Add Maths Project 2006 by Chin Wynn

8/14/2019 Add Maths Project 2006 by Chin Wynn

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SEKOLAH MENENGAH KEBANGSAANBUKIT KEMUNING

40460 SHAH ALAM

ADDITIONAL MATHEMATICS PROJECT

ALUMINUM TIN

Students Name : CHIN WYNN

Form : 4 SCIENCE 1I.C. No. : 910715-14-5397Teachers Name : MADAM ONG LIN LIN


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CHIN WYNN

CONTENTS

NO. TITLE PAGE1. Acknowledgement 2

2. Introduction 3

3. Conjecture 6

4. Discussion 6

5. Identifying Information 8

6. Strategy 9

7. Results 41

8. Conclusion 42

9. Appendix 44

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CHIN WYNN

ACKNOWLEDGEMENT

Firstly, I would like to thank our AdditionalMathematics Teacher, Madam Ong Lin Lin for guiding usthroughout this project. She explained and showed us everycontents of this project clearly.

Next, I would like to thank my friends for givingassistance and advice about this project. Besides that, they

also gave me some mental support to doing this project.

Last but not least, I appreciate that my parents fullybelieved and supported me. They sacrificed their time tosend me to my friends house in order to complete thisproject. They also contributed money for me to carry outthis assignment.

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CHIN WYNN

INTRODUCTION

I am doing a project work to calculate on how to reduce the production

cost of a cylindrical aluminum can company when its production cost isproportional to the area of the aluminum sheet used.

History of the Aluminum Can

The earliest kind of metal beverage can was made out of steel (similar to a tincan) and had no pull tab. Instead, it was opened by using a tool called a can

opener or bottle opener (colloquially, a church key). The opener resembled abottle opener but had a sharp point. The can was opened by punching two holesin the lid, a large one for drinking through, and a smaller one that allowed air into replace the displaced fluid. Further advancements saw the end pieces of thecan made out of aluminum instead of steel.

In North America, the standard can size is 12 fluid ounces (335 ml). In most ofEurope, standard cans are 330 ml, which is approximately 1/3 of a liter. InAustralia, the standard can size is 375 ml. In South Africa, standard cans are

340 ml. In India, the standard can size is 300 ml.

An empty aluminum can weigh approximately 15 grams, or 0.5 ounce. Therefore,there are roughly 30 empty aluminum cans to an avoirdupois pound.

The Benefits of Aluminum Can Recycling

Environmental BenefitsRecycling aluminum cans save precious natural resources, energy, time andmoney - all for a good cause - helping out the earth, as well as the economy andlocal communities.

Aluminum cans are unique in that in 60 days a can is recycled, turned into a newcan & back on store shelves. Aluminum is a sustainable metal and can be recycledover and over again. In 2003, 54 billion cans were recycled, saving the energyequivalent of 15 million barrels of crude oil - America's entire gas consumptionfor one day.

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CHIN WYNN

Economic Benefits

The aluminum can is the most valuable container to recycle and is the mostrecycled consumer product in the U.S. today. Each year, the aluminum industry

pays out over $800 million dollars for empty aluminum cans - that's a lot ofmoney that can go to organizations, like Habitat for Humanity, the Boy or GirlScouts of America, or even a local school. Money earned from recycling canshelps people help themselves and their communities. Recycling helps build newhomes, pays for a group trip, supports a project or buys a lunch! Today it ischeaper, faster and more energy-efficient to recycle aluminum than everbefore. The aluminum can is 100 percent recyclable and can be recycledindefinitely. The can remains the most recyclable of all materials. Usedaluminum beverage cans are the most recycled item in the U.S., but other types

of aluminum, such as siding, gutters, car components, storm window frames, andlawn furniture can also be recycled. Aluminum has a high market value andcontinues to provide an economic incentive to recycle. When aluminum cans arerecycled curbside, they help pay for community services.

Community Benefits

Aluminum can recycling enables charitable organizations and groups to earnfunds to further local projects. The money earned enhances programs,communities and improves the quality of people's lives. From a local can drive toraise money for school improvements, to a Boy or Girl Scout troop "Cans IntoCash" competition to pay for camp, recycling is used all over the country to helpothers.

