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Additional MathematicsProject Work 2

: ALVIN SOO CHUN KIT I/C Num: Angka Giliran: School :

Date :

Alvin Soo Chun Kit 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

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

PART I

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 Dimensions Calculation of Baking Times Modification of Recipe according to scale

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

PART II

1) 1 kg = 3800 cm3

h = 7 cm

5 kg = 3800 x 5= 19000 cm3

V = Πr2h 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 cmh = 45.0 cm

a)

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

h/cm d/cm

1155.5262519

2109.9736674

389.79312339

477.76312594

569.5534543

663.49332645

758.78339783

854.98683368

951.84208396

1049.18171919

1146.89292932

1244.89656169

1343.13522122

1441.56613923

1540.15670556

h/cm d/cm

1638.88156297

1737.72065671

1836.65788912

1935.68016921

2034.77672715

2133.93861056

2233.15830831

2332.42946528

2431.74666323

2531.10525037

2630.50120743

2729.93104113

2829.39169891

2928.88049994

3028.39507881

h/cm d/cm31 27.9333394432 27.4934168433 27.0736453734 26.6725321535 26.288734736 25.9210419837 25.5683583138 25.229689639 24.9041315840 24.5908595941 24.2891198342 23.9982216743 23.7175310644 23.4464646645 23.18448477

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

approximately 29 cm in diameter. This is because a cake with these dimensions is more symmetrical and easier to decorate.

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

c) i) V = Πr2h

V = 19000 cm3

r = d/2

19000 = 3.142 x (d/2)2 x h

d 2 = 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

log10 h log10 d1 1.6918142 1.1918143 0.6918144 0.191814

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

30 cm

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

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

1 cm

15.44 cm

Diagram 1: Cake without Cream

14.44 cm

Diagram 2: Cake with Cream

1 cm

29 cm

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

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

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

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

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

h/cmvolume of cream

used/cm3 h/cmvolume of cream

used/cm3 h/cmvolume of cream

used/cm3

1 19983.61 18 3303.66 35 3629.542 10546.04 19 3304.98 36 3657.463 7474.42 20 3310.62 37 3685.674 5987.37 21 3319.86 38 3714.135 5130.07 22 3332.12 39 3742.816 4585.13 23 3346.94 40 3771.677 4217.00 24 3363.92 41 3800.678 3958.20 25 3382.74 42 3829.799 3771.41 26 3403.14 43 3859.01

10 3634.38 27 3424.89 44 3888.3011 3533.03 28 3447.80 45 3917.6512 3458.02 29 3471.71 46 3947.0413 3402.96 30 3496.47 47 3976.4614 3363.28 31 3521.98 48 4005.8815 3335.70 32 3548.12 49 4035.3116 3317.73 33 3574.81 50 4064.7217 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

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

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 / 3.142r2

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 = 0dx

dy = 2Πr – (38000/r2)dx 0 = 2Πr – (38000/r2) 0 = 6.284r3 – 38000 38000 = 6.284r3

6047.104 = r3

18.22 = rWhen 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.

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

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 = 0.8118116.772 14676.585

. 9627.995 = 0.8111886.414

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

15 kg = 57000 cm3

57000 > 18116.772(1-0.8 n ) 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

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

n = 4Verification 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 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

Alvin Soo Chun Kit Additional Mathematics Project Work 2 2011

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.