add math's project 2011

8/6/2019 Add Math's Project 2011

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Additional MathematicsProject Work 2

Name : Wong Xin Hui

I/C Num : 940406-14-5016

Angka Giliran : 43School : SMK Seri Kembangan

Date : 22/6/2011


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TABLE OF CONTENTS

Num. Question Page

1 Part I 3

2

Part II 4 - 8

~ Question 1

~ Question 2 (a)

~ Question 2 (b)

~ Question 2 (c)

~ Question 3 (a)

~ Question 3 (b)

~ Question 3 (c)

3 Part III 9 - 10

4 Further Exploration 11


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PART I

His

y ofc ke baking and decorating

Although clear e amples of the difference between cake andbread areeasy to find, the

preciseclassification has always been elusive For e ample,banana bread may be properly

considered either a quick bread or a cake The Greeks invented beer as a leavener, frying

fritters in olive oil, and cheesecakes using goat's milk In ancient Rome, basic bread dough

wassometimesenriched with butter, eggs, and honey, which produced a sweet and cake-

like baked good. Latin poet Ovid refers to the birthday of him and his brother with party and

cake in his first book ofe ile,Tristia.Earlycakes in England were also essentially bread: the

most obvious differences between a "cake" and "bread" were the round, flat shape of the

cakes, and thecooking method, which turned cakes over once whilecooking, while bread

was left upright throughout the baking process. Spongecakes, leavened with beaten eggs,

originated during the Renaissance, possibly in Spain.

Cake decorating is one of thesugar arts requiring mathematics that usesicing or frosting

and other edible decorativeelements to make otherwise plaincakes more visually

interesting. Alternatively, cakescan be moulded and sculpted to resemble three-

dimensional persons, places and things. In many areas of the world, decorated cakes are

often a focal point of a special celebration such as a birthday, graduation, bridal shower,

wedding, or anniversary.

Mathematics are often used to bake and decoratecakes, especially inthe following actions:

y Measurement of Ingredientsy Calculation of Price and Estimated Costy Estimation of Dimensionsy Calculation ofBaking Timesy Modification of Recipe according to scale


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PART II

1) 1g = 3800 cm

3

h = 7 cm

5 g = 3800 x 5

= 19000 cm3

V = r2h

19000 = 3.142 x r2

x 7

r2

= 19000 .

3.142 x 7

r2

= 863.872

r = 29.392 cm

d = 2r

d = 58.783 cm

2) Maximum dimensions of ca

e:

d = 60.0 cm

h = 45.0 cm

a)

b) i) h < 7 cm , h > 45 cm

This is because any heights lower than 7 cm will result in the diameter of the ca

e

being too big to fit into the ba ing oven while any heights higher than 45 cm will

cause the ca

e being too tall to fit into the baing oven

h/cm d/cm

1 155.5262519

2 109.9736674

3 89.79312339

4 77.763125945 69.5534543

6 63.49332645

7 58.78339783

8 54.98683368

9 51.84208396

10 49.18171919

11 46.89292932

12 44.89656169

13 43.13522122

14 41.56613923

15 40.15670556

h/cm d/cm

16 38.88156297

17 37.72065671

18 36.65788912

19 35.6801692120 34.77672715

21 33.93861056

22 33.15830831

23 32.42946528

24 31.74666323

25 31.10525037

26 30.50120743

27 29.93104113

28 29.39169891

29 28.88049994

30 28.39507881

h/cm d/cm

31 27.93333944

32 27.49341684

33 27.07364537

34 26.67253215

35 26.2887347

36 25.92104198

37 25.56835831

38 25.2296896

39 24.90413158

40 24.59085959

41 24.28911983

42 23.99822167

43 23.71753106

44 23.44646466

45 23.18448477


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b) ii) I would suggest the dimensions of the cae to be 29 cm in height and approximately

29 cm in diameter. This is because a ca

e with these dimensions is more

symmetrical and easier to decorate.

c) i) V = r2h

V = 19000 cm3

r = d/2

19000 = 3.142 x (d/2)

2x h

d2

= 19000 .

4 3.142 x (d2/4)

d2

= 76000 .3.142 x h

d = 155.53 x h-1/2

log10 d = -1/2 log10 h + log10 155.53

c) ii) a) When h = 10.5 cm, log10 h = 1.0212

According to the graph, log10 d = 1.7 when log10 h = 1.0212

Therefore, d = 50.12 cm

b) When d = 42 cm, log 10 d = 1.6232According to the graph, log10 h = 1.2 when log10 d = 1.6232

Therefore, h = 15.85 cm

log10 h log10 d

1 1.691814

2 1.1918143 0.691814

4 0.191814


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3) a) h = 29 cm

r = 14.44 cm

To calculate volume of cream used, the cream is symbolised as the larger cylinder

and the ca e is symbolised as the smaller cylinder.

