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mathematics

MX3021 apply linear programming methods in solving problems ncea level 3

2013/1

mathematics ncea level 3

Expected time to complete work This work will take you about eight hours to complete. You will work towards the following standard: Achievement Standard 91574 (Version 1) Mathematics and Statistics 3.2 Apply linear programming methods in solving problems Level 3, Internal 3 credits In this booklet you will focus on these learning outcomes: •• showing the solution to linear inequalities using graphs •• representing a feasible region for the solution to a given problem using graphs •• optimising a situation subject to the constraints of a problem. All the work for this standard is in this booklet.

Copyright © 2013 Board of Trustees of Te Aho o Te Kura Pounamu, Private Bag 39992, Wellington Mail Centre, Lower Hutt 5045, New Zealand. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means without the written permission of Te Aho o Te Kura Pounamu.

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contents 1

Simultaneous equations

2

Straight line graphs

3

4

Simultaneous linear inequalities and shading out

5

Using linear programming to solve unique answer problems

6

Using linear programming to solve non-unique answer problems

7

Review activity

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how to do the work When you see:

1A Complete the activity.

Your Mathematics and Statistics teacher will assess this work.

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simultaneous equations learning intention In this lesson you will:

• • revise your previous work on solving problems that have simultaneous linear equations.

introduction This lesson revises NCEA Level 1 and 2 content on solving simultaneous equations. In this lesson, and throughout the booklet, you will study systems of linear simultaneous equations that contain two unknowns. Two methods for solving simultaneous equations – elimination and substitution – will be revised in this lesson.

method 1: elimination

The steps are: 1. Multiply each equation by a number that will give the coefficients of either the x terms or the y terms the same absolute value but the opposite sign (you might need to multiply only one equation, or sometimes neither). 2. Add the two equations to eliminate either the x terms or the y terms. 3. Solve the resulting equation. 4. Substitute the value for the variable found in one of the original equations to find the other unknown. 5. Check your solution in the other equation. 6. Write out your solution. Example Solve by elimination: 3x + 8y = 37 (1) 2x + 7y = 31 (2) (Note: We usually number these equations for easy reference.) Answer We could eliminate either x or y, but this time we will eliminate x. There is no reason for this decision other than dealing with smaller numbers. To eliminate x, the coefficients of x need to have the same absolute value but the opposite sign. Multiplying equation (1) by 2 and equation (2) by –3 results in the coefficient of x in each equation having the same absolute value, but opposite signs. (1) × 2 6x + 16y = 74 (3) – – (2) × 3 6x − 21y = –93 (4) To eliminate x, we add (3) and (4): – (3) + (4) 5y = –19 y = 3.8

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simultaneous equations

To find x, substitute y = 3.8 in (1) or (2): (1) 3x + 8y 3x + 8 × 3.8 3x + 30.4 3x x

= = = = =

37 37 37 6.6 2.2

Because we found the value of x by substituting in (1), check (2) by using x = 2.2 and y = 3.8: (2) 2x + 7y = 2 × 2.2 + 7 × 3.8 = 4.4 + 26.6 = 31 This confirms that the solution is x = 2.2 and y = 3.8.

method 2: substitution

The steps are: 1. Rearrange one equation to obtain an expression for one of the unknowns. 2. Substitute this expression in the other equation. 3. Solve this equation − it will have only one variable. 4. Substitute the value for the variable found in either of the original equations or the rearranged equation to find the other unknown value. 5. Check your solution. 6. Write out your solution in context. Example Solve by substitution:

3x − y = 8 (1) 4x + 3y = 15 (2)

Answer We will rearrange equation (1) because it is the easier equation to write in the form y = ax + b. From (1): y = 3x − 8 Substitute y = 3x − 8 in (2): 4x + 3y = 4x + 3(3x − 8) = 4x + 9x − 24 = 13x = x =

15 15 15 39 3

Substitute x = 3 in (1): 3x − y = 8 3 × 3 − y = 8 y = 1

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simultaneous equations

Because we found the value of y by substituting in (1), check using (2) with x = 3 and y = 1. (2) 4x + 3y = 4 × 3 + 3 × 1 = 15 This confirms that the solution is x = 3 and y = 1. 1. Solve the following pairs of simultaneous equations by using your preferred method. a.

4x + 2y = 20 b. 5x + 6y = 39

c. x − y − 4 = 0 d. 2x + 3y – 33 = 0 e. 3x = 4y + 5 f. 6y = 2x − 8

8x − y = 10 3x + 5y = 36 7x − 4y = 13 5x + 2y = 22

0.5x + 3.5y = 19 1.5x − 2.5y = 5

solving problems using simultaneous linear equations One of the main uses of simultaneous equations is problem solving. Some people have an intuitive approach to problem solving, but we recommend a systematic approach.

The steps are: 1. Read through the problem carefully and record the important mathematical information. 2. Introduce two letters (variables) to represent the two unknowns and state clearly what each represents. 3. Form two linear equations from the information in the question – read the question again to check that you have used of all the information. 4. Solve the equations simultaneously by substitution or elimination. 5. Check your answer with the conditions stated in the problem. 6. Write a sentence answering the question. Example

Peter’s very generous grandpa has given him \$60 to spend on toy cars. He knows that a Kingway car costs \$4.50 and that a Noble car costs \$7.00. Peter would like to spend all of his money and buy 10 cars. How many of each type of car should he buy? 17049612

1A

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simultaneous equations

Total money = \$60 A Kingway costs \$4.50.

Total number of cars = 10 A Noble costs \$7.00.

2. Introduce two letters to represent the two unknowns. Let k be the number of Kingways that Peter buys. Let n be the number of Nobles that Peter buys. 3. Form two linear equations.

Amount spent on Kingways = \$4.50 × k. Amount spent on Nobles = \$7.00 × n. Total to spend on both types of cars = \$60. Therefore, 4.5k + 7n = 60. Total number of cars bought = 10. Therefore, k + n = 10.

4. Solve the equations. 4.5k + 7n = 60 (1) k + n = 10 (2) Rearrange (2) to make n the subject: n = 10 − k

Substitute in (1):

4.5k + 7(10 − k) = 4.5k + 70 − 7k = – 2.5k = k =

60 60 – 10 4

Substitute k = 4 in (2): k + n = 10 4 + n = 10 n = 6

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simultaneous equations

5. Check your answer. Substitute k = 4 and n = 6 in (1): 4.5k + 7n = 4.5 × 4 + 7 × 6 = 18 + 42 = 60 This is correct, so the solution is k = 4 and n = 6. 6. Write a sentence answering the question.

1B

To meet the conditions set out in the problem, Peter should buy 4 Kingway and 6 Noble cars.

1.

Littlemore Theatre runs a pantomime show every year. The theatre manager knows that last year they sold 260 tickets for a total of \$2 910 but she doesn’t know how many of the tickets were sold as adult tickets. Adults’ tickets sold for \$15 and children’s tickets sold for \$6. How many of each type of ticket were sold at last year’s pantomime?

2. Mr Brown is six times as old as his son. The difference in their ages is 40 years. How old are Mr Brown and his son? 3. Tama is a new postage clerk working at Dally Ltd. On his first day at work, Tama forgets to record how many of each type of letter the company sends out. However, he does know that the total cost of postage for sending the letters was \$54 and that the total cost of the envelopes was \$20.60. He also has a table showing the costs of the envelopes and postage. This table is as follows: Envelope size Small Medium

Cost of envelope \$0.20 \$0.50

Cost of postage \$0.60 \$1.20

How many of each size of letter were sent on Tama’s first day?

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straight line graphs learning intention In this lesson you will:

• • revise previous learning on drawing straight line graphs.

introduction This lesson revises NCEA Level 1 and 2 content on drawing straight line graphs. In this lesson, and throughout the booklet, you will be expected to draw only graphs that are linear in nature. In this lesson you will revise how to draw graphs of the type ax + by = c and y = mx + c.

drawing graphs of the type ax + by = c

The easiest way to draw a graph of this type is to identify where it crosses the x-axis and the y-axis. Then draw a line through these two points to create the graph. This is best shown by an example like the one below. Example Draw the graph of 3x + 2y = 12. Answer The steps are: Step 1: Find where the line crosses the x-axis. It crosses x-axis when y = 0, 3x + 2 × 0 = 12 x = 4 Step 2: Find where the line crosses the y-axis. It crosses y-axis when x = 0,

3 × 0 + 2y = 12 y = 6 8 y

Step 3: Plot the points on the axes. The points (4, 0) and (0, 6) are now plotted on the axes and a line drawn through and beyond them to the edge of the grid, as seen to the right.

