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This is Additional Mathematics Project Work 2/2011 with questions and full answer. This is the outcome after combined some others' project work. Credit shall be given to them.
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ADDITIONAL MATHEMATICS
PROJECT WORK 2/2011
MATH IN CAKE BAKING
1
NAME:
NRIC:
TEACHER:
NG CHI SAN
941021-14-5896
PN. ONG KAH YAN
CONTENTS
Appreciation 3
Introduction 4
Objectives 5
PART I 6
PART II 8
PART III 16
Further Exploration 18
Conclusion 21
Reflection 22
2
APPRECIATION
First and foremost, I would like to manifest my appreciation in given an opportunity to carry
out such a challenging project work. A very special thank you to Puan Ong Kah Yan, our Additional
Mathematics teacher for her guidance in helping me to finish this project. Furthermore, a high
gratitude to my parents who have given me the support in both emotion and finance.
Thank you Joseph Carl Robnett Licklider the person which invented the internet, without the
help of internet I will not be able to finish this project. I’d also like to express my thanks to all my
fellow friends especially the 5 Arts 1 students. They are my classmates who involved in this project
as we work together as a secondary family. We do practice teamwork.
Not forgetting those people who had helped me with or without my knowledge or even by
coincidences.
Thank you.
3
INTRODUCTION
In the course of this project, the below mathematical principles explained will be applied.
A volume is the amount of 3-dimensional space enclosed by a closed boundary, usually, the
space that a substance or shape occupies or contains. The volume of a container is generally taken as
the capacity of the container, or the inner dimensions of the container.
In Mathematics, a geometric progressions is also known as a geometric sequence. A
geometric progression is a sequence of numbers where each term after the first is found by
multiplying the previous one by a number which is not zero. That number is referred to as the
common ratio of the sequence. We use geometric progressions to calculate a value in a specific
sequence. For example, if a ball is dropped from 100cm, what will the height of the ball be at the 5 th
bounce if the second bounce is ¾ of the first bounce and so on? From this, the total vertical distance
can also be calculated.
Similarly, linear equations are also greatly used in application of mathematics. A linear
equation is an algebraic equation where each term is a constant, or a product of constant, and the
first power of a single variable. Linear equations are very useful as many non linear equations can
be reduced to linear equations, which will form a straight line on a Cartesian plane.
Differentiation is a branch of Calculus that uses derivatives to measure how a function
changes as its input changes. Differentiation is commonly taken as the inverse of integration.
Differentiation is used in industry to calculate the maximum amount or quantity required as a means
of maximizing efficiency and goods production.
4
OBJECTIVES
This project will serve as a training stage for me to prepare myself for the demands for my
future undertakings in the university and work life. I will apply mathematics in everyday situations
and situations and appreciate the importance and beauty of mathematics. I aim to improve my
problem solving skills, thinking skills, reasoning and mathematical communication skills. This
project shall stimulate a learning environment which enhances effective learning, inquiry-based and
teamwork. This project will develop my mathematical knowledge such that it increases my interest
and confidence.
5
PART I
Cakes come in a variety of forms and flavours and are among favourite desserts served
during special occasions such as birthday parties, Hari Raya, weddings and etc. Cakes are treasured
not only because of their wonderful taste but also in the art of cake baking and cake decorating. Find
out how mathematics is used in cake baking and cake decorating and write about your findings.
Answer:
One very important aspect of cake baking is the amount of ingredients required. This is
because a cake should be of reasonable pricing, at the same time, beauty is also a priority. In cake
baking, ingredients must not be wasted unnecessarily. Therefore, Calculus can be applied in cake
baking. Particularly, the second derivative is greatly used. This is because the second derivative
allows bakeries to calculate the maximum or minimum amount of ingredients needed to increase
profit and efficiency. This allows positive growth in business. At the same time, the bakeries can
avoid under-order, or over-order the ingredients.
Cake decorating is an art. Many bakeries strive to produce cakes of beauty and of substance.
In art, geometry is given importance. Geometry allows cakes of many interesting shapes and sizes
to be created. Geometry also defines the ideal dimensions of a cake to be baked in ovens. Geometry
will determine the popularity of the cake besides the price. We calculate the surface area and volume
of the cake to determine the price per kilogram and also the area available for decorating and writing
words.
In the baking of more complex cakes, such as multi-storey cakes or multilayered cakes,
progressions are applied. Progressions allow us to calculate the size or volume of a subsequent
layer. Also, it allows us the estimate the quantity of ingredients needed. Usually, geometric
progressions are used.
Lastly, ratios are used in cake baking. More often than not bakers need to estimate the
amount of ingredients used or substitute the ingredient with another if that ingredient is not
available. For example, we often use cookbooks guiding us to use 3 parts of water for 1 part of flour.
This is ratio of water to flour 3:1, allows us to bake a cake of different sizes. Although we may bake
a smaller or larger cake, the flour and water used still obeys the proportion set. We are then allowed
to creatively bake cakes.