Vcream = 3.142 x 15.442 x 30 19000

= 22471 19000

= 3471 cm3

1 cm

15.44 cm

Diagram 1: Ca e without Cream

14.44 cm

Diagram 2: Ca e with Cream

1 cm

30 cm

29 cm


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3 b) i) Squareshaped cake

Estimated volume ofcream used

=30 x 27.6 x 27.6-19000

=22852.819000

=3852.8cm

3

b) ii) Triangleshaped cake


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Estimated volume ofcream used

= x 39.7 x 39.7 x 3019000

=23641.419000=4641.4cm3

b) iii) Trapezium shaped cake


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Estimated volume of cream used

= x (28+42.5) x 22 x 30 - 19000

= 23265 19000

= 4265 cm3

* All estimations in the values are based on the assumption that the layer of cream is

uniformly thic

at 1 cm

c) Based on the values I have obtained, the round shaped ca e requires the least amount

of fresh cream (3471 cm3)

PART III

Method 1: By comparing values of height against volume of cream used

h/cm

volume of cream

used/cm3 h/cm

volume of cream

used/cm3 h/cm

volume of cream

used/cm3

1 19983.61 18 3303.66 35 3629.54

2 10546.04 19 3304.98 36 3657.46

3 7474.42 20 3310.62 37 3685.67

4 5987.37 21 3319.86 38 3714.13

5 5130.07 22 3332.12 39 3742.81

6 4585.13 23 3346.94 40 3771.67

7 4217.00 24 3363.92 41 3800.67

8 3958.20 25 3382.74 42 3829.79

9 3771.41 26 3403.14 43 3859.01

10 3634.38 27 3424.89 44 3888.3011 3533.03 28 3447.80 45 3917.65

12 3458.02 29 3471.71 46 3947.04

13 3402.96 30 3496.47 47 3976.46

14 3363.28 31 3521.98 48 4005.88

15 3335.70 32 3548.12 49 4035.31

16 3317.73 33 3574.81 50 4064.72

17 3307.53 34 3601.97

According to the table above, the minimum volume of cream used is 3303.66 cm3

when h =

18cm.

When h = 18cm, r = 18.3 cm


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Therefore, the above conjecture is proven to be true.

FURTHER EXPLORATION

a) Volume of ca e 1 Volume of ca e 2

= r2h = r2h

= 3.142 x 31 x 31 x 6 = 3.142 x (0.9 x 31)2 x 6

= 18116.772 cm3

= 3.142 x (27.9)2

x 6

= 14676.585 cm3

Volume of ca e 3 Volume of ca e 4

= r2h = r

2h

= 3.142 x (0.9 x 0.9 x 31)2

x 6 = 3.142 x (0.9 x 0.9 x 0.9 x 31)2

x 6

= 3.142 x (25.11)2

x 6 = 3.142 x (22.599)2

x 6

= 11886.414 cm3

= 9627.995 cm3

The values 118116.772, 14676.585, 11886.414, 9627.995 form a number pattern .

The pattern formed is a geometrical progression.

This is proven by the fact that there is a common ratio between subsequent numbers, r =

0.81.

14676.585 = 0.81 11886.414 = 0.81

18116.772 14676.585

. 9627.995 = 0.81

11886.414

b) Sn = a(1-rn) = 18116.772 ( 1-0.8n)

1-r 1-0.8

15 g = 57000 cm3

57000 > 18116.772(1-0.8n)

0.2

11400 > 18116.772(1-0.8n)

0.629 > 1-0.8n

-0.371 > - 0.8n

0.371 < 0.8n

log 0.371 < n log 0.8


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log 0.371 < n

log 0.8

4.444 < n

n = 4

Verification of answer

If n = 4

Total volume of 4 ca es

= 18116.772 cm3

+ 14676.585 cm3

+ 11886.414 cm3

+ 9627.995 cm3

= 54307.766 cm3

Total mass of ca es

= 14.29 g

If n = 5

Total volume of 5 ca es

= 18116.772 cm3

+ 14676.585 cm3

+ 11886.414 cm3

+ 9627.995 cm3

+ 7798.676 cm3

= 62106.442 cm3

Total mass of ca

es

= 16.34

g

Total mass of ca es must not exceed 15 g.Therefore, maximum number of ca

es needed to be made = 4


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Reflection

In the process of conducting this project, I have learnt that perseverance pays off, especially

when you obtain a just reward for all your hard wor. For me, succeeding in completing this

project wor has been reward enough. I h ave also learnt that mathematics is used

everywhere in daily life, from the most simple things li e ba ing and decorating a ca e, to

designing and building monuments. Besides that, I have learned many moral values that I

practice. This project wor

had taught me to be more confident when doing something

especially the homewor

given by the teacher. I also learned to be a more disciplined

student who is punctual and independent.

add math's project 2011

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