7 6 5 4

3x + 2y = 12

3 2 1 3

2

x 0 1 0 1 2 3 4 5 6 7 – 1

2

8

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straight line graphs

drawing graphs of the type y = mx + c

The easiest way to draw a graph of this type is to identify the y-axis intercept (where x = 0) and the gradient, and then use these values to plot two points. From the equation of the line, y = mx + c, the y-intercept is given by the value of c and the gradient is given by the value of m. The y-intercept is first plotted on the axes, and then the gradient is applied from this point to identify the second point for plotting. Again, this is best shown by an example like the one below. Example Draw the graph of y = 3x – 4. Answer The steps are: Step 1: Identify the y-intercept. The equation means the same thing as y = 3x + (–4), and so in this case c = –4. Therefore, the y-intercept is at the point (0, –4). Step 2: Identify the gradient. From the equation, the gradient is 3. Therefore:rise 3 Gradient = = =3 run 1 This means that the second point is plotted 1 across and 3 up from the y-intercept. Therefore, the co-ordinates for the second point are (1, –1). Step 3: Plot the points on the axes. The points (0, –4) and (1, –1) are now plotted and a line drawn through and beyond them to the edge of the grid, as seen to the right.

4

y

3 y = 3x – 4

2 1 3

2

x 0 1 0 1 2 3 4 5 6 7 – 1

2

3

3 up

4

5 1 across

6

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straight line graphs

2A

1. Draw the following graphs on the grid provided below. 6 y

a. x + 3y = 6 b. 3x + 4y = 12 c. y = 4x + 2

5 4 3 2 1 –

6

5

4

3

2

x 0 1 0 1 2 3 4 5 6 – 1 –

2

3

4

5

6

2. Draw the following graphs on the grid provided below and identify all the points of intersection that can be seen within the grid. 6 y

a. 3y − 4x = 12 b. x = –3 c. y = –4 3 d. y= x−1 2

5 4 3 2 1 –

6

5

4

3

2

x 0 1 0 1 2 3 4 5 6 – 1 –

2

3

4

5

6

10

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linear inequalities and shading in learning intention

In this lesson you will learn to: • • show a graphical solution to a linear inequality by using the method of shading in.

introduction

So far in this booklet you have looked at equations and graphs that use the equal sign (=). In this lesson you will learn how to show inequalities which use the following signs.

> ≥ < ≤

greater than greater than or equal to less than less than or equal to.

linear inequalities with one variable

To solve an inequality, you use the same methods that you have applied to solving equations. However, if you need to multiply or divide by a negative number you need to reverse the inequality sign. Example 1 Solve the inequation 5x + 4 > 19. Answer 5x + 4 > 19 5x > 15 x > 3

(subtracting 4 from both sides of the inequation) (dividing both sides of the inequation by 5)

Therefore, the solution to the inequality is x > 3.

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Example 2 Solve the inequation 18 – 4x ≤ –10. Answer 18 – 4x ≤ –10 – 4x ≤ –28 (subtracting 18 from both sides of the inequation) x ≥ 7 (inequality sign is reversed because both sides are divided by –4) Therefore, the solution is x ≥ 7. Alternatively: 18 – 4x ≤ 18 ≤ 28 ≤ 7 ≤

10 10 + 4x 4x x – –

(adding 4x to both sides of the inequation) (adding 10 to both sides of the inequation) (dividing both sides of the inequation by 4)

Note that x ≥ 7 and 7 ≤ x are the same mathematical statements. Example 3 Solve the inequation 3(4 – x) < 18 and show the solution graphically. Answer 3(4 – x) < 12 – 3x < – 3x < x >

18 18 (multiplying out the brackets) 6 (subtracting 12 from both sides of the inequation) – 2 (inequality sign is reversed because both sides are divided by –3)

This solution of x > –2 can then be shown graphically by shading in the area we want to show as the solution. Note that the line x = –2 is dashed to show that it is not included as part of the solution. 4 y 3 2 1 4

3

2

x > –2

x 0 1 0 1 2 3 4 – 1

2

3

4

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(Note: The solution x > –2 is written inside the shaded region.) © te ah o o te k u ra p ou n a mu

Example 4 Solve 1 < 2x + 3 ≤ 8 and show the solution graphically. Answer 1 < 2x + 3 ≤ 8 means that the value of 2x + 3 is greater than 1 and the value of 2x + 3 is less than or equal to 8. The solution is found at the intersection of these two separate equations. 2x + 3 > 1 and 2x + 3 ≤ 8 2x > –2 2x ≤ 5 x > –1 x ≤ 2.5 This means that the solution of –1 < x ≤ 2.5 can be shown by shading in the area we want to show as the solution. Note that the line x = 2.5 is solid to show that it is part of the solution and that x = –1 is dashed because it is not part of the solution. 4 y 3 2 1 < x ≤ 2.5

1 –

4

3

2

x 0 1 0 1 2 3 4 – 1

2

3

4

3A

1. Solve the following inequalities and show the solution graphically by shading in. a. 2(x + 3) ≤ 5

b. 3 – 4x > 7

– c. 3 < 2x + 7 < 5

d.

2 – x 3 – 2x > 3 5

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linear inequalities with two variables

A linear inequality with two variables will probably use x and y although, in fact, any variables can be used. An example of a linear inequality with two variables could be 6x – 2y < 12. Because it can be difficult to interpret this as it currently stands, it is best to show it graphically. To graph an inequality: 1. Draw the related straight line: –– using a solid line for ≤ and ≥ –– using a dashed line for < and >. 2. Decide which side of the line to shade by testing a point in the inequality (do not choose a point on the line). The point (0, 0) is often used because it is an easy coordinate point to use for substitution. –– If you get a true statement when you substitute for x and y, then the point lies in the required region and you can shade that side of the line. –– If you get a false statement you can shade the other side of the line. Example 5 Graph the inequality 3x + 4y < 12. Answer 1. Draw the related straight line.

5 y 4

If x = 0 then y = 3, so plot the point (0, 3). If y = 0 then x = 4, so plot the point (4, 0).

Because the inequality is <, we can draw a dashed line through these points and to the edge of the grid.

3B

In this situation we can test the point (0, 0).

3 × 0 + 4 × 0 < 12 is true, so we can shade the side containing the point (0, 0).

1. Graph the following inequalities by shading in. a. 7x – 2y ≥ 14

b. y < 2x + 3

c. x – y – 4 ≤ 0

d.

2 1

5

2. Test a point that is not on the line.

3

3x + 4y < 12

4

3

2

x 0 1 0 1 2 3 4 5 – 1

2

3

4

5

x y + >1 2 3

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simultaneous linear inequalities and shading out learning intention

In this lesson you will learn to: • • show a graphical solution to simultaneous linear inequalities by using the method of shading out.

introduction

In the previous lesson, you dealt with fairly simple linear inequalities that could be shown graphically by the method of shading in. Shading in can be hard to interpret when it is used to solve simultaneous linear inequalities. In these situations the method of shading out is applied; that is, the region that is not required is shaded. The required region is left non-shaded, and the equations of the lines used to create the region are written on the diagram for more detail. For both methods (shading in and shading out), the boundary conventions are the same. A solid line is therefore still used for inequalities containing ≤ and ≥, and a dashed line is used for < and >. Example 1 Show graphically the solutions of the system: y ≤ 2x + 3 x + y > 1 Answer First you draw the graphs of y = 2x + 3 (use a solid line because the original inequality was ≤) and x + y = 1 (use a dashed line because the original inequality was >).

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simultaneous linear inequalities and shading out

5 y 4 x+y=1

y = 2x + 3

3 2 1

5

4

3

2

x 0 1 0 1 2 3 4 5 – 1

2

3

4

5

Next, test the point (0, 0) on both of the inequalities because you can see that it doesn’t lie on either of the lines. For y ≤ 2x + 3:

5 y

0 ≤ 2 × 0 + 3 is true, so for shading out we shade the region without (0, 0).

