depot available
D1 160
D2 115
Transport costs per computer can be summarised
as shown:Transport costs per
Journey computer (dollars)
From D1 to R1 2.20
From D1 to R2 2.80
From D2 to R1 1.50
From D2 to R2 1.70
The importer wishes to decide the most
economical way to supply exactly the right
number of computers to each retail outlet.
Let x be the number of computers moved
from D1 to R1 and let y be the number of
computers moved from D1 to R2.
a Write down, in simplest form, the constraints on x and y.
b Write down, in simplest form, the formula for the transport cost, $C, in terms of x and y.
c Use a graphical method to find the most economical way to solve the problem.
16 A mining company is required to move 200 workers and 36 tonnes of equipment by air. It is
able to charter two aircraft: a Hawk, which can accommodate 20 workers and 6 tonnes of
equipment, and an Eagle, which can accommodate 40 workers and 4 tonnes of equipment.
Let x denote the number of trips made by the Hawk aircraft and y denote the number of
trips made by the Eagle aircraft.
a Write down the constraints that x and y must satisfy.
b Indicate on a graph the values of x and y that satisfy the constraints of a.
c If the respective costs of the Hawk and the Eagle aircraft are $3000 and $4000 per trip,
find the number of trips that should be made by each aircraft in order to minimise the
total cost, and find the minimum cost.
SAMPLE
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502 Essential Further Mathematics – Module 3 Graphs and relations
17 A produce grower decides to buy fertiliser containing three nutrients A, B and C to spread
on her paddocks. The minimum needs are 160 units of A, 200 units of B and 80 units of C.
There are two popular brands of fertiliser on the market: Fast Grow, costing $4 a bag,
contains three units of A, five units of B and one unit of C. Easy Grow, costing $3 a bag
contains two units of each nutrient.
If the grower wishes to minimise her costs while still maintaining the nutrients required,
how many bags of each brand should she buy?
18 An outdoor clothing manufacturer has 520 metres of polarfleece material. The
manufacturer will use it to make jackets of two types, Polarbear and Polarfox, to sell to
retailers. For each jacket of either type, 2.0 metres of material is required. However, the
Polarbear is simpler in design, requiring 2.4 hours each in the production process while
each Polarfox requires 3.2 hours. There are 672 hours available.
From past experience of demand, the manufacturer has decided to make no more than
half as many Polarfox jackets as Polarbear jackets. If the profit on each Polarbear jacket is
$36 and the profit on each Polarfox jacket is $42, use a graphical method to find how many
of each type should be made in order to maximise profit. What is this maximum profit?
19 The army is required to airlift 450 people and 36 tonnes of baggage by helicopter. There
are 9 Redhawk helicopters and 6 Blackjet helicopters available. Each Redhawk can carry
45 passengers and 3 tonnes of baggage, while each Blackjet can carry 30 passengers and
4 tonnes of baggage. Running costs per hour are $1800 for each Redhawk and $1600 for
each Blackjet. If the army wishes to minimise the cost of the airlift per hour, use a
graphical method to find how many of each helicopter should be used.
20 At the end of summer a small surfboard maker has 150 litres of polyester resin and
proposes to use it up making two types of surfboard for the next season. He can make
full-length boards, which require 5 litres of polyester resin each, and boogie boards, which
require 3 litres of resin each. He has at hand 200 metres of glassfibre cloth, and knows he
will need 6 metres of it for each full-length board and 4 metres for each boogie board. The
surfboard maker can obtain polyurethane blanks from a nearby supplier, but has found that
the supplier has only enough blanks for 25 full-length boards and 20 boogie boards. He is
confident that, as long as he makes more than two-thirds as many boogie boards as
full-length boards, he can sell all the boards he makes at a profit of $180 on each
full-length board and $110 on each boogie board.
Use a graphical method to find how many of each type of board he should make in order
to use up as much of the resin and cloth as possible while maximising profit. (You will need
to consider ‘integer points’ near the corner points of the feasible region.) How much resin
or cloth, if any, is left over with this optimal production scheme?SAMPLE
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Review
Chapter 18 — Linear programming 503
Key ideas and chapter summary
Procedure for sketching The procedure for representing an inequality in a plane is as
follows:� Sketch the boundary line. The line is dotted if the
boundary line is not included and continuous if the
boundary line is included.
� Test with a point to determine which side of the line to
shade.
Example: x + y < 10
For the point (0, 0) 0 + 0 < 0
∴ shade underneath the boundary line.
x
y
x + y = 1010
10
0
(0, 0)(1, 8)
(–1, 0)
an inequality
Procedure for sketching The procedure for representing a system of inequalities for the
plane is as follows:� Graph each of the inequalities using a different shading
for each one.
ande.g� The overlap determines the required region.