A perfect example of this is the Cans for Habitatprogram. Through a nationalpartnership between the Aluminum Association and Habitat for HumanityInternational, aluminum cans are recycled via a network of drop-off locations toraise money for Habitat for Humanity to build decent, affordable housing withlow-income families. To think, just by recycling a can once destined for thelandfill, you are keeping our local environment clean, providing a needed resourcefor the aluminum recycling process, and most importantly, helping provide localhousing to those in need. It's a win-win for the individual, community, business,industry and the environment.

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CHIN WYNN

ADDITIONAL MATHEMATICS PROJECT WORK

FORM 4 YEAR 2006

PACKAGE: SCIENCE & TECHNOLOGY

Aluminum Can

Problem Being InvestigatedThe Muhibbah Company is a manufacturer of cylindrical aluminum tins. The managerplans to reduce the cost of production. The production cost is proportional to thearea of the aluminum sheet used. The volume that each tin can hold is 1000 cm3 (1liter).

1) Determine the value of h, rand hence calculate the ratio h/r when the totalsurface area of each tin is minimum. Here, hcm denotes the height and rcm theradius of the tin.

2) The top and bottom pieces of the tin of height hcm are cut from square-shapedaluminum sheets.Determine the value for r, hand hence calculate the ratio h/ r so that the totalarea of the aluminum sheets used for making the tin is minimum. (Refer to thediagram below.)

rcm hcm rcm

Top Surface Curved Surface Bottom Surface

3) Investigate cases where the top and bottom surfaces are cut from(i) Equilateral triangle.(ii) Regular hexagon.

Find the ratio of h/r for each case.

Further InvestigationInvestigate cases where the top and bottom faces of the tin are being cut fromaluminum sheets consisting shapes of polygons. From the results of yourinvestigation, what conclusion can you derive from the relationship of the ratio ofh/r and the number of sides of a regular polygon?

Wastage occurs when circles are cut from aluminum sheet, which is not round in

shape. Suggest the best possible shape of aluminum sheets to be used so as toreduce the production cost.

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CHIN WYNN

CONJECTURE

Polygons with more sides have smaller ratio of h/r and smaller surface area.

DISCUSSION

By using mathematical facts, formulae and methods such as differentiation,trigonometry and trial and improvement, the minimum areas of aluminum differaccording to the shape cut from it.

Below are some mathematical facts and formulae used for determining theanswers for each question. Furthermore, diagrams, tables and graphs are alsoinserted to help presenting the data and information. Also, all answers in thisfolio are in at least 4 significant figures.

Mathematical Facts and Formulae that are Used

a) Volume of the tin = 1000cm3

r2h = 1000h = 1000

r2

b) Total surface area of tin,A = 2(Area of top surface) + Area of curved surface

= 2r2 + 2rh= 2r2 + 2r(1000/r2)= 2r2 + 2000/r

c) When A is minimum, dA = 0dr

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CHIN WYNN

d)r (cm) A = 2r2 + 2000/r (cm2)

e)

The Graph of A (cm2) against r (cm)

Check the answer from the graph by getting the minimum point.

8

r (cm)

A(

cm2)


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CHIN WYNN

IDENTIFYING INFORMATION

To find the minimum area of the cylinder tin The production cost is proportional to the surface area of the aluminum

sheets used The volume of each tin is equivalent to 1000cm3 or 1 liter

Question 1

Determine the values of h, r and hence calculate the ratio of h/r when the totalsurface area of each tin is minimum.

Question 2

The top and bottom pieces of the tin are cut from square-shaped aluminumsheets. Determine the values of h, r and hence calculate the ratio of h/r whenthe total surface area of each tin is minimum.

Question 3

The top and bottom surfaces are cut from equilateral triangle and regularhexagon. Determine the ratio of h/r for each case.

Further Investigation

Aluminum sheets are cut from regular polygons. Based on the results of theinvestigation, derive a conclusion from the relationship of the ratio of h/r andthe number of sides of the regular polygon. Suggest the best shape of aluminumsheets to be used so as to reduce the production cost.

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CHIN WYNN

STRATEGY

1) Determine the value of h, r and hence calculate the ratio h/r when the total

surface area of each tin is minimum. Here, h cm denotes the height and r cmthe radius of the tin.