4 x+y=1

3

For x + y > 1:

2

0 + 0 > 1 is false, so for shading out we shade the region with (0, 0). The solution to the system of x + y > 1 and y ≤ 2x + 3 is shown here as the non-shaded region.

y = 2x + 3

1 0 –

5

4

3

2

x

1 0 1 2 3 4 5 – 1

2

3

4

5

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simultaneous linear inequalities and shading out

Example 2 Show graphically the solutions of the system: x + 3y < 3 x > –4 y > –2 x – y ≤ 2 Answer First you draw the graphs of x + 3y = 3, x = –4 and y = –2 (all dashed) and x – y = 2 (solid).

2

y

1 4

3

2

x–y=2

x 0 1 0 1 2 3 4 5 6 7 8 9 – 1

x + 3y = 3

2

x = –4

3

y = –2

4

5

6

7

8

Next, test the point (0, 0) for the two more complex graphs. For x + 3y < 3: 0 + 3 × 0 < 3 is true, so for shading out we shade the region without (0, 0). For x – y ≤ 2: 0 – 0 ≤ 2 is true, so for shading out we shade the region without (0, 0).

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simultaneous linear inequalities and shading out

2

y x–y=2

1 0 4

3

2

x = –4

x

1 0 1 2 3 4 5 6 7 8 9 – 1

2

3

4

5

6

7

x + 3y = 3 y = –2

The non-shaded region that is left after shading out, including the points on the solid line, make up the graphical solution to the given system. This region is called the feasible region.

4A

1. Show a graphical solution of the following systems by shading out. a. x ≥ 2 b. x < 2 x + 2y ≤ 4 y < x + 2 y > –x + 2 2. Identify the feasible region for the following systems by shading out. 1 a. y > – x + 2 b. 2<x≤5 2 y < x + 2 –1 < y ≤ 3 y < –2x + 8 Check your answers.

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using linear programming to solve unique answer problems learning intentions

In this lesson you will learn to: •• show the feasibility region on a graph that satisfies a set of linear inequalities •• maximise or minimise a required quantity •• solve real problems with a unique answer by applying the process of linear programming.

introduction It is sometimes possible to find a system of linear inequalities to describe the limiting factors when an optimum decision is required. For example, a business might want to know which is the least expensive of several courses of action. A market gardener might want to know where he should plant different crops to make maximum profit. A mathematical approach called linear programming is often the best way to solve such problems. Linear programming involves: •• describing the problem with two or more linear inequalities •• identifying the feasible region by graphing the inequalities and shading out •• finding the point on the graph that gives the maximum or minimum value for an expression (this is often called the optimum value) •• presenting a solution to the original problem. The solution to a linear programming problem relies on a theorem that will be studied and identified in two examples in this lesson.

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using linear programming to solve unique answer problems

general theorem of linear programming Example 1 Draw the feasible region for the system 0 ≤ x ≤ 3 and –1 ≤ y ≤ 1. Answer 3 y 2 1 x 0 1 0 1 2 3 4

1

2

The grid above show the feasible region (non-shaded area) for the system 0 ≤ x ≤ 3 and 1 ≤ y ≤ 1. The dots identify all the points in the feasible region that have integer coefficients.

What happens to the value of 3x + y for these points? The table below shows these values. Point (0, 1) (0, 0) (0, –1)

3x + y 1 0 – 1

Point (1, 1) (1, 0) (1, –1)

3x + y 4 3 2

Point (2, 1) (2, 0) (2, –1)

3x + y 7 6 5

Point (3, 1) (3, 0) (3, –1)

3x + y 10 9 8

If you look at the values of 3x + y, you can see that the maximum value is 10 and it occurs at (3, 1), which is one of the vertices of the feasible region. The minimum value, –1, occurs at the point (0, –1), which is also one of the vertices of the feasible region. What happens to the value of y – 2x for these points? Again, a table is used to show these values. Point (0, 1) (0, 0) (0, –1)

20

y – 2x 1 0 – 1

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Point (1, 1) (1, 0) (1, –1)

y – 2x – 1 – 2 – 3

Point (2, 1) (2, 0) (2, –1)

y – 2x – 3 – 4 – 5

Point (3, 1) (3, 0) (3, –1)

y – 2x – 5 – 6 – 7

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using linear programming to solve unique answer problems

Although the maximum and minimum values of y – 2x occur at different points to those of 3x + y, they are still vertices of the feasible region. If you were to try this with other linear expressions of the form ax + by, you would find that for all points in the feasible region it would have both maximum and minimum values at a vertex (corner) of the feasible region. Example 2 Draw the feasible region for the system y ≤ x + 1, y ≥ 3x – 3 and x + y ≥ 1. Answer 3 y 2 1

2

x 0 – 1 0 1 2 3 1

2

In this example, the grid shows the feasible region for the system y ≤ x + 1, y ≥ 3x – 3 and x + y ≥ 1. All of the points with integer coefficients within the feasible region have been identified with a dot. What happens to the value of 4x + 5y for these points? The table below shows these values. Point (0, 1) (1, 0) (1, 1) (1, 2) (2, 3)

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4x + 5y 5 4 9 14 23

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using linear programming to solve unique answer problems

If you look at these values of 4x + 5y, you can see that the maximum value is 23 and it occurs at (2, 3), which is one of the vertices of the feasible region. The minimum value, 4, occurs at the point (1, 0), which is also one of the vertices of the feasible region. What happens to the value of –2x – y for these points? The table now becomes: 2x – y – 1 – 2 – 3 – 4 – 7

Point (0, 1) (1, 0) (1, 1) (1, 2) (2, 3)

Again, both the maximum and minimum values occur at the vertices of the feasible region. The results of these two examples lead us to a general theorem of linear programming that states:

If a linear expression in two variables P = ax + by is evaluated for all points in a feasible region defined by a convex polygon, the maximum value of P will occur at a vertex of the region, and the minimum value of P will also occur at a vertex.

finding maximum and minimum values Example 1 The points A (1, 2), B (3, 6), C (7, 8) and D (10, 0) are the vertices of a feasible region which lies inside the polygon ABCD. Draw the polygon that represents the feasible region, identify the maximum and minimum values for P = 2x + 3y, and the points at which they occur. Answer

8 y

C

7 B

6 5 4 3 2

A

1 0 1

D x 0 1 2 3 4 5 6 7 8 9 10

2

22

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using linear programming to solve unique answer problems

For each of the vertices the value of P is as follows:

A (1, 2) C (7, 8)

P = 2 × 1 + 3 × 2 = 8 P = 2 × 7 + 3 × 8 = 38

B (3, 6) D (10, 0)

P = 2 × 3 + 3 × 6 = 24 P = 2 × 10 + 3 × 0 = 20

The maximum value for P is therefore 38, at point C, and the minimum value is 8, at point A. Example 2 Find the maximum and minimum values of P = 2x – 3y for the following system: x + y ≥ 5

2y ≤ x + 10

x ≤ 8

y≥x–5

B

8

6

x=8

2y = x + 10

7 A

5 4

x+y=5

3 y=x–5

2 1

C

D

0

x

0 1 2 3 4 5 6 7 8 9 10 By graphing each equation and shading out, you will discover that the vertices of the feasible region are A (0, 5), B (8, 9), C (8, 3) and D (5, 0). For each of the vertices, the value of P is as follows: A (0, 5) C (8, 3)

P = 2 × 0 − 3 × 5 = –15 P = 2 × 8 − 3 × 3 = 7

B (8, 9) D (5, 0)

P = 2 × 8 − 3 × 9 = –11 P = 2 × 5 − 3 × 0 = 10

The maximum value for P is therefore 10, at the point (5, 0), and the minimum value is –15, at the point (0, 5).

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using linear programming to solve unique answer problems

5A

1. The points A (3, 1), B (5, –2), C (1, –6), D (–2, –3) and E (0, 1) are vertices of a feasible region. Draw the polygon that represents the feasible region and identify the maximum and minimum values for P = 3x − y and the points at which they occur. 2. Find the maximum and minimum values of P = 5x + y for the following system: y ≤ x + 5

y ≥ –x − 3

y ≥ –1 y ≤ –2x

solving real problems

The techniques of linear programming can be applied to real situations and used to find optimal solutions. To solve a real problem, break the linear programming process into the following steps: 1. 2. 3. 4. 5. 6.

