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 ADDITIONAL MATHEMATICS PROJECT WORK 2 Name: Class: Serial No:

Add Maths Project 2

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ADDITIONAL MATHEMATICS

PROJECT WORK 2

Name:

Class:

Serial No:

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TABLE OF CONTENTS

1

PART I:

Num. Question Page

1 Part I 2

2

Part II 3-8

~ Question 1 3

~ Question 2 (a) 3

~ Question 2 (b) 4

~ Question 2 (c) 4-5

~ Question 3 (a) 6

~ Question 3 (b) 6-7 

~ Question 3 (c) 8

3 Part III 8-9

4 Further Exploration 10-11

5 Reflection 11

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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.Early cakes 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 DimensionsCalculation of Baking Times

Modification of Recipe according to scale

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PART II:

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1) 1 kg = 3800 cm3 

h = 7 cm

5 kg = 3800 x 5

= 19000 cm3 

V = πr 2h 

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 cake:d = 60.0 cm

h = 45.0 cm

a)

3

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

h/cm d/cm

1 155.5262519

2 109.9736674

3 89.79312339

4 77.76312594

5 69.55345436 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 35.68016921

20 34.7767271521 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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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.

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, log10 d = 1.6232

 According to the graph, log10

h = 1.2 when log10

d = 1.6232

Therefore, h = 15.85 cm

4

log10 h  log10 d 

1 1.691814

2 1.191814

3 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 cake is symbolised as the smaller cylinder.

Vcream= 3.142 x 15.442 x 30 – 19000

= 22471 – 19000

= 3471 cm3 

3) b) i) Square shaped cake

6

1 cm

15.44 cm

Diagram 1: Cake without Cream

14.44 cm

Diagram 2: Cake with Cream

1 cm

30 cm

29 cm

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Estimated volume of cream used 

= 30 x 27.6 x 27.6 - 19000

= 22852.8 – 19000

= 3852.8 cm3 

b) ii) Triangle shaped cake

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 

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* 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 

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 3888.30

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= r 

2+ 2rh

h = 19000 / 3.142r2

Surface area in contact with cream

= r 2

+ 2r(19000 / 3.142r2 )

= r 2

+ (38000/r)

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

differentiation.

=0

= 2r – (38000/r 

2 )

0 = 2r – (38000/r 2 )

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 

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~ 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.

FURTHER EXPLORATION:

a) Volume of cake 1 Volume of cake 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 cake 3 Volume of cake 4

= r2h = r2h

= 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 cm

3

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 = 0.8111886.414

b) Sn = a(1-rn) = 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 > 1-0.8

n

-0.371 > - 0.8n 

0.371 < 0.8n 

log 0.371 < n log 0.8

log 0.371 < nlog 0.8 10

4.444 < n

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n = 4

Verification of answer: 

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 cm

= 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

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, from the 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.

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