Volume of the tin = 1000cm3

r2h = 1000

h = 1000r2

r cm h cm r cm


Total surface area of tin, A = 2r

2

+ 2r(1000/r

2

)= 2r2 + 2000r

Method 1: Trial and Improvement

r (cm) A = 2r2 + 2000/r (cm2)

1 2006.282 1025.133 723.224 600.535 557.086 559.537 593.598 652.129 731.16

From the table above, A is minimum when r = 5 cm.The Graph of A (cm2) against r (cm)

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2006.28

1025.13

723.22

600.53

557.08

559.53

593.59

652.12

731.16

0

50 0

1000

1500

2000

2500

1 2 3 4 5 6 7 8 9

r (cm)

A

(cm2)

CHIN WYNN

A More Comprehensive Tabler (cm) A = 2r2 + 2000/r (cm2)

5.0 557.085.1 555.585.2 554.515.3 553.855.4 553.595.5 553.705.6 554.185.7 555.015.8 556.19

From the table above, A is minimum when r = 5.4 cm.


11

557.08

555.58

554.51

553.85

553.59

553.70

554.18

555.01

556.19

551

552

553

554

555

556

557

558

5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

r (cm)

A

(cm2)


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CHIN WYNN

A More Comprehensive Tabler (cm) A = 2r2 + 2000/r (cm2)5.40 553.5885.41 553.583

5.42 553.5815.43 553.5835.44 553.5895.45 553.5995.46 553.6125.47 553.6295.48 553.650



The Most Comprehensive Tabler (cm) A = 2r2 + 2000/r (cm2)

5.418 553.5810755.419 553.581046

5.420 553.5810555.421 553.5811025.422 553.5811865.423 553.581308

5.424 553.5814685.425 553.5816655.426 553.581900


12

553.588

553.583

553.581

553.583

553.589

553.599

553.612

553.629

553.650

553.54

553.56

553.58

553.60

553.62

553.64

553.66

5 .4 0 5 .4 1 5 .4 2 5 . 43 5 .4 4 5 .4 5 5 .4 6 5 .4 7 5 .4 8

r (cm)

A

(cm2)


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CHIN WYNN


r = 5.419 cm

h = 1000r2

= 1000(5.419)2

= 10.84 cm

h = 10.84r 5.419= 2.000

13

553.581075

553.581046

553.581055

553.581102

553.581186

553.581308

553.58146

8

553.58

1665

553

.581900

553.5806

553.5808

553.5810

553.5812

553.5814

553.5816

553.5818

553.5820

5 .4 18 5 .4 19 5 .4 20 5 .4 21 5 .4 22 5 .4 23 5 .4 24 5 .4 2 5 5 .4 26

r (cm)

A

(cm2)


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CHIN WYNN

Method 2: DifferentiationVolume = r2h

1000 = r2h

h = 1000r2

Total surface area of tin, A = 2r2 + 2r (1000/r2)= 2r2 + 2000

r

dA = 4r 2000dr r2

When A is minimum,dA = 0dr

4r 2000 = 0r2

4r = 2000r2

4r3 = 2000r3 = 2000

4r = 5.419 cm

h = 1000r2

h = 1000(5.419)2

= 10.84 cm

h/r = 10.845.419

= 2.000

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CHIN WYNN

2) The top and bottom pieces of the tin of height h cm are cut from square-shaped aluminum sheets.Determine the value for r, h and hence calculate the ratio h/ r so that thetotal area of the aluminum sheets are used for making the tin is minimum.

2r

rcm 2r rcmh cm


V = 1000r2h = 1000

h = 1000r2

Total Surface Area,A = 4r2 + 4r2 + 2rh

= 8r2 + 2r(1000)

r2

= 8r2 + 2000r


r (cm) A = 8r2 + 2000/r (cm2)1 2008.002 1032.003 738.67

4 628.005 600.006 621.337 677.71

8 762.00

9 870.22From the table above, A is minimum when r = 5 cm.