Choose letters (variables) for the unknown numbers and state clearly what each represents. Form inequalities from the conditions in the problem. Graph the inequalities by the method of shading out. Find an expression for the quantity that is to be optimised. Find the value of this expression at each vertex. Write a sentence to answer the problem.

Example A supermarket manager is preparing her weekly order of dog food. She stocks two brands – Mongrel Mash and Pedigree Pellets. Both come in bags of the same size, and she has storage space for 800 bags. From her knowledge of her customers, she expects to sell at least twice as many bags of Mongrel Mash as Pedigree Pellets. The buying price and profit are as follows. Buying Price Profit

Pedigree Pellets \$4.50 \$2.50

Mongrel Mash \$1.50 \$0.75

If the manager has up to \$1 500 to spend, how much profit can she expect to make as a maximum and how much of each should she order? Answer 1. Choose letters for the unknown numbers and state clearly what each represents. Let x be the number of bags of Pedigree Pellets ordered. Let y be the number of bags of Mongrel Mash ordered. 2. Form inequalities from the conditions in the problem. x ≥ 0 y ≥ 0 x + y ≤ 800 y ≥ 2x 4.5x + 1.5y ≤ 1 500 24

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(she can’t order negative numbers of bags) (she can’t order negative numbers of bags) (there is storage space for 800 bags) (she expects to sell at least twice as many bags of mash as pellets) (the cost of her order must not exceed \$1 500) © te ah o o te k u ra p ou n a mu

using linear programming to solve unique answer problems

3. Graph the inequalities by the method of shading out. y 1000 y = 2x 800 A B 600

400

C

x + y = 800

200

0

4.5x + 1.5y = 1 500

D

x 0 200 400 600 800 1000

4. Find an expression for the quantity that is to be optimised. The manager wants to maximise the profit where the profit (P) = 2.5x + 0.75y 5. Find the value of this expression at each vertex.

From the diagram we can see that A = (0, 800), C = (200, 400) and D = (0, 0). It is not entirely clear what the value of point B is so this will need to be calculated.

B is the intersection of x + y = 800 and 4.5x + 1.5y = 1 500, so these need to be solved simultaneously. x + y = 800 (1) 4.5x + 1.5y = 1 500 (2)

Multiply (1) by –1.5:

1.5x – 1.5y = –1 200 (3)

Eliminate y by adding (2) and (3): 3x = 300 x = 100 (this seems right when compared with the graph) Substitute x = 100 in (1): 100 + y = 800 y = 700 (this seems right when compared with the graph)

Therefore, point B must be (100, 700).

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using linear programming to solve unique answer problems

Point A (0, 800) B (100, 700) C (200, 400) D (0, 0)

P = 2.5x + 0.75y \$600 \$775 \$800 \$0

6. Write a sentence to answer the problem. 5B

The maximum profit of \$800 is made from an order of 200 bags of Pedigree Pellets and 400 bags of Mongrel Mash.

1. A local theatre can hold 450 people and tickets for the next show cost \$8 and \$12. The theatreâ&#x20AC;&#x2122;s policy is that it will have to sell at least 150 cheap seats for the show to go ahead and at least two cheap seats must be sold for every expensive seat. The total production costs are \$1 920. What is the maximum profit the theatre can expect? 2. A market gardener grows lettuces and cabbages. He has permanent orders of at least two boxes of lettuces and at least three boxes of cabbages each week. It costs \$6 to produce a box of lettuces and \$4 to produce a box of cabbages and he has, at most, \$72 per week to spend on this production. He cannot handle more than 15 boxes of these vegetables each week. If the profit is \$10 on each box of lettuces and \$14 on each box of cabbages, what is the maximum profit he can expect to make? 3. A company manufactures computers that are sold locally or exported. The company has contracts to provide at least four computers per month locally and at least six computers per month for export. It has the capacity to produce no more than 20 computers per month. The costs associated with producing each computer are \$3 000 for those sold locally and \$4 000 for those that are exported. A maximum of \$72 000 is available per month for production costs. If the profit on each computer sold locally is \$400 and the profit on each computer sold for export is \$500, find the most profitable combination for the company, and the maximum profit that can be obtained per month. 4. A company needs to transport 360 construction workers and 36 tonnes of equipment. They can use a mixture of large trucks (each of which carries up to 24 people and a tonne of equipment), and small trucks (each of which carry up to 12 people and two tonnes of equipment). What is the least number of trucks required?

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using linear programming to solve unique answer problems

5. A factory employs six trained and twelve untrained workers to make deluxe and standard model television sets. Producing each standard model uses 1 hour of trained and 4 hours of untrained labour. Each deluxe model requires 2 hours of trained and 2 hours of untrained labour. In each 8-hour period, the number of standard models produced cannot be more than twoand-a-half times the number of deluxe models produced. How many of each kind of model should be produced in each 8-hour period if the profits are \$250 for a deluxe model and \$100 for a standard model? Check your answers.

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6

using linear programming to solve non-unique answer problems learning intentions

In this lesson you will learn to: •• show the feasibility region on a graph that satisfies a set of linear inequalities •• maximise or minimise a required quantity •• s olve real problems with non-unique answers by applying the process of linear programming.

introduction

So far, all of the problems in the examples and activities have given a unique solution. In this lesson, you will see that it is sometimes possible to have multiple scenarios that give the same maximum (or minimum) value.

problems with multiple solutions Example 1 Jo, who suffers from a disease, needs a daily supplement of 4 milligrams of vitamin A, 11 milligrams of vitamin B and 100 milligrams of vitamin C. She can buy two brands of pills, Vitafix at 6 cents a pill or Myovit at 8 cents a pill. The following table shows the vitamin content of each pill in milligrams. Vitamin type A B C

Vitafix 2 3 25

Myovit 1 4 50

How many pills of each brand should she take each day in order to satisfy her needs most economically, and what is the most economical cost? Answer Identify the variables: Let x be the number of Vitafix. Let y be the number of Myovit. Inequalities are: x ≥ 0 and y ≥ 0 2x + y ≥ 4 3x + 4y ≥ 11 25x + 50y ≥ 100

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(cannot have negative number of pills) (needs 4 mg of vitamin A) (needs 11 mg of vitamin B) (needs 100 mg of vitamin C)

© te ah o o te k u ra p ou n a mu

using linear programming to solve non-unique answer problems

Graphing the situation: 5 y

4 P

3

2

2x + y = 4

Q 3x + 4y = 11

1 R

0

25x + 50y = 100 S 0 1 2 3 4 5

x

Expression for optimisation: Jo wants to minimise the cost, where the cost (C) is C = 6x + 8y. Values for the expression at each vertex: From the diagram we can see that P is (0, 4), Q is (1, 2) and S is (4, 0). The y value for R is not entirely clear, but the x value is 3. Substitute x = 3 in 25x + 50y = 100: 25 Ă&#x2014; 3 + 50y = 100 50y = 25 y = 0.5 Therefore, R must be (3, 0.5): Point P (0, 4) Q (1, 2) R (3, 0.5) S (4, 0)

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C = 6x + 8y 32 cents 22 cents 22 cents 24 cents

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using linear programming to solve non-unique answer problems

Conclusion: There is an equal minimum cost for two different scenarios. Jo can either take three Vitafix and half a Myovit, or one Vitafix and two Myovit. However, it might be difficult to split one of the Myovit pills into two halves. The practical solution is to recommend that Jo take one Vitafix and two Myovit pills. The cost for this recommendation is the minimum cost at 22 cents per day.

parallel expressions and inequalities

In Example 1, the points Q (1, 2) and R (3, 0.5) were both on the line 3x + 4y = 11. Also, the expression to be minimised was C = 6x + 8y and the minimum value for this was 22. What do you notice about 3x + 4y = 11 and 6x + 8y = 22? You should notice that the second expression is double the first. This means that any point on the line 3x + 4y = 11 between point Q and R on the graph gives rise to the same minimum value of 22 cents. In general, this means that if the optimised value occurs at two different vertices, then any point on the line that connects these two points may also give rise to a solution. Careful consideration must then be given to the context of the question to determine whether any further solutions are sensible or not. For the situation of the pills in Example 1, it is sensible to only consider whole numbers of pills, and so the answer given still stands. Example 2 A farmer has to decide how many hectares of carrots and peas to grow. He has the following information to help him make this decision: •• The farm consists of 90 hectares in total. •• The farmer is able to grow up to three times as many peas as carrots. •• The farmer has 112 tonnes of fertiliser to use. Carrots need 1.6 tonnes and peas need 0.7 tonnes of fertiliser per hectare. The farmer can expect \$4 800 profit per hectare of carrots and \$2 100 profit per hectare of peas. What area of peas and carrots should the farmer grow to maximise his profits? Identify the variables: Let x be the number of hectares of carrots. Let y be the number of hectares of peas.