Example: The region of points which satisfy y ≤ x
and x − y ≤ 10
x
y
x – y = 10
10
–10
0
y = x requiredregion
the region defined by asystem of inequalities
Feasible region The feasible region is the set of points that can be considered for
the given inequalities. For example, for the set of inequalities:
y ≥ 0
2x − y ≤ 0
x + y ≤ 30
SAMPLE
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504 Essential Further Mathematics – Module 3 Graphs and relations
The feasible region is as shown.
y
x
feasibleregion
(0, 0) (30, 0)
(0, 30)
(10, 20)
2x – y = 0
x + y = 30
Linear programming Linear programming involves maximising or minimising a
linear quantity subject to the constraints represented by a set of
linear inequalities. The constraints define the feasible region in
which the quantity is to be maximised or minimised.
Objective function The objective function is a linear expression representing the
quantity to be maximised or minimised in a linear programming
problem. For example, P = 3x + 5y could be an objective
function representing profit.
The corner point principle The corner point principle states that in linear programming
problems, the maximum or minimum value of a linear objective
function will occur at one of the corners of the feasible region, or
on line on the boundary of the feasible region joining two of the
corners.
Maximum or minimum The maximum (or minimum) value of the objective function is
found by evaluating its values at the vertices (or along
boundaries) of the feasible region. (See the corner point
principle.)
Skills check
Having completed this chapter you should be able to:
sketch the region defined by inequalities
determine the feasible region defined by several inequalities
find the maximum or minimum value of an objective functionSAMPLE
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Review
Chapter 18 — Linear programming 505
Multiple-choice questions
1 In the graph shown, the shaded region with
boundary included is the feasible region for
a linear programming problem that involves an
objective function P = 2x + 3y. The maximum
value of P for this feasible region occurs at a point
with coordinates:
A (0, 40) B (16, 40) C (36, 40)
D (48, 20) E (60, 0)x
y
(36, 40)
(48, 20)
(60, 0)
(0, 40)
0
(16, 40)
2 In which of the following graphs does the shaded region represent the points (x, y)
such that 5x + 3y ≤ 30? (In each case the boundary is included.)
A
x
y
06
10
B
0
3
5x
y C
x
y
0 3
5
D
6
0 10x
y E
x
y
0
3
5
3 The shaded region of the graph is described by:
A y + 3x ≥ 0 B y ≥ x + 1 C 3y − x ≥ 0
D 3y + x ≥ 3 E 3x − y ≥ 3
x
y
0
1
3
4 For the graph shown, the shaded region (boundaries
included) is the feasible region for a linear programming
problem that involves an objective function T = 2x + 3y.
The maximum value of T for this region occurs at the
point with coordinates:
A (0, 60) B (0, 50) C (20, 40)
D (30, 0) E (10, 60)x
y
(10, 50)
(0, 60)
(20, 40)
(30, 0)0
5 The region LMNO is shown in the diagram. If (x, y)
is a point in the region LMNO (boundaries included),
then the greatest value of P = 2x + 3y is:
A 8 B 12 C 13 D 15 E 16
x
y
O
L
M
N
1
123456
2 3 4 5 6
SAMPLE
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506 Essential Further Mathematics – Module 3 Graphs and relations
6 Which of the following sets of inequalities describes
the shaded region in the diagram?
A x, y ≥ 0 2x + y ≥ 2 y ≤ x x + y ≤ 6
B x, y ≥ 0 2x + y ≤ 2 y ≥ x x + y ≤ 6
C x, y ≥ 0 2x + y ≥ 2 y ≥ x x + y ≤ 6
D x, y ≥ 0 2x + y ≤ 2 y ≤ x x + y ≤ 6
E x, y ≥ 0 2x + y ≥ 2 y ≥ x x + y ≥ 6x
y
1 6
6
2
y = x
0
7 The shaded region is described by:
A x ≤ 3, y ≤ 3, x ≥ 0
B x + y ≥ 3, x ≥ y, x ≥ 0
C x + y ≥ 3, x ≤ y, x ≥ 0
D x + y ≤ 3, x ≥ y, x ≥ 0
E x + y ≤ 3, x ≤ y, x ≥ 0x
y
y = x
x + y = 3
8 For point (x, y) in the region LMNO (boundaries
included) as shown, the smallest value
of 3x − 4y + 25 is:
A 4 B 9 C 10 D 25 E 46
x
y
O
L (0, 4)
(7, 0)
5
5
10N
M (3, 6)
9 Which one of the following sets of inequations describes
the shaded region (boundaries included) on the graph
shown?