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CHIN WYNN



5.0 600.0005.1 600.2375.2 600.9355.3 602.0785.4 603.6505.5 605.6365.6 608.0235.7 610.797



16

2008.00

1032.00

738.67

628.00

600.00

621.33

677.71

762.00

870.22

0

500

1000

1500

2000

2500

1 2 3 4 5 6 7 8 9

r (cm)

A

(cm2)

600.000

600.237

600.935

602.078

603.650

605.636

608.023

610.797

590

595

600

605

610

615

5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7

r (cm)

A

(cm2)


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600.00000

600.00002

600.00010

600.00022

600.000

38

600

.00060

599.9996

599.9998

600.0000

600.0002

600.0004

600.0006

600.0008

5.000 5.001 5.002 5.003 5.004 5.005

r (cm)

A

(cm2)

CHIN WYNN


r = 5.000 cm

Substitute into h = 1000,r2

h = 1000(5.000)2

= 12.73 cm

h = 12.73r 5.000

= 2.546

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CHIN WYNN

Method 2: Differentiation

Volume of the tin = 1000 cm3

r2h = 1000

h = 1000r2

Total surface area of tin,A = 4r2 + 4r2 + 2rh

= 8r2 + 2r(1000/r2)= 8r2 + (2000/r)

dA = 16r 2000

dr r2

When A is minimum,dA = 0dr

16r (2000/r2) = 016r = (2000/r2)16r3 = 2000

r3 = 2000/16

= 125r = 5 cm

h = 1000(5)2

= 12.73 cm

h = 12.73r 5

= 2.546

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CHIN WYNN

3) Investigate cases where the top and bottom surfaces are cut from(i) Equilateral triangle(ii) Regular hexagonFind the ratio of h/r for each case.

(i) Equilateral triangle

Method 1: Trigonometry and Differentiation

60o h

rx 30o

Top and Bottom Surface Curved Surface

tan 60o = x/r x= r tan 60o

Area of the triangle = 0.5 x (2x)(r) x 3= 3xr= 3(r tan 60)(r)= 3r23 (tan 60 = 3)

Total Surface Area,A = 2 x (3r23) + 2rh

= 6r23 + 2r(1000) (h = 1000)r2 r2

= 6r23 + 2000r

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CHIN WYNN

dA = 12r3 2000dr r2

When A is minimum,

dA = 0dr

12r3 2000 = 0r2

12r3 = 2000r2

r3 = 2000123

r = 4.582 cm

h = 1000r2

= 1000(4.582)2

= 15.16 cm

h/r = 15.164.582

= 3.309


r (cm) A = 6r23 + 2000/r (cm2)1 2010.3922 1041.5693 760.197

4 666.2775 659.8086 707.4567 794.937

8 915.108


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CHIN WYNN



4.4 655.740

4.5 654.8894.6 654.6844.7 655.0984.8 656.105

4.9 657.6835.0 659.8085.1 662.4615.2 665.623



22

20

10.3

92

1041.5

69

760.1

97

666.2

77

659.8

08

707.4

56

794.9

37

915.1

08

1063.9

99

0

500

1000

1500

2000

2500

1 2 3 4 5 6 7 8 9

r (cm)

A

(cm2)

655.740

654.889

654.684

655.098

656.105

657.683

659.808

662.461

665.623

645

650

655

660

665

670

4.4 4.5 4.6 4.7 4.8 4.9 5.0 5.1 5.2

r (cm)

A

(cm2)


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CHIN WYNN

A More C omprehensive Tabler (cm) A = 6r23 + 2000/r (cm2)4.55 654.7074.56 654.690

4.57 654.6794.58 654.6744.59 654.6764.60 654.6844.61 654.6984.62 654.7184.63 654.744

From the tables above, A is minimum when r = 4.58 cm.


The Most Comprehensive Table

r (cm) A = 6r23 + 2000/r (cm2)4.580 654.67436614.581 654.67424554.582 654.67418744.583 654.67419164.584 654.67425824.585 654.67438714.586 654.67457834.587 654.6748317


23

654.707

654.690

654.679

654.674

654.676

654.684

654.698

654.718

654.744

654.62

654.64

654.66

654.68

654.70

654.72

654.74

654.76

4.55 4.56 4.57 4.58 4.59 4.60 4.61 4.62 4.63

r (cm)

A

(cm2)