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Inequalities are: x ≥ 0 y ≥ 0 x + y ≤ 90 y ≤ 3x 1.6x + 0.7y ≤ 112

(cannot have a negative area for growing) (cannot have a negative area for growing) (total farm area is 90 hectares) (can grow up to three times as many peas) (maximum amount of fertiliser is 112 tonnes)

Graphing the situation: 160 y 140 y = 3x

120 100 80 60

1.6x + 0.7y = 112 B

40 20

C x + y = 90

D x 0 A 0 20 40 60 80 100 120 140 Expression for optimisation: The farmer wants to maximise his profit, where the profit (P) is P = \$4 800x + \$2 100y. Values for the expression at each vertex: From the diagram we can see that A is (0, 0) and D is (70, 0). The values for B and C are not entirely clear, so these will need to be calculated by solving simultaneous equations.

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using linear programming to solve non-unique answer problems

For B: For C: Solve x + y = 90 and y = 3x: Solve x + y = 90 and 1.6x + 0.7y = 112: x + y = 90 (1) x + y = 90 y = 3x (2) 1.6x + 0.7y = 112

(1) (2)

Substitute (2) in (1): Multiply (1) by –1.6: x + 3x = 90 x = 22.5

1.6x – 1.6y = –144

(3)

Add (2) and (3): Substitute x = 22.5 in (2): –0.9y = –32 5 y = 3 × 22.5 y = 35 9 = 67.5 Therefore B is (22.5, 67.5).

Substitute in (1):

5 = 90 9 4 x = 54 9 x + 35

Therefore, C is (54

Point

P = 4 800x + 2 100y

A (0, 0)

\$0

B (22.5, 67.5)

\$249 750

C (54

4 5 , 35 ) 9 9

D (70, 0)

4 5 , 35 ) . 9 9

\$336 000 \$336 000

Conclusion: Because the maximum profit of \$336 000 occurs for the points C and D, we must also consider all of the points on the line between them. Because the amount of land in hectares doesn’t need to be whole numbers, the farmer has the option of any ratio of land split for carrots and peas that 4 satisfies the equation 1.6x + 0.7y = 112, where 54 ≤ x ≤ 70. 9

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using linear programming to solve non-unique answer problems

6A

1. A small town cinema can seat 80 people, and the tickets sell for \$9 for adults and \$6 for children. There cannot be more children than adults at any one showing of a movie. For a movie to be shown, the income from the ticket sales must be at least \$360. If the profit from an adult’s ticket is \$3 and the profit from a child’s ticket is \$2, what is the minimum profit the cinema can expect when a show goes ahead, and what combination of seats are needed to create this profit? 2. Grimsdale Glass has two different-sized vans for delivering sheets of glass from the factory to the warehouse. The two-tonne van can carry 20 large sheets and 80 small sheets, whereas the three-tonne van can carry 30 large sheets and 60 small sheets. During any given week the vans need to transport 1 200 large sheets and 2 400 small sheets to keep the warehouse stock levels up. The vans will only deliver full loads and together they can make no more than 50 trips in a week. The cost of fuel for each trip with the two-tonne van is \$5 and the cost of fuel for each trip with the three-tonne van is \$7.50. What is the minimum total fuel cost for a week of glass deliveries for Grimsdale Glass, and how many trips should each van make to meet this minimum total fuel cost? Check your answers.

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review activity

7

learning intentions In this lesson you will:

15117024

•• revise your work on solving problems by using linear programming.

introduction Make sure you show all the steps in your working for these problems. Read the questions carefully and check that your answers give the full solution to the problems. 7A

You will mark this section of the review activity yourself. 1. Solve the following simultaneous equations: a.

2x − 5y + 19 = 0 b. 4x = 5 + 3y 3x + 4y − 29 = 0 2y = 10 + x

2. Draw the following graphs on the grid provided below: a. 2x + 5y = 10 b. 3x − 5y = 15 c. y = –3x + 4

6 y 5 4 3 2 1 –

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6

5

4

3

2

x 0 1 0 1 2 3 4 5 6 – 1 –

2

3

4

5

6

© te ah o o te k u ra p ou n a mu

review activity

3. Solve the following linear inequalities and show a graphical solution:

a.

6 – 5x ≤ 12

b.

11 < 6x – 5 < 19

4. Identify the feasible region for the following systems: a. y ≤ 3x + 4 b. 2 < x ≤ 5 y ≥ x − 5 –4 < y ≤ x + 3 y < –4x + 7

5. Growquik and Beanstalk are two kinds of fertiliser. A bag of Growquik contains 2 kg of superphosphate and 1 kg of potash. A bag of Beanstalk contains 1 kg of superphosphate and 2 kg of potash. The school garden needs at least 8 kg of superphosphate and at least 10 kg of potash. Growquik costs \$6 a bag and Beanstalk costs \$9 a bag. Find the minimum cost of fertilising the garden and the number of bags of each kind of fertiliser that must be purchased. Check your answers.

7B

Your teacher will mark this part of the review activity. This is the type of question you can expect as an assessment task for Achievement Standard (91574) Apply linear programming methods in solving problems. Set out your working (including any graphs) clearly on paper, and also state your answer in a sentence.

practice assessment: the coffee shop

The manager of a new coffee shop, ‘Roasty Toasty’, needs to order two different types of coffee bean – Arabica and Robusta beans. Her experience has shown that she will need to order at least two bags of Arabica beans and six bags of Robusta beans to prevent running out. Her sales forecast predicts that she will use at least twice as many Robusta beans as Arabica. The table below shows the cost and profit, per bag, for each type of bean. Cost per bag Profit per bag

Arabica \$300 \$120

Robusta \$200 \$80

If the manager has a total budget of \$2 800 for purchasing coffee beans, what is the maximum profit that she can expect to make from the first order of coffee beans, and how many bags of each type will she need to order? Your Mathematics and Statisics teacher will assess this work.

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8

simultaneous equations

1. a.

4x + 2y = 20 (1) b. 8x − y = 10 (1) 5x + 6y = 39 (2) 3x + 5y = 36 (2)

Multiply (1) by –3: Multiply (1) by 5:

–12x – 6y = –60 (3) 40x − 5y = 50 (3) Add (2) and (3): Add (2) and (3): – 7x = –21 43x = 86 x = 3 x = 2

Substitute x = 3 in (1):

Substitute in (1):

4 × 3 + 2y = 20 8 × 2 − y = 10 2y = 8 y = 6 y = 4

The solution is x = 3 and y = 4.

c. x − y − 4 = 0 2x + 3y − 33 = 0

The solution is x = 2 and y = 6.

(1) d. (2)

7x − 4y = 13 (1) 5x + 2y = 22 (2)

Multiply (1) by 3: Multiply (2) by 2: – 3x − 3y − 12 = 0 (3) 10x + 4y = 44 (3) Add (2) and (3): Add (1) and (3): – 5x – 45 = 0 3x = 57 x = 9 x = –19

Substitute in (1): Substitute x = –19 in (1): 9 – y – 4 = 0 7 × –19 – 4y = 13 y = 5 y = –36.5

36

The solution is x = 9 and y = 5.

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The solution is x = –19 and y = –36.5.