x
y
0
6
8
5 7
A x ≥ 0, y ≥ 0
5x + 8y = 40
7x + 6y = 42
B x ≥ 0, y ≥ 0
8x + 5y = 40
6x + 7y = 42
C x ≥ 0, y ≥ 0
5x + 8y ≤ 40
7x + 6y ≤ 42
D x ≥ 0, y ≥ 0
8x + 5y ≤ 40
6x + 7y ≤ 42
E x ≥ 0, y ≥ 0
8x + 5y ≥ 40
6x + 7y ≥ 42
10 In the graph shown, the shaded region, with boundary
included, is the feasible region for a linear programming
problem. The equation of the objective function is
P = 2x + y. The maximum value of P for this
feasible region occurs at the point with coordinates:
A (0, 30) B (40, 80) C (40, 90)
D (20, 60) E (40, 0) x
y
(0, 0)
(0, 30)
(0, 40)(20, 60)
(40, 90)
(40, 80)
(40, 0)
SAMPLE
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Review
Chapter 18 — Linear programming 507
Extended-response questions
1 Adam opens a new business. He employs several people to produce carved models
of two famous mathematicians, Descartes and Pythagoras. Each model is made of
wood and Adam employs people to:� carve the models (wood carvers)� paint the models (painters).
During one particular week,
Adam employs five wood
carvers and two painters. Each
worker works for a total of
40 hours from Monday to Friday.
The time spent for the carving and painting of each model is shown in the table.
Trade Pythagoras model Descartes model
Wood carver 4 hours 2 hours
Painter 1 hour 1 hour
Let x represent the number of Pythagoras models made in a week and y represent the
number of Descartes models made in a week.
a What is the maximum number of hours that can be spent carving the models in the
week?
b If each of the wood carvers only carve Pythagoras models, how many models
would they carve altogether in the week?
c The business has a permanent weekly order for 10 Pythagoras models and 20
Descartes models. These facts give the inequalities:
x ≥ 10
y ≥ 20
The inequality formed from the total time available to the wood carvers is
4x + 2y ≤ 200
Write down the corresponding inequality formed from the total time available to
the painters.
d Draw the graphs of the four inequalities in part c.
e Clearly show the feasible region defined by the four inequalities. [VCAA 2000]
2 Adam pays each of his workers an
hourly wage as shown in the table.Trade Payment per hour
Wood carver $24
Painter $20The cost of raw materials for each model is $10.
The profit on the Pythagoras model is $30 and
the profit on the Descartes model is $10.
a The selling price of a Pythagoras model is $156. Calculate the selling price of a
Descartes model.
b All the models produced in the week are sold. Write down an expression in terms
of x and y for the total profit $A that Adam will receive.
(cont’d.)
SAMPLE
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iew
508 Essential Further Mathematics – Module 3 Graphs and relations
c Use the graph from Question 1 and the expression from Question 2b, to
determine the number of models of each type that should produced in the week
to result in the largest possible profit for Adam.
d Find the maximum profit that Adam could make in the week. [VCAA 2000]
3 A small canning company produces two types of canned tuna with additional chilli.
The Super Tuna with fried chilli and the Elite Tuna with dried chilli. A can of Super
Tuna requires 200 g of tuna and 30 g of chilli. A can of Elite Tuna requires 300 g of
tuna and 20 g of chilli. The company can produce 800 cans of tuna a day. There is
200 kg of tuna available to them every day and no limit on the chilli. Let x be the
number of cans of Elite Tuna produced in a day and let y the number of cans of
Super Tuna produced in a day. The inequalities produced by this information are:
x ≥ 0
y ≥ 0
x + y ≤ 800
0.3y + 0.2x ≤ 200
a Sketch the graphs of these inequalities.
b Find the coordinates of the vertices of the region defined by these inequalities.
c If the company makes $1.00 profit on the Elite and $0.80 profit on the Super,
write an expression for the daily profit in terms of x and y.
d How many cans of each type of tuna should be produced to maximise the profit?
4 The Victory Vineyard makes both a
red wine and a white wine. The
table summarises the production
costs and sale prices per bottle of
the white wine and the red wine.
Type of wine Production cost Sale price
White $7 $15
Red $10 $20
The following conditions apply to the production of wine at the Victory Vineyard.� The maximum total number of bottles of red wine and white wine
produced by the Victory Vineyard in any day is 700.� The total production cost of red and white wine cannot exceed $6400 per day.� The maximum number of bottles of red wine that can be produced is 570 per
day.
a Use R to represent the number of bottles of red wine produced each day and W to
represent the number of bottles of white wine produced each day. These
conditions are then expressed algebraically by:
Constraint 1 R + W ≤ 700
Constraint 2 10R + 7W ≤ 6400
Constraint 3 R ≤ 570
SAMPLE
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Review
Chapter 18 — Linear programming 509
The lines bordering regions of the R-W plane defined by these constraints are
shown and labelled on the graph. Identify the line representing each constraint
equation.
R0 200 400 600
200
400
600
800
1000
A BC
Y
W
800
b Find the coordinates of the point Y.
c Given that the profit per bottle is $8 for white wine and $10 for red wine, write
down an expression for $P, the total profit for the sales of the wine, in terms of
W and R.
d Find the maximum daily profit that can be earned by the Victory Vineyard from
selling their wine. [VCAA 2001]
SAMPLE
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