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Additional MathematicsProject Work 2
Name : Wong Xin Hui
I/C Num : 940406-14-5016
Angka Giliran : 43School : SMK Seri Kembangan
Date : 22/6/2011
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TABLE OF CONTENTS
Num. Question Page
1 Part I 3
2
Part II 4 - 8
~ Question 1
~ Question 2 (a)
~ Question 2 (b)
~ Question 2 (c)
~ Question 3 (a)
~ Question 3 (b)
~ Question 3 (c)
3 Part III 9 - 10
4 Further Exploration 11
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PART I
His
y ofc ke baking and decorating
Although clear e amples of the difference between cake andbread areeasy to find, the
preciseclassification has always been elusive For e ample,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
wassometimesenriched 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 ofe ile,Tristia.Earlycakes in England were also essentially bread: the
most obvious differences between a "cake" and "bread" were the round, flat shape of the
cakes, and thecooking method, which turned cakes over once whilecooking, while bread
was left upright throughout the baking process. Spongecakes, leavened with beaten eggs,
originated during the Renaissance, possibly in Spain.
Cake decorating is one of thesugar arts requiring mathematics that usesicing or frosting
and other edible decorativeelements to make otherwise plaincakes more visually
interesting. Alternatively, cakescan 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 decoratecakes, especially inthe following actions:
y Measurement of Ingredientsy Calculation of Price and Estimated Costy Estimation of Dimensionsy Calculation ofBaking Timesy Modification of Recipe according to scale
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PART II
1) 1g = 3800 cm
3
h = 7 cm
5 g = 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 ca
e:
d = 60.0 cm
h = 45.0 cm
a)
b) i) h < 7 cm , h > 45 cm
This is because any heights lower than 7 cm will result in the diameter of the ca
e
being too big to fit into the ba ing oven while any heights higher than 45 cm will
cause the ca
e being too tall to fit into the baing oven
h/cm d/cm
1 155.5262519
2 109.9736674
3 89.79312339
4 77.763125945 69.5534543
6 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 35.6801692120 34.77672715
21 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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b) ii) I would suggest the dimensions of the cae to be 29 cm in height and approximately
29 cm in diameter. This is because a ca
e 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, log 10 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 1.1918143 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 ca e 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: Ca e without Cream
14.44 cm
Diagram 2: Ca e with Cream
1 cm
30 cm
29 cm
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3 b) i) Squareshaped cake
Estimated volume ofcream used
=30 x 27.6 x 27.6-19000
=22852.819000
=3852.8cm
3
b) ii) Triangleshaped cake
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Estimated volume ofcream used
= x 39.7 x 39.7 x 3019000
=23641.419000=4641.4cm3
b) iii) Trapezium shaped cake
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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 thic
at 1 cm
c) Based on the values I have obtained, the round shaped ca e 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 3888.3011 3533.03 28 3447.80 45 3917.65
12 3458.02 29 3471.71 46 3947.04
13 3402.96 30 3496.47 47 3976.46
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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Therefore, the above conjecture is proven to be true.
FURTHER EXPLORATION
a) Volume of ca e 1 Volume of ca e 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 ca e 3 Volume of ca e 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 = 0.81
18116.772 14676.585
. 9627.995 = 0.81
11886.414
b) Sn = a(1-rn) = 18116.772 ( 1-0.8n)
1-r 1-0.8
15 g = 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
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log 0.371 < n
log 0.8
4.444 < n
n = 4
Verification of answer
If n = 4
Total volume of 4 ca es
= 18116.772 cm3
+ 14676.585 cm3
+ 11886.414 cm3
+ 9627.995 cm3
= 54307.766 cm3
Total mass of ca es
= 14.29 g
If n = 5
Total volume of 5 ca es
= 18116.772 cm3
+ 14676.585 cm3
+ 11886.414 cm3
+ 9627.995 cm3
+ 7798.676 cm3
= 62106.442 cm3
Total mass of ca
es
= 16.34
g
Total mass of ca es must not exceed 15 g.Therefore, maximum number of ca
es needed to be made = 4
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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 wor. For me, succeeding in completing this
project wor has been reward enough. I h ave also learnt that mathematics is used
everywhere in daily life, from the most simple things li e ba ing and decorating a ca e, to
designing and building monuments. Besides that, I have learned many moral values that I
practice. This project wor
had taught me to be more confident when doing something
especially the homewor
given by the teacher. I also learned to be a more disciplined
student who is punctual and independent.