© te ah o o te k u ra p ou n a mu

e. 3x = 4y + 5 (1) f. 6y = 2x – 8 (2)

(1) becomes (2) becomes

Multiply (3) by 2 and (4) by 3:

3x – 4y – 5 = 0 2x + 6y + 8 = 0

6x – 8y – 10 = 0 – 6x + 18y + 24 = 0

0.5x + 3.5y = 19 (1) 1.5x − 2.5y = 5 (2)

(3) (4) Multiply (1) by –3:

(5) (6)

1.5x – 10.5y = –57 (3)

Add (5) and (6): Add (2) and (3): 10y + 14 = 0 –13y = –52 7 y = – y = 4 5 7 Substitute y = – in (1) Substitute y = 4 in (1): 5 7 3x = 4 × – + 5 0.5x + 3.5 × 4 = 19 5 1 3x = – 0.5x = 5 5 7 x = – x = 10 5 1B

The solution is x = –

1 7 and y = – . The solution is x = 10 and y = 4. 5 5

1. Let x be the number of adults’ tickets sold and let y be the number of children’s tickets sold. x + y = 260 (total number of tickets sold is 260) 15x + 6y = 2 910 (total income from ticket sales is \$2 910) Solving x + y = 260 15x + 6y = 2 910

Multiply (1) by –6:

(1) (2)

6x – 6y = –1 560 (3)

9x = 1 350 x = 150 Substitute x = 150 in (1): 150 + y = 260 y = 110

Therefore, there were 150 adults’ and 110 children’s tickets sold.

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2. Let x be Mr Brown’s age and y be his son’s age. x = 6y (Mr Brown is six times as old as his son) x – y = 40 (the difference in their ages is 40) Solving x = 6y (1) x – y = 40 (2) Substitute x = 6y in (2): 6y – y = 40 y = 8 Substitute y = 8 in (1): x = 6 × 8 x = 48

Therefore, Mr Brown is 48 years old and his son is 8 years old.

3. Let x be the number of small letters sent and y be the number of medium-sized letters sent. 0.2x + 0.5y = 20.6 0.6x + 1.2y = 54

(the total cost for the envelopes) (the total cost for postage)

Solving

(1) (2)

Multiply (1) by –3:

0.2x + 0.5y = 20.6 0.6x + 1.2y = 54

0.6x – 1.5y = –61.8 (3)

–0.3y = –7.8 y = 26 Substitute y = 26 in (1): 0.2x + 0.5 × 26 = 20.6 0.2x = 7.6 x = 38

38

Therefore, Tama sends out 38 small letters and 26 medium-sized letters.

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2. 2A

straight line graphs

1. a. x + 3y = 6

6 y

If x = 0 then y = 2, so (0, 2). If y = 0 then x = 6, so (6, 0).

5 4

b. 3x + 4y = 12

3

If x = 0 then y = 3, so (0, 3). If y = 0 then x = 4, so (4, 0).

x + 3y = 6

2 1

c. y = 4x + 2 The y intercept is at (0, 2) The gradient is 4, so the next point is at (1, 6).

6

5

4

3

2

x 0 1 0 1 2 3 4 5 6 – 1 3x + 4y = 12 – 2

3

4

5

6

2. a. 3y – 4x = 12

6 y

If x = 0 then y = 4, so (0, 4). If y = 0 then x = –3, so (–3, 0).

5

b. x = –3 is a straight line up through (–3, 0).

3

6

5

A 4 –3

2

The y intercept is at (0, 1). 3 The gradient is so the 2 next point is at (2, 2).

Points of intersection are:

A = (–3, 0);

B = ( 6, 4);

C = (–3, –4); D = (–2, –4);

E = (–3, –5.5).

© te aho o t e k ur a p o un a m u

3 x–1 2

1

y=

2

3 x–1 2

3y – 4x = 12

4

x = –3

c. y = –4 is a straight line across through (0, –4). d. y=

y = 4x + 2

x 0 1 0 1 2 3 4 5 6 – 1

2

3

D

C B

y = –4

4

5

E

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3. 3A

1.

linear inequalities and shading in 2(x + 3) ≤ 5

4

2x + 6 ≤ 5 2x ≤

y

3

1 1 x ≤ – 2 –

2(x + 3) ≤ 5

2 1

4

3

2

x 0 1 0 1 2 3 4 – 1 2

3

4

2.

3 – 4x > 7

4 y

4x > 4

3

x < –1

3 – 4x > 7

2 1

5

4

3

2

0

1

1

0

1

2

x 3

2

3

4

3. 2x + 7 < 5

and

4

2x + 7 > –3

2x < –2 2x > –10

3

x < 1 x > 5 –

y

3 < 2x + 7 < 5

2 1

6

5

4

3

2

x 0 1 0 1 2 – 1

2

3

4

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2–x 3 – 2x > 3 5 5(2 – x) > 3(3 – 2x)

10 – 5x > 9 – 6x

4.

4 y 3 2 1

x > –1 –

3

2

2 – x 3 – 2x > 3 3

x 0 1 0 1 2 3 4 5 – 1 2

3

4

3B

1. Need to draw 7x – 2y = 14 with a solid line because the ≥ sign is used. If x = 0 then y = –7, so (0, –7). If y = 0 then x = 2, so (2, 0).

Check the point (0, 0).

7x – 2y ≥ 14 becomes 7 × 0 – 2 × 0 ≥ 14.

This is false, so shade the area not containing (0, 0).

2 y 7x – 2y ≥ 14

1

x 0 1 0 1 2 3 4 5 6 7 8 – 1

2

3

4

5

6

7

2. Need to draw y = 2x + 3 with a dashed line because the < sign is used.

5 4

The y intercept is 3, so it is at (0, 3).

The gradient is 2, so the next point is at (1, 5).

Check the point (0, 0).

y < 2x + 3 becomes 0 < 2 × 0 + 3.

This is true, so shade the area containing (0, 0).

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y

3 2 1 2

y < 2x + 3

x 0 1 0 1 2 3 4 5 – 1

2

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3. Need to draw x – y – 4 = 0 with a solid line because the ≤ sign is used. If x = 0 then y = 4, so (0, 4). If y = 0 then x = 4, so (4, 0). –

2 y 1

Check the point (0, 0).

x – y – 4 ≤ 0 becomes 0 – 0 – 4 ≤ 0.

This is true, so shade the area containing (0, 0).

2

x–y–4≤0

x 0 1 0 1 2 3 4 5 6 – 1

2

3

4

5

6

x y + = 1 with a dashed line 2 3 because the > sign is used.

4. Need to draw

4 y 3

If x = 0 then y = 3, so (0, 3). If y = 0 then x = 2, so (2, 0).

Check the point (0, 0): 0 0 x y + > 1 becomes + > 1. 2 3 2 3

4

3

2

x 0 1 0 1 2 3 4 – 1 –

2

3

4

simultaneous linear inequalities and shading out

1. Solutions to the systems are the non-shaded regions.

3 y 2

a. For x ≥ 2, draw x = 2 as a solid line vertically up through the point (2, 0). For x + 2y ≤ 4, draw x + 2y = 4 as a solid line.

1 –

3

2

x 0 1 0 1 2 3 4 5 6 – 1 2

Check the point (0, 0):

3

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x=2

4

x + 2y ≤ 4 becomes 0 + 2 × 0 ≤ 4. This is true, so shade the area not containing (0, 0).

x + 2y = 4

If x = 0 then y = 2, so (0, 2). If y = 0 then x = 4, so (4, 0).

42

x y + >1 2 3

1

This is false, so shade the area not containing (0, 0).

4. 4A

2

6

© te ah o o te k u ra p ou n amu

b. For x < 2, draw x = 2 as a dashed line vertically up through the point (2, 0).

For y < x + 2, draw y = x + 2 as a dashed line.

6 y

The y intercept is 2 so it is at (0, 2).

5

The gradient is 1 so the next point is at (1, 3).

Check the point (0, 0):

3

This is true, so shade the area not containing (0, 0).

x=2

2

y < x + 2 becomes 0 < 0 + 2.

y=x+2

4

1 2

For y > –x + 2, draw y = –x + 2 as a dashed line.

x 0 1 0 1 2 3 4 5 6 – 1 y = –x + 2 – 2

The y intercept is 2, so it is at (0, 2).

The gradient is –1 so the next point is at (1, 1).

Check the point (0, 0):

y > –x + 2 becomes 0 > –0 + 2.

This is false, so shade the area containing (0, 0).

2. Feasible regions for the systems are the non-shaded regions. a. For y < x + 2, draw y = x + 2 as a dashed line.

8 y

The y intercept is 2, so it is at (0, 2).

7

The gradient is 1, so the next point is at (1, 3).

6 5

y = –1 x + 2 2

Check the point (0, 0):

3

This is true, so shade the area not containing (0, 0).

