folio add math 2011 (ain)

8/6/2019 Folio Add Math 2011 (Ain)

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SEKOLAH MENENGAH KEBANGSAAN SERI INDAH

43300 SERI KEMBANGAN, SELANGOR.

Additional

Mathematics____________________________________________________________________________________

Project Work 2 / 2011

Name : Nur Ain Binti Omar

Class : 5 SN


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Number Question Page

1. Part 1 2

2. Part 2 3

Question 1 3

Question 2 (a) 4

Question 2 (b) 5

Question 2 (c) 5

Question 3 (a) 7

Question 3 (b) 9

Question 3 (c) 13

1. Part 3 14

2. Further Exploration 17

3. Reflection 19


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1

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

properlyconsidered 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 doughwas 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 cakein his first book of exile,Tristia. .Early cakes in England

were also essentially bread: the mostobvious differences between a "cake" and "bread" were the round,

flat shape of the cakes, andthe cooking method, which turned cakes over once while cooking, while bread

was leftupright throughout the baking process. Sponge cakes, leavened with beaten eggs,

originatedduring the Renaissance, possibly in Spain.Cake decorating is one of the sugar arts requiring

mathematics that uses icing or frosting andother 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 aspecial celebration such as a birthday, graduation, bridal shower, wedding,

or anniversary.

Uses of mathematics in cake baking and decorating:

Measurement of Ingredient Calculation of Price and Estimated Cost Estimation of Dimensions Calculation of Baking Times


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2

1) Volume 1 kg of cake = 3800 cm3

Height of the cake = 7 cm

Volume 5 kg of cake = 3800 x 5= 19000 cm3

Diameter of the cake :

Volume of cylinder = r2h

19000 = 3.142 x r2 x 7

r2 = 19000 (3.142 7)

r

2

= 863.872

r = 29.392 cm

d = 2r

d = 29.392 2

d = 58.783 cm


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3

2)

maximum dimension of cake :

d = 60.0 cm

h = 45.0 cm

(a)


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Height, h (cm) Diameter, d

(cm)

23.0 32.429

24.0 31.747

25.0 31.105

26.0 30.501

27.0 29.931

28.0 29.392

29.0 28.880

30.0 28.395

31.0 27.933

32.0 27.493

33.0 27.074

34.0 26.673

35.0 26.289

36.0 25.921

37.0 25.568

38.0 25.230

39.0 24.904

40.0 24.591

41.0 24.289

42.0 23.998

43.0 23.718

44.0 23.446

45.0 23.184

Height, h (cm) Diameter, d

(cm)

1.0 155.526

2.0 109.974

3.0 89.793

4.0 77.763

5.0 69.553

6.0 63.493

7.0 58.783

8.0 54.987

9.0 51.842

10.0 49.182

11.0 46.893

12.0 44.897

13.0 43.135

14.0 41.566

15.0 40.157

16.0 38.882

17.0 37.721

18.0 36.658

19.0 35.680

20.0 34.777

21.0 33.939

22.0 33.158


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4

(b) (i) the range of height that is not suitable for the cake is :

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.

(ii) I would suggest the dimensions of the cake to be 29 cm in height and approximately29 cm in

diameter. This is because a cake with these dimensions is moresymmetrical and easier todecorate.

(c) (i) V = r2h , V = 19000 cm3 , r = d/2

19000 = 3.142 (d/2)2 h

(d2 / 4) = 19000 [3.142 (d2/4)]

d2 = 76000 (3.142 h)

d = 155.53 h-1/2

log10 d = 1/2 log

10 h + log10 155.53

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

1.0212

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

log10 d= 1.7

d= 101.7

d= 50.12 cm

log10 h log10 d

0 2.191814

1 1.691814

2 1.191814

3 0.691814

4 0.191814


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(b) When d= 42, log10 d= 1.6232

According to the graph, log10 h= 1.18 when log10 d= 1.6232

log10 h= 1.18

h= 101.18

h= 15.14 cm 5

6

3. (a) h= 29 cm

r = 14.44 cm

14.44 cm

29 cm

Diagram 1 - Cake without cream

15.44 cm

1 cm

30 cm


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1 cm

Diagram 2- Cake with cream

7

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

symbolised as the smaller cylinder.

