57347904 Additional Maths Project Work 2 2011
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Transcript of 57347904 Additional Maths Project Work 2 2011
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AdditionalMathematicsProject Work 2
Written by: ALVIN SOO CHUN KIT
I/C Num :Angka Giliran:
School :
Date :
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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
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) a) h = 29 cmr = 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
30 cm
29 cm
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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
Estimated volume of cream used
= 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
= 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.