For y < –2x + 8, draw y = –2x + 8 as a dashed line.

y = –2x + 8

2

y < x + 2 becomes 0 < 0 + 2.

y=x+2

4

1 –

4 –3 –2

x 0 1 0 1 2 3 4 5 6 – 1 –

2

The y intercept is 8, so it is at (0, 8). © te aho o t e k ur a p o un a m u

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The gradient is –2, so the next point is at (1, 6).

Check the point (0, 0):

y < –2x + 8 becomes 0 < –2 × 0 + 8.

This is true, so shade the area not containing (0, 0).

1 1 For y > – x + 2, draw y = – x + 2 as a dashed line. 2 2 The y intercept is 2, so it is at (0, 2).

1 The gradient is – , so the next point is at (2, 1). 2

Check the point (0, 0):

1 1 y > – x + 2 becomes 0 > – × 0 + 2. 2 2

This is false, so shade the area containing (0, 0).

b. For 2 < x ≤ 5, draw x = 2 as a dashed vertical line through (2, 0) and x = 5 as a solid vertical line through (5, 0). For –1 < y ≤ 3, draw y = –1 as a dashed horizontal line through (0, –1) and y = 3 as a solid horizontal line through (0, 3).

6 y 5 4

x=2

x=5

3 2

y=3

1 x 0 1 0 1 2 3 4 5 6 7 – 1 y = –1 – 2

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5. 5A

using linear programming to solve unique answer problems

1. For each of the vertices, the value of P is:

A (3, 1) B (5, –2) C (1, –6) D (–2, –3) E (0, 1)

Therefore, the maximum value for P is 17 at point B, and the minimum value is –3 at point D.

2 y

P = 3 × 3 – 1 = 8 P = 3 × 5 – (–2) = 17 P = 3 × 1 – (–6) = 9 P = 3 × –2 – (–3) = –3 P = 3 × 0 – 1 = –1

1 E 2

A

x 0 1 0 1 2 3 4 5 6 – 1 B

2

D

3

4

5

6

C

2. For y ≥ –1, draw y = –1 as a solid horizontal line through (0, –1).

5 y = –2x

For y ≤ x + 5, draw y = x + 5 as a solid line. The y intercept is 5, so it is at point (0, 5).

The gradient is 1, so the next point is at (1, 6).

Check the point (0, 0):

y ≤ x + 5 becomes 0 ≤ 0 + 5.

This is true, so shade the area not containing (0, 0).

y

4 B

y = –x – 3

3 2 1

A –

7

6

5

4

3

2

D y=x+5

x 0 1 0 1 2 3 C – 1 y = –1 – 2

3

For y ≥ –x – 3, draw y = –x – 3 as a solid line.

4

The y intercept is –3, so it is at (0, –3).

The gradient is –1, so the next point is at (1, –4).

Check the point (0, 0):

y ≥ –x – 3 becomes 0 ≥ –0 – 3.

This is true, so shade the area not containing (0, 0).

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For y ≤ –2x, draw y = –2x as a solid line. The y intercept is 0, so it is at (0, 0).

The gradient is –2 so the next point is at (1, –2).

Check the point (1, 0) because the point (0, 0) is on the line:

y ≤ –2x becomes 0 ≤ –2 × 1.

This is false, so shade the area containing (1, 0).

The easily identifiable vertices of the feasible region are A (–4, 1), C (0.5, –1) and D (–2, –1).

To find the co-ordinates of B, you need to solve y = x + 5 and y = –2x:

y = x + 5 (1) y = –2x (2) Substitute y = –2x in (1) 2x = x + 5

3x = 5 x = – 5 3

Substitute x = – 5 in (2) 3

y = –2 × – 5 3 10 y = 3 –

5 10 , ). 3 3

Therefore, point B = (

For each of the vertices the value of P is:

A (–4, 1)

C (0.5, –1) P = 5 × 0.5 + (–1) = 1.5

Therefore the minimum value for P is –19 at the point A (–4, 1), and the maximum value is 1.5 at the point C (0.5, –1).

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P = 5 × –4 + 1 = –19 B (

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5 10 , ) 3 3

D (–2, –1)

P = 5×

5 10 – + = 5 3 3

P = 5 × –2 + (–1) = –11

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5B

1. Let x be the number of cheap seats and y be the number of expensive seats.

Inequalities are:

x ≥ 0 y ≥ 0 x ≥ 150 x + y ≤ 450 2y ≤ x

(can’t have negative tickets sold) (can’t have negative tickets sold) (must sell at least 150 cheap seats) (theatre can hold 450 people) (must sell at least two cheap seats for every expensive seat)

Graphing:

500 y x = 150

400 300

2y = x

200 C

100 0

B

x + y = 450

D x A 0 100 200 300 400 500

Quantity for optimisation:

The profit is to be maximised where the profit (P) = revenue – production costs = 8x + 12y – 1 920

Value of expression at each vertex:

A = (150, 0);

B = (150, 75);

Point A (150, 0) B (150, 75) C (300, 150) D (450, 0)

C = (300, 150);

and D = (450, 0).

P = 8x + 12y − 1 920 – \$720 \$180 \$2 280 \$1 680

The maximum profit that the theatre can expect is \$2 280. To achieve this, they need to sell 300 cheap seats and 150 expensive seats.

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2. Let x be the number of boxes of lettuces sold and y be the number of boxes of cabbages sold. Inequalities are: x ≥ 2 y ≥ 3 6x + 4y ≤ 72 x + y ≤ 15 Graphing:

(due to permanent orders) (due to permanent orders) (production costs cannot be higher than \$72) (cannot handle more than 15 boxes of vegetables) 16 y 14

B

12

6x + 4y = 72

10

C

8 x=2

6 4

x + y = 15

A

D

y=3

2 0

x 0 2 4 6 8 10 12 14 16

Quantity for optimisation:

The profit (P), in dollars, is given by P = 10x + 14y.

Value of expression at each vertex:

A = (2, 3);

B = (2, 13); Point A (2, 3) B (2, 13) C (6, 9) D (10, 3)

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C = (6, 9);

and D = (10, 3). P = 10x + 14y \$62 \$202 \$186 \$142

The maximum profit that the gardener can expect is \$202. To achieve this, he needs to sell 2 boxes of lettuces and 13 boxes of cabbages.

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3. Let x be the number of computers sold locally and y be the number sold for export. Inequalities are: x ≥ 4 y ≥ 6 3x + 4y ≤ 72 x + y ≤ 20 Graphing:

(due to contracts) (due to contracts) (simplified version of 3 000x + 4 000y ≤ 7 200 for production costs) (cannot produce more than 20 computers per month)

16 y

x + y= 20 B

14 12

C

10

3x + 4y= 72

8

x=4

6

y=6

A

4

D

2 0

x 0 2 4 6 8 10 12 14 16

Quantity for optimisation: The profit (P), in dollars, is given by P = 400x + 500y. Value of expression at each vertex: A = (4, 6):

B = (4, 15): Point A (4, 6) B (4, 15) C (8, 12) D (14, 6)

C = (8, 12):

and D = (14, 6).

P = 400x + 500y \$4 600 \$9 100 \$9 200 \$8 600

The maximum profit that the company can expect is \$9 200. To achieve this, they need to sell 8 computers locally and 12 computers for export.

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4. Let x be the number of large trucks used and y be the number of small trucks used.