6
Numerous shapes of cakes by using Geometry skill:
Half Sphere-shaped
Other shapes :
Multi-storey cakes design by using the Progressions method:
7
Round-shaped Square-shaped
PART II
Best Bakery shop received an order from your school to bake a 5 kg of round cake as shown in
Diagram 1 for the Teachers’ Day celebration. (Diagram 11)
1) If a kilogram of cake has a volume of 3800cm3, and the height of the cake is to be 7.0cm,
calculate the diameter of the baking tray to be used to fit the 5 kg cake ordered by your school.
[Use π = 3.142]
Answer:
Volume of 5kg cake = Base area of cake x Height of cake ( πrh)
3800 x 5 = (3.142)(d2
)² x 7
190007
(3.142) = (d2
)²
863.872 = (d2
)²
d2
= 29.392
d = 58.7834 cm
2) The cake will be baked in an oven with inner dimensions of 80.0 cm in length, 60.0 cm in width
and 45.0 cm in height.
8
h
d
Diagram 11
a) If the volume of cake remains the same, explore by using different values of heights, hcm, and the
corresponding values of diameters of the baking tray to be used, d cm. Tabulate your answers
Answer:
First, form the formula for d in terms of h by using the above formula for volume of cake, V =
19000, that is:
19000 = (3.142)(d/2)²h
19000(3.142)h
= d ²4
24188.415h
= d²
d = 155 .5263
√h
Height, h (cm) Diameter, d(cm)
1.0 155.5263
2.0 109.9737
3.0 89.7931
4.0 77.7631
5.0 68.5535
6.0 63.4833
7.0 58.7834
8.0 54.9868
9.0 51.8421
10.0 49.1817
(b) Based on the values in your table,
9
(i) State the range of heights that is NOT suitable for the cakes and explain your answers.
Answer:
h< 7cm is NOT suitable, because the resulting diameter produced is too large to fit into the
oven. Furthermore, the cake would be too short and too wide, making it less attractive.
(ii) Suggest the dimensions that you think most suitable for the cake. Give reasons for your answer.
Answer:
h = 8cm, d = 54.99cm, because it can fit into the oven, and the size is suitable for easy handling.
(c)
(i) Form an equation to represent the linear relation between h and d. Hence, plot a suitable graph
based on the equation that you have formed. [You may draw your graph with the aid of computer
software.]
Answer:
19000 = (3.142)(d2
)²h
19000/(3.142)h = d ²4
24188.415h
= d²
d = 155.53
√ h
d = 155.53 h−12
log d = log 155.53 h−12
10
log d = −12
log h + log 155.53
Log h 0 1 2 3 4
Log d 2.19 1.69 1.19 0.69 0.19
11
(ii)
(a) If Best Bakery received an order to bake a cake where the height of the cake is 10.5 cm, use your
graph to determine the diameter of the round cake pan required.
Answer:
h = 10.5cm, log h = 1.021, log d = 1.680, d = 47.86cm
(b) If Best Bakery used a 42 cm diameter round cake tray, use your graph to estimate the height of
the cake obtained.
Answer:
d = 42cm, log d = 1.623, log h = 1.140, h = 13.80cm
3) Best Bakery has been requested to decorate the cake with fresh cream. The thickness of the cream
is normally set to a uniform layer of about 1cm.
(a) Estimate the amount of fresh cream required to decorate the cake using the dimensions that you
have suggested in 2(b)(ii).
Answer:
h = 8cm, d = 54.99cm
Amount of fresh cream = VOLUME of fresh cream needed (area x height)
Amount of fresh cream = Vol. of cream at the top surface + Vol. of cream at the side surface
Vol. of cream at the top surface
= Area of top surface (πr2) x Height of cream
12
= (3.142)(54.99
2)² x 1
= 2375 cm³
Vol. of cream at the side surface
= Area of side surface x Height of cream
= (Circumference of cake x Height of cake) x Height of cream
= 2(3.142)(54.99/2)(8) x 1
= 1382.23 cm³
Therefore, amount of fresh cream = 2375 + 1382.23 = 3757.23 cm³
(b) Suggest three other shapes for cake, that will have the same height and volume as those
suggested in 2(b)(ii). Estimate the amount of fresh cream to be used on each of the cakes.