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CHIN WYNN


r = 4.582 cm

Substitute into,h = 1000

r2

= 1000(4.582)2

= 15.16 cm

h/r = 15.164.582

= 3.309

24

654.6743661

654.6742455

654.6741874

654.6741916

654.6742582

654.6743871

654.6745

783

654

.6748317

654.6738

654.6740

654.6742

654.6744

654.6746

654.6748

654.6750

4.580 4.581 4.582 4.583 4.584 4.585 4.586 4.587

r (cm)

A

(cm2)


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CHIN WYNN

d A = 24r 2000dr 3 r2

When A is minimum,

dA = 0dr

24r 2000 = 03 r2

24r = 20003 r2

r3 = 2000324

r = 5.246 cm

h = 1000r2

= 1000(5.246)2

= 11.57 cm

h/r = 11.575.246

= 2.205


r (cm) A = 12r2 + 2000 (cm2)3 r

1 2006.9282 1027.7133 729.020

4 610.8515 573.2056 582.7497 625.196

8 693.405


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CHIN WYNN


A More Comprehensive Tabler (cm) A = 12r2 + 2000 (cm2)

3 r5.0 573.2055.1 572.359

5.2 571.9545.3 571.972

5.4 572.3975.5 573.215

5.6 574.4115.7 575.975



27

2006.928

1027.713

729.020

610.851

573.205

582.749

625.196

693.405

783.407

0

500

1000

1500

2000

2500

1 2 3 4 5 6 7 8 9

r (cm)

A

(cm2)

573.205

572.359

571.954

571.972

572.397

573.215

574.411

575.975

569

570

571

572

573

574

575

576577

5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7

r (cm)

A

(cm2)


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CHIN WYNN

A More Comprehensive Tabler (cm) A = 12r2 + 2000 (cm2)

3 r5.21 571.9370

5.22 571.92425.23 571.9156

5.24 571.91125.25 571.91105.26 571.91495.27 571.9229

5.28 571.9351From the table above, A is minimum when r = 5.25 cm.


The Most Comprehensive Tabler (cm) A = 12r2 + 2000 (cm2)3 r

5.245 571.91058275.246 571.91057955.247 571.91061805.248 571.91069805.249 571.91081955.250 571.9109825


28

571.9370

571.9242

571.9156

571.9112

571.9110

571.9149

571.9229

571.9351

571.890

571.900

571.910

571.920

571.930

571.940

5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28

r (cm)

A

(cm2)


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CHIN WYNN


r = 5.246 cm

Substitute into,h = 1000

r2

= 1000(5.246)2

= 11.567

h = 11.57r 5.246= 2.205

29

571.9105827

571.9105795

571.9106180

571.9106980

571.91081

95

571.910

9825

571.9111870

571.9102

571.9104

571.9106

571.9108

571.9110

571.9112

571.9114

5.245 5.246 5.247 5.248 5.249 5.250 5.251

r (cm)

A

(cm2)


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CHIN WYNN


Investigate cases where the top and bottom faces of the tin are being cut fromaluminum sheets consisting shapes of polygons. From the results of yourinvestigation, what conclusion can you derive from the relationship of the ratio

of h/r and the number of sides of a regular polygon?

Wastage occurs when circles are cut from aluminum sheet, which is not round inshape. Suggest the best possible shape of aluminum sheets to be used so as toreduce the production cost.

Answer

To determine the relationship between the ratio of h/r and the number of sidesof a regular polygon, different types of regular polygons are used, such as anequilateral triangle, a square, a regular pentagon, a regular hexagon and aregular octagon.

From the answers of Question 2, 3(i) and 3(ii), we already calculated the ratiosof h/r of a square, an equilateral triangle and a regular hexagon. However, westill have to calculate the ratios of a regular pentagon and a regular octagon.

The calculations for determining the ratios of a regular pentagon and a regularoctagon are shown below.