Inequalities are: x ≥ 0 y ≥ 0 24x + 12y ≥ 360 x + 2y ≥ 36 Graphing:

(cannot have negative number of large trucks) (cannot have negative number of small trucks) (360 workers who must be transported) (36 tonnes of equipment that must be transported)

35 y 30 A 25

24x + 12y = 360

20 15

B

10 x + 2y = 36

5 0

Cx 0 5 10 15 20 25 30 35

Quantity for optimisation: The number of trucks required is given by x + y. Value of expression at each vertex: From the diagram we can see that A = (0, 30) and C = (36, 0) are easily identifiable, but point B will need to be evaluated. To find B: Solve:

24x + 12y = 360

(1)

x + 2y = 36

(2)

Multiply (2) by –6: –6x – 12y = –216

(3)

Add (1) and (3): 18x = 144 x = 8 50

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Substitute x = 8 in (2): 8 + 2y = 36 y = 14 Therefore, point B = (8, 14). P=x+y 30 22 36

Point A (0, 30) B (8, 14) C (36, 0)

The minimum number of trucks to be used is 22. This can be achieved by using 8 large trucks and 14 small trucks. 5. Let x be the number of deluxe televisions produced and y be the number of standard televisions produced. Inequalities are: x ≥ 0 y ≥ 0 x + 2y ≤ 48 4x + 2y ≤ 96 x ≤ 2.5y Graphing:

(cannot produce negative number of televisions) (cannot produce negative number of televisions) (trained staff hours available is a maximum of 6 × 8 = 48) (untrained staff hours available is a maximum of 12 × 8 = 96) (number of standard sets is no more than 2.5 times the number of deluxe sets) 50 y

40

4x + 2y = 96

30 B 20

x = 2.5y

C 10

D

x + 2y = 48

0 A

x 0 10 20 30 40 50

Quantity for optimisation: The profit (P), in dollars, is given by P = 100x + 250y. © te aho o t e k ur a p o un a m u

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Value of expression at each vertex: From the diagram we can see that A = (0, 0) and B = (0, 24) are easily identifiable, but points C and D will need to be evaluated. To find C: To find D: Solve:

4x + 2y = 96 (1) Solve: x = 2.5y x + 2y = 48 (2) 4x + 2y = 96

(1) (2)

Subtract (2) from (1): Substitute x = 2.5y in (2): 3x = 48 4 × 2.5y + 2y = 96 x = 16 y = 8. Substitute x = 16 in (2): Substitute y = 8 in (1): 16 + 2y = 48 x = 2.5 × 8 y = 16 x = 20 Therefore, point C = (16, 16). Point A (0, 0) B (0, 24) C (16, 16) D (20, 8)

Therefore, point D = (20, 8).

P = 100x + 250y \$0 \$6 000 \$5 600 \$4 000

The maximum profit that the company can expect is \$6 000. To achieve this, they would need to produce zero standard sets and 24 deluxe sets.

6. 6A

using linear programming to solve non-unique answer problems

1. Let x be the number of adults’ tickets sold and y be the number of children’s tickets sold. x ≥ 0 y ≥ 0 x + y ≤ 80 x ≥ y 9x + 6y ≥ 360

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(number of tickets sold cannot be negative) (number of tickets sold cannot be negative) (theatre holds 80 people) (cannot be more children than adults) (income needs to be at least \$360 for show to go ahead)

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Graphing:

80

y

70 60 y=x

50 40

C

9x + 6y = 360

30 x + y = 80

B

20 10

A

0

Dx

0 10 20 30 40 50 60 70 80

Quantity for optimisation: The profit (P), in dollars, is given by P = 3x + 2y. Value of expression at each vertex: From the diagram we can see that A = (40, 0), C = (40, 40) and D = (80, 0) are easily identifiable, but the point B will need to be evaluated. To find B: Solve y = x

(1)

9x + 6y = 360 (2)

Substitute y = x in (2): 9x + 6x = 360 x = 24 Substitute x = 24 in (1): y = 24 Therefore, the point B = (24, 24).

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Point A (40, 0) B (24, 24) C (40, 40) D (80, 0)

P = 3x + 2y \$120 \$120 \$200 \$240

The minimum profit that the theatre can expect if a show goes ahead is \$120. This profit is achieved at points A (40, 0) and B (24, 24), which implies that any point on the line connecting the two will also create the minimum profit. As the data is discrete (must be whole numbers of people), then any point with integer values on the line 9x + 6y = 360 for 24 ≤ x ≤ 40 satisfies the minimum profit. 2. Let x be the number of two-tonne vans and y be the number of three-tonne vans. x ≥ 0 y ≥ 0 20x + 30y ≥ 1 200 80x + 60y ≥ 2 400 x + y ≤ 50 Graphing:

60

(cannot have a negative number of vans) (cannot have a negative number of vans) (vans must deliver 1 200 large sheets) (vans must deliver 2 400 small sheets) (no more than 50 deliveries can be made during a week)

y

50 B 40 A

x + y = 50

30 20

C

20x + 30y = 1 200

10 80x + 60y = 2 400 x

0 0 10 20 30 40 50 60

Quantity for optimisation: The total delivery cost P (in dollars) is given by P = 5x + 7.5y. Value of expression at each vertex: From the diagram we can see that the points are A = (0, 40), B = (0, 50) and C = (30, 20).

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Point A (0, 40) B (0, 50) C (30, 20)

P = 5x + 7.5y \$300 \$375 \$300

The minimum delivery cost for Grimsdale Glass is \$300. This delivery cost is achieved at points A (0, 40) and C (30, 20), which implies that any point on the line connecting the two will also create the minimum delivery cost. As the data is discrete (must be whole numbers of deliveries), then any point with integer values on the line 20x + 30y = 1 200 for 0 ≤ x ≤ 30 satisfies the minimum delivery costs.

7. 7A

review activity

1. a.

2x − 5y +19 = 0 3x + 4y − 29 = 0

(1) b. (2)

Multiply (1) by 4 and (2) by 5:

4x = 5 + 3y (1) 2y = 10 + x (2)

Substitute x = 2y – 10 in (1):

8x – 20y + 76 = 0 (3) 4(2y – 10) 15x + 20y – 145 = 0 (4) 8y – 40 5y Add (3) and (4): y

= = = =

5 + 3y 5 + 3y 45 9

23x – 69 = 0 x = 3 Substitute x = 3 in (1): Substitute y = 9 in (2): 2 × 3 – 5y + 19 = 0 2 × 9 = 10 + x –5y = –25 x = 8 y = 5

The solution is x = 3 and y = 5.

2.

The solution is x = 8 and y = 9.

6 y 5 2x + 5y = 10

4 3 2 1

6

5

4

3

2

x 0 1 0 1 2 3 4 5 6 – 1

2

3

3x – 5y = 15

4

y = –3x + 4

5

6

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3. a. 6 – 5x ≤ 12 – 5x ≤ 6 x ≥ –1.2

4

y

3 2

6 – 5x ≤ 2

1 4

3

2

x 0 1 0 1 2 3 4 – 1

2

3

4

– b. 11 < 6x − 5 < 19

4

so –11 < 6x – 5 and 6x – 5 < 19 –6 < 6x 6x < 24 –1 < x x < 4 x > –1

3

1 4

3

2

4. a. b. 10

0 x 5 0 5 10

5

56

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10

2

3

4

5

y=x–5

10

x 0 1 0 1 2 3 4 – 1

10 y

y = 3x + 4

5

11 < 6x – 5 < 19

2

y

y

x=5

y=x+3 –

10

x 0 5 0 5 10 x=2

y = –4x + 7

y = –4

5

10

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5. Let x be the number of bags of Growquik and y be the number of bags of Beanstalk. Inequalities are: x ≥ 0 y ≥ 0 2x + y ≥ 8 x + 2y ≥ 10

(cannot have negative number of bags) (cannot have negative number of bags) (needs at least 8kg of superphosphate) (needs at least 10kg of potash)

Graphing: 10

y

8 D 6

2x + y = 8

E

4

x + 2y = 10

2 0

F x 0 2 4 6 8 10

Quantity for optimisation: The cost (C), in dollars, is given by C = 6x + 9y. Value of expression at each vertex: D = (0, 8);

E = (2, 4); Point D (0, 8) E (2, 4) F (10, 0)

and F = (10, 0) P = 6x + 9y \$72 \$48 \$60

The minimum cost for fertilising is \$48. To achieve this result, the school needs to buy two bags of Growquick and four bags of Beanstalk.

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acknowledgements Every effort has been made to acknowledge and contact copyright holders. Te Aho o Te Kura Pounamu apologies for any omissions and welcomes more accurate information. Graph diagrams created using GeoGebra program from www.geogebra.org/cms. Used in any medium for education and its promotion by permission. Photos: istock: Cover: Grid – 3376474 Auto accident – 17049612 Magnifying object – 15117024

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Show the solution to linear inequalities using graphs. Represent a feasible region for the solution to a given problem using graphs.

Optimise a situation subject to the constraints of a problem.

Student comment Show the solution to linear inequalities using graphs.

Represent a feasible region for the solution to a given problem using graphs.

Optimise a situation subject to the constraints of a problem.

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Maths NCEA Level 3