project 2 (tukar nama settle)

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AdditionalMathematic

sProject Work 2

Written by: Muhamad Hariz Bin Rumainor

I/C Num : 940622 -01-6251Angka Giliran: JK013 A064School : SMK Agama Johor Bahru


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Additional Mathematics Project Work 2 2011

Muhamad Hariz Bin Rumainor

TABLE OF CONTENTS

Num. Question

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

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, fryingfritters 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:

y Measurement of Ingredientsy Calculation of Price and Estimated Costy Estimation of Dimensionsy Calculation of Baking Timesy 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 = 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 cake:d = 60.0 cm

h = 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 cakebeing 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.

h/cm d/cm

1 155.5262519

2 109.9736674

3 89.79312339

4 77.76312594

5 69.5534543

6 63.493326457 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 34.77672715

21 33.9386105622 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 25.9210419837 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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c) i) V = r2h

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

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

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

Therefore, d = 50.12 cm

b) When d = 42 cm, log10 d = 1.6232According 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 1.191814

3 0.691814

4 0.191814


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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


= x 39.7 x 39.7 x 30 19000

= 23641.4 19000

= 4641.4 cm3

b) iii) Trapezium shaped cake


= 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)


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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 3917.6512 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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Method 2: Using differentiation

Assuming that the surface area of the cake is proportionate to the amount of fresh creamneeded to decorate the cake.*

Formula for surface area

= r2 + 2rh

h = 19000 / 3.142r2

Surface area in contact with cream

= r2

+ 2r(19000 / 3.142r2)

= r2

+ (38000/r)

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

through differentiation.

dy = 0

dx

dy = 2r (38000/r2)

dx

0 = 2r (38000/r2)

0 = 6.284r3 38000

38000 = 6.284r3

6047.104 = r3

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

= r2h = r

2h

= 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 = 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 = 0.8118116.772 14676.585

. 9627.995 = 0.81

11886.414

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

1-r 1-0.8

15 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 = 4


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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 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, especiallywhen 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. Thisproject 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.

project 2 (tukar nama settle)

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