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DANARAJ CHELATHURAI ADDITIONAL MATHEMATICS PROJECT WORK 2 2011

TABLE OF CONTENTS

 Num. Question Page

1 Part I

2 Part II

~ Question 1

~ Question 2 (a)

~ Question 2 (b)

~ Question 2 (c)

~ Question 3 (a)

~ Question 3 (b)

~ Question 3 (c)

3 Part III

4 Further Exploration

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PART I

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History of cake baking and decorating

Although clear examples of the difference between cake and bread are easy to find, the precise

classification has always been elusive. For example, 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 was sometimes enriched 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 of exile, 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 the cooking method, which turned cakes over 

once while cooking, while bread was left upright throughout the baking process. Sponge cakes,

leavened with beaten eggs, originated during the Renaissance, possibly in Spain.

Cake decorating is one of the sugar arts requiring mathematics that uses icing or frosting and other 

edible decorative elements to make otherwise plain cakes more visually interesting. Alternatively,

cakes can 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 decorate cakes, especially in the following actions:

• Measurement of Ingredients

• Calculation of Price and Estimated Cost

• Estimation of Dimensions

• Calculation of Baking Times

• Modification of Recipe according to scale

 

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PART II

1) 1 kg = 3800 cm3

  h = 7 cm

5 kg = 3800 x 5

= 19000 cm3

V = Πr2h19000 = 3.142 x r 2 x 7

r 2 = 19000 . 

3.142 x 7

r 2 = 863.872

r = 29.392 cm

d = 2r d = 58.783 cm

2) Maximum dimensions of cake:

d = 60.0 cm

h = 45.0 cm

h/cm d/cm

1

155.526251

9

2

109.973667

4

3

89.7931233

9

4

77.7631259

4

5 69.5534543

6

63.4933264

5

7

58.7833978

3

8 54.98683368

9

51.8420839

6

10

49.1817191

9

11

46.8929293

2

12

44.8965616

9

13

43.1352212

2

14

41.5661392

3

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15

40.1567055

6

a)

h/cm d/cm

16

38.8815629

7

1737.72065671

18

36.6578891

2

19

35.6801692

1

20

34.7767271

5

21

33.9386105

6

22

33.1583083

1

23

32.4294652

8

24

31.7466632

3

25

31.1052503

7

26

30.5012074

3

27 29.93104113

28

29.3916989

1

29

28.8804999

4

30

28.3950788

1

h/cm d/cm

31

27.9333394

4

32

27.4934168

4

3327.07364537

34

26.6725321

5

35 26.2887347

36

25.9210419

8

37

25.5683583

1

38 25.2296896

39

24.9041315

8

40

24.5908595

9

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41 24.28911983

42

23.9982216

7

43

23.7175310

6

44

23.4464646

6

45

23.1844847

7

 b) i) h < 7 cm , h > 45 cm

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

too big to fit into the baking oven while any heights higher than 45 cm will cause the cake

 being too tall to fit into the baking oven

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

29 cm in diameter. This is because a cake with these dimensions is more symmetrical

and easier to decorate.

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c) i) V = Πr2h

V = 19000 cm3

r =d

/2

19000 = 3.142 x (d/2)2 x 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

log10

h log10

d

1 1.691814

2 1.191814

3 0.691814

4 0.191814

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, log10

d = 1.6232

According to the graph, log10

h = 1.2 when log10

d = 1.6232

Therefore, h = 15.85 cm

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3) a) h = 29 cm

r = 14.44 cm

14.44 cm

29 cm

Diagram 1: Cake without Cream

1 cm

15.44 cm

1 cm

Diagram 2: Cake with Cream

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

the cake is symbolised as the smaller cylinder.

Vcream

= 3.142 x 15.442 x 30 – 19000

= 22471 – 19000

= 3471 cm3

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3) b) i) Square shaped cake

Estimated volume of cream used

= 30 x 27.6 x 27.6 - 19000

= 22852.8 – 19000

= 3852.8 cm3

 b) ii) Triangle shaped cake

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Estimated volume of cream used= ½ x 39.7 x 39.7 x 30 – 19000

= 23641.4 – 19000

= 4641.4 cm3

 b) iii) Trapezium shaped cake

Estimated volume of cream used

= ½ 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

thick at 1 cm

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

of fresh cream (3471 cm3)

PART III

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

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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 3685.674 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 3888.30

11 3533.03 28 3447.80 45 3917.65

12 3458.02 29 3471.71 46 3947.04

13 3402.96 30 3496.47 47 3976.4614 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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Method 2: Using differentiation

Assuming that the surface area of the cake is proportionate to the amount of fresh cream needed to

decorate the cake.*

Formula for surface area

= Πr2 + 2Πrh

h = 19000 / 3.142r 2

Surface area in contact with cream

= Πr2 + 2Πr(19000 / 3.142r2)= Πr2 + (38000/r)

The values, when plotted into a graph will from a minimum value that can be obtained through

differentiation.

dy = 0

dx

dy = 2Πr – (38000/r2)dx

0 = 2Πr – (38000/r2)0 = 6.284r 3 – 38000

38000 = 6.284r 3

6047.104 = r 3

18.22 = r When r = 18.22 cm, h = 18.22 cm

The dimensions of the cake that requires the minimum amount of fresh cream to decorate is

approximately 18.2 cm in height and 18.2 cm in radius.

I would bake a cake of such dimensions because the cake would not be too large for the cutting or 

eating of said cake, and it would not be too big to bake in a conventional oven.

* The above conjecture is proven by the following

When r = 10,

~ the total surface area of the cake is 4114.2 cm2

~ the amount of fresh cream needed to decorate the cake is 4381.2 cm3

~ the ratio of total surface area of cake to amount of fresh cream needed is 0.94

When r = 20,

~ the total surface area of the cake is 3156.8 cm2

~ the amount of fresh cream needed to decorate the cake is 3308.5 cm3

~ the ratio of total surface area of cake to amount of fresh cream needed is 0.94

Therefore, the above conjecture is proven to be true.

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FURTHER EXPLORATION

a) Volume of cake 1 Volume of cake 2

= Πr2h = Πr2h= 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 cake 3 Volume of cake 4

= Πr2h = Πr2h= 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-r n) = 18116.772 ( 1-0.8n)

1-r 1-0.8

15 kg = 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

log 0.371 < n

log 0.8

4.444 < n

n = 4Verification of answer 

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If n = 4

Total volume of 4 cakes

= 18116.772 cm3 + 14676.585 cm3 + 11886.414 cm3 + 9627.995 cm3

= 54307.766 cm3

Total mass of cakes= 14.29 kg

If n = 5

Total volume of 5 cakes

= 18116.772 cm3 + 14676.585 cm3 + 11886.414 cm3 + 9627.995 cm3 + 7798.676 cm3

= 62106.442 cm3

Total mass of cakes

= 16.34 kg

Total mass of cakes must not exceed 15 kg.

Therefore, maximum number of cakes 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 work. For me, succeeding in completing this project work 

has been reward enough. I have also learnt that mathematics is used everywhere in daily life, fromthe most simple things like baking and decorating a cake, to designing and building monuments.

Besides that, I have learned many moral values that I practice. This project work had taught me to

 be more confident when doing something especially the homework given by the teacher. I also

learned to be a more disciplined student who is punctual and independent.