Finding the Value of h/r of A Regular Pentagon


h

rx x


x/r = tan 36

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CHIN WYNN

x= r tan 36

= 0.7265r

Area of the small triangle = 0.5 x 2xx r= xr

= 0.72654r x r

= 0.7265r2

Total surface area,

A = 2[5(0.7265r2)] + 2rh

= 7.265r2 + 2r(1000)

r2

= 7.265r2 + 2000/r

dA = 14.53r 2000/r2

dr

When A is minimum,

dA = 0

dr

14.53r 2000/r2 = 0

14.53r3 2000 = 0

14.53r3 = 2000

r3 = 2000

14.53

r = 5.163 cm

h = 1000

r2

= 1000

(5.163)2

= 11.94 cm

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CHIN WYNN

h/r = 11.94

5.163

= 2.313


r (cm) A = 10(tan36r2) + 2000/r (cm2)

1 2007.272 1029.063 732.064 616.255 581.646 594.897 641.728 714.999 810.72

From the table above, A is minimum when r = 5 cm.


32

2007.2

7

1029.0

6

732.0

6

616.2

5

581.6

4

594.8

9

641.7

2

714.9

9

810.7

2

0

500

1000

1500

2000

2500

1 2 3 4 5 6 7 8 9

r (cm)

A(cm

2)


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CHIN WYNN

A More Comp rehensive Table

r (cm) A = 10(tan36r2) + 2000/r (cm2)5.0 581.636

5.1 581.1315.2 581.0725.3 581.4445.4 582.2305.5 583.4155.6 584.9875.7 586.931



A More Comprehensive Table

r (cm) A = 10(tan36r2) + 2000/r (cm2)5.15 581.04685.16 581.04325.17 581.04405.18 581.0492

5.19 581.05875.20 581.07255.21 581.09065.22 581.1130


33

581.6

36

581.1

31

581.0

72

581.4

44

582.2

30

583.4

15

584.9

87

586.9

31

578

580

582

584

586

588

5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7

r (cm)

A(cm2)


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CHIN WYNN



r (cm) A = 10(tan36r2) + 2000/r (cm2)5.160 581.04320665.161 581.04309195.162 581.04302095.163 581.0429934

5.164 581.04300965.165 581.04306935.166 581.04317265.167 581.04331955.168 581.0435099



34

581.0

468

581.0

432

581.0

440

581.0

492

581.0

587

581.0

725

581

.0906

581.1

130

581.00

581.02

581.04

581.06

581.08

581.10

581.12

5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22

r (cm)

A

(cm2)

581.0

432066

581.0

430919

581.0

430209

581.0

429934

581.0

430096

581.0

430693

581.0

431726

581.0

433195

581.0

435099

581.0426

581.0428

581.0430

581.0432

581.0434

581.0436

5.160 5.161 5.162 5.163 5.164 5.165 5.166 5.167 5.168

r (cm)

A

(cm2)


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CHIN WYNN

r = 5.163 cm

V = 1000

r2h = 1000

h = 1000

r2

= 1000

(5.163)2

= 11.94 cm

h/r = 11.94

5.163

= 2.313

Finding the Value of h/r of A Regular Octagon


r h

x x


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CHIN WYNN

x/r = tan 22.5

x= r tan 22.5

= 0.4142r

Area of small triangle = 0.5 x 2xx r

= xr

= 0.4142r x r

= 0.4142r2

Total surface area,

A = 2[8(0.4142r2)] + 2rh= 6.6272r2 + 2r(1000)

r2

= 6.6272r2 + 2000

r

dA = 13.2544r - 2000

dr r2

When A is minimum,

dA = 0

dr

13.2544r 2000/r2 = 0

13.2544r3 2000 = 0

13.2544r3 = 2000

r3 = 2000/13.2544

r = 5.324 cm

h = 1000

r2

= 1000

(5.324)2

= 11.23 cm

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CHIN WYNN

h/r = 11.23

5.324

= 2.109


r (cm) A = 16(tan 22.5r2) + 2000/r (cm2)1 2006.632 1026.513 726.314 606.045 565.696 571.927 610.468 674.15

From the table above, A is minimum when r = 5 cm.