Vcream = [ r2h] 19000

= [3.142 (15.44)2 30] 19000

= 22471 19000

= 3471 cm3


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8

(b) (i) Square shaped cake

25.6 cm

25.6 cm

29 cm

Diagram 3 (a)-Cake without cream

1 cm 1 cm

1 cm

27.6 cm


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1 cm

27.6 cm

Diagram 3 (b)- Cake with cream (plan view)

9

1 cm

30 cm

Diagram 3(c)- Cake with cream (side view)

Estimated volume of cream used = (30 27.6 27.6) 19000

= 22852.8 19000

= 3852.8 cm3

(ii) Triangle shaped cake


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29.0 cm

36.2 cm 36.2 cm

Diagram 4(a)- Cake without cream

1

0

1 cm

2.5 cm

39.7 cm

1 cm

1 cm 2.5 cm

39.7 cm

Diagram 4(b)- Cake with cream (plan view)

1 cm


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30 cm


1

1

Estimated volume of cream used = ( 39.7 39.7 30) 19000

= 23641.35 19000

= 4641.35 cm3

(iii) Trapezium shaped cake

26 cm

29 cm

20cm 39.5 cm

Diagram 5(a)- Cake without cream

28 cm

1 cm 1 cm

1 cm


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22 cm

1 cm

1 cm 2 cm

42.5 cm

Diagram 5(b)- Cake with cream (plan view) 12

1 cm

30 cm


Estimated volume of cream used = [ (28 + 42.5) 22 30] 19000

= 23265 19000

= 4265 cm3

* All estimations in the values are based on the assumption that the layer of cream isuniformly thick at 1 cm.

(c) Based on the values I have obtained, the round shaped cake requires the least amountof fresh

cream (3471 cm3

).


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1

3

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


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height, cm Volume of

cream

used, cm

1 19983.61

2 10546.04

3 7474.42

4 5987.37

5 5130.07

6 4585.13

7 4217.00

8 3958.20

9 3771.41

10 3634.38

11 3533.03

12 3458.02

13 3402.96

14 3363.28

15 3335.70

16 3317.73

17 3307.53

18 3303.66

19 3304.98

20 3310.62

21 3319.86

22 3332.12

23 3346.94

24 3363.92

25 3382.74

Height, cm Volume of

cream

used, cm3

26 3403.14

27 3424.89

28 3447.80

29 3471.71

30 3496.47

31 3521.98

32 3548.12

33 3574.8134 3601.97

35 3629.54

36 3657.46

37 3685.67

38 3714.13

39 3742.81

40 3771.67

41 3800.67

42 3829.79

43 3859.01

44 3888.30

45 3917.65

46 3947.04

47 3976.46

48 4005.88

49 4035.31

50 4064.72


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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

14

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 obtainedthrough differentiation.

dydx = 0

dydx = 2r (38000 r2)

0 = 2r (38000 r2)

0 = 6.284r3 38000

38000 = 6.284r3

6047.104 = r3

r = 18.22

When r = 18.22 cm, h = 18.22 cm

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

isapproximately 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 thecutting or eating of said

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


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15

* 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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1

6

(a) Volume of the first cake = r2h

= 3.142 312 6

= 18116.772 cm3

Volume of the second cake = r2h

= 3.142 (0.9 31)2 6

= 14674.585 cm3

Volume of the third cake = r2h

= 3.142 (0.9 0.9 31)2 6

= 11886.414 cm3

Volume of the fourth cake = r2h

= 3.142 (0.9 0.9 0.9 31)2 6

= 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.

14676.58518116.772= 0.81

9627.99511886.414= 0.81

11886.41414674.585= 0.81


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(b) Sn = a(1 rn)

1 r

15 = 18116.772 (1 0.8n)

1 0.81

15 = 5700057000 > 18116.772 (1 0.8n)

0.2

0.6293 > 1 0.8n

-0.371 > -0.8n

0.371 < 0.8n

log10 0.371 < n log10 0.8

log10 0.371 < n

log10 0.8

4.444 < n

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 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

1

8


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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

everywherein 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 thehomework

given by the teacher. I also learned to be a more disciplined student who is punctual and independent.

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