37

2006.6

3

1026.5

1

726.3

1

606.0

4

565.6

9

571.9

2

610.4

6

674.1

5

0

500

1000

1500

2000

2500

1 2 3 4 5 6 7 8

r (cm)

A

(cm2

)


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CHIN WYNN


r (cm) A = 16(tan 22.5r2) + 2000/r (cm2)5.0 565.685

5.1 564.5365.2 563.8215.3 563.5235.4 563.6265.5 564.1165.6 564.9795.7 566.202




r (cm) A = 16(tan 22.5r2) + 2000/r (cm2)5.30 563.52265.31 563.51515.32 563.51175.33 563.51215.34 563.51665.35 563.52505.36 563.5374


38

565.6

85

564.5

36

563.8

21

563.5

23

563.6

26

564.1

16

564.9

79

566.2

02

562

563

564

565

566

567

5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7

r (cm)

A

(cm2)


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CHIN WYNN



r (cm) A = 16(tan 22.5r2) + 2000/r (cm2)5.320 563.5116565

5.321 563.51152675.322 563.51143685.323 563.51138665.324 563.5113762

5.325 563.51140565.326 563.51147475.327 563.51158355.328 563.5117321



39

563.5

151

563.5

117

563.5

121

563.5

166

563.5

250

563.5

374

563.5

536

563.5

226

563.49

563.50

563.51

563.52

563.53563.54

563.55

563.56

5.30 5.31 5.32 5.33 5.34 5.35 5.36 5.37

r (cm)

A

(cm2)

563.5116565

563.5115267

563.5114368

563.5113866

563.5113762

563.5114056

563.5114747

563.5115835

563.511732

1

563.5111

563.5112

563.5113

563.5114

563.5115

563.5116

563.5117

563.5118

5 .3 20 5 .3 21 5 .3 22 5 .3 23 5 .3 24 5 .3 25 5 .3 26 5 .3 27 5 .3 28

r (cm)

A

(cm2)


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CHIN WYNN

r = 5.324 cm

V = 1000

r2h = 1000

h = 1000

r2

= 1000

(5.324)2

= 11.23 cm

h/r = 11.23

5.324

= 2.109

A table is drawn to conduct comparisons on the ratio of h/r of different regularpolygons, so as to see the relationship of the ratio of h/r and the number ofsides of a regular polygon.

Regular Polygons h r h/r

Circle

10.84 5.419 2.000

Octagon

11.23 5.324 2.109

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CHIN WYNN

Hexagon

11.57 5.246 2.205

Pentagon

11.94 5.163 2.313

Square

12.73 5.000 2.546

Equilateral Triangle

15.16 4.582 3.309

From the table above, we can conclude that polygons with more sides havesmaller value of h/r and it is near to 2. Also, the more the sides of a polygon,the more minimum the area is. Therefore, aluminum sheets with more sides likeregular octagons are used to save the production cost of aluminum can.

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CHIN WYNN

RESULTS

Question 1Minimum radius, r = 5.42 cmMinimum height, h = 10.84 cmRatio of h/r = 2Minimum area, A = 553.581 cm2

Question 2

Minimum radius, r = 5.00 cmMinimum height, h = 12.73 cm

Ratio of h/r = 2.546Minimum area, A = 600.000 cm2

Question 3(i)

Minimum radius, r = 4.582 cmMinimum height, h = 15.161 cmRatio of h/r = 3.309Minimum area, A = 654.674 cm2

Question 3(ii)Minimum radius, r = 5.246 cmMinimum height, h = 11.566 cmRatio of h/r = 2.205Minimum area, A = 571.911 cm2


Polygons with more sides have smaller ratio of h/r and it is near to 2.

The best possible shape of aluminum sheets to be used so as to reduce theproduction cost is a regular octagon.

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CHIN WYNN

CONCLUSION

1. Wastage is reduced when the number of sides of the aluminum sheetsused is increased. (Regular polygon-shaped)

2. Types of shapes that can be used are as below:a) equilateral trianglesb) squaresc) regular hexagonsd) regular octagons

3. So, circles should be cut from those shapes stated above.

a)

Wastage is very high when aluminumsheets of equilateral triangles are used.

b)

W astage is still high when aluminumsheets of squares are used.

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CHIN WYNN

c)

Wastage is low when aluminum sheets of regular hexagons are used.

d)

Wastage is very low when aluminum sheets of regular octagons are used.

Therefore, aluminum sheets of regular octagons are the best solution forminimum wastage so as to reduce the production cost.

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CHIN WYNN

A PPENDIX

1. http://nextlevel.com.sg/question/10742.Exploring Additional Mathematics Project
http://nextlevel.com.sg/question/1074http://nextlevel.com.sg/question/1074

Documents

Add Maths Project 2006 by Chin Wynn