Answer:
1 – Rectangle-shaped base (cuboid)
19000(V) = base area x 8 (h)
base area = 19000
8
length x width = 2375
By trial and improvement, 2375 = 50 x 47.5 (length = 50cm, width = 47.cm5, height = 8cm)
Therefore, volume of cream
13
= 2(Area of left/right side surface)(Height of cream) + 2(Area of front/back side surface)(Height of
cream) + Area of top surface(Height of cream)
= 2(8 x 50)(1) + 2(8 x 47.5)(1) + 2375(1) = 3935 cm³
2 Triangle-shaped base
19000(V) = base area x 8 (h)
base area = 19000
8
base area = 2375
12
x length x width = 2375
length x width = 4750
By trial and improvement, 4750 = 95 x 50 (length = 95cm, width = 50cm, height = 8cm)
Slant length of triangle = √(95² + 25²)= 98.23
Therefore, amount of cream
= Area of rectangular front side surface(Height of cream) + 2(Area of slant rectangular left/right side
surface)(Height of cream) + Area of top surface(Height of cream)
= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375(1) = 4346.68 cm³
1 Pentagon-shaped base (cube)
14
19000 = base area x height
base area = 19000
8
base area = 2375 = area of 5 similar isosceles triangles in a pentagon
therefore:
2375 = 5(length x width)
475 = length x width
By trial and improvement, 475 = 25 x 19 (length = 25cm, width = 19cm, height = 8cm)
Therefore, amount of cream
= 5(area of one rectangular side surface)(height of cream) + Area of top surface(Height of cream)
= 5(8 x 19) + 2375(1) = 3135 cm³
(c) Based on the values that you have found which shape requires the least amount of fresh
cream to be used?
Answer:
Pentagon-shaped cake, since it requires only 3135 cm³ of cream to be used.
15
PART III
Find the dimension of a 5 kg round cake that requires the minimum amount of fresh cream to
decorate. Use at least two different methods including Calculus.State whether you would choose to
bake a cake of such dimensions. Give reasons for youranswers.
Answer:
Method 1: Differentiation
Use two equations for this method: the formula for volume of cake (as in Q2/a), and the formula for
amount (volume) of cream to be used for the round cake (as in Q3/a).
19000 = (3.142)r²h → (1)
V = (3.142)r² + 2(3.142)rh → (2)
From (1): h = 19000
(3.142)r ² → (3)
Sub. (3) into (2):
V = (3.142)r² + 2(3.142)r(19000
(3.142)r ²)
V = (3.142)r² + (38000
r)
V = (3.142)r² + 38000r-1
(dVdr
) = 2(3.142)r – (38000
r ²)
0 = 2(3.142)r – (38000
r ²) -->> minimum value, therefore
dVdr
= 0
16
38000r ²
= 2(3.142)r
380002(3.142)
= r³
6047.104 = r³
r = 18.22
Sub. r = 18.22 into (3):
h = 19000
(3.142)(18.22)²
h = 18.22
therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cm
Method 2: Quadratic Functions
Use the two same equations as in Method 1, but only the formula for amount of cream is the main
equation used as the quadratic function.
Let f(r) = volume of cream, r = radius of round cake:
19000 = (3.142)r²h → (1)
f(r) = (3.142)r² + 2(3.142)hr → (2)
From (2):
f(r) = (3.142)(r² + 2hr) -->> factorize (3.142)
= (3.142)[ (r + 2h2
)² – (2h2
)² ] -->> completing square, with a = (3.142), b = 2h and c = 0
= (3.142)[ (r + h)² – h² ]
= (3.142)(r + h)² – (3.142)h²
(a = (3.142) (positive indicates min. value), min. value = [f(r) = –(3.142)h²], corresponding value of
x = [r = -h]
Sub. r = -h into (1):
19000 = (3.142)(-h)²h
h³ = 6047.104
h = 18.22
Sub. h = 18.22 into (1):
19000 = (3.142)r²(18.22)
r² = 331.894
r = 18.22
17
therefore, h = 18.22 cm, d = 2r = 2(18.22) = 36.44 cm
I would choose not to bake a cake with such dimensions because its dimensions are not
suitable (the height is too high) and therefore less attractive. Furthermore, such cakes are
difficult to handle.
FURTHER EXPLORATION
Best Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, as shown in
Diagram 2.
The height of each cake is 6.0 cm and the radius of the largest cake is 31.0 cm. The radius of the
second cake is 10% less than the radius of the first cake, the radius of the third cake is 10% less than
the radius of the second cake and so on.(a)
Find the volume of the first, the second, the third and the fourth cakes. By comparing all these
values, determine whether the volumes of the cakes form a number pattern? Explain and elaborate
on the number patterns.
Answer:
height, h of each cake = 6cm
radius of largest cake = 31cm
radius of 2nd cake = 10% smaller than 1st cake
radius of 3rd cake = 10% smaller than 2nd cake
31, 27.9, 25.11, 22.599…
a = 31, r = 9
10
18
V = (3.142)r²h
Radius of 1st cake = 31, volume of 1st cake = (3.142)(31)²(6) = 18116.772
Radius of 2nd cake = 27.9, vol. of 2nd cake = 14674.585
Radius of 3rd cake = 25.11, vol. of 3rd cake = 11886.414
Radius of 4th cake = 22.599, vol. of 4th cake = 9627.995
18116.772, 14674.585, 11886.414, 9627.995, …
a = 18116.772, ratio, r = T2/T1 = T3 /T2 = … = 0.81
(b) If the total mass of all the cakes should not exceed 15 kg, calculate the maximum number of
cakes that the bakery needs to bake. Verify your answer using other methods.
Answer:
Sn =(a (1– rn))
(1−r)
Sn = 57000, a = 18116.772 and r = 0.81
57000 =(18116.772 (1– (0.81)n))
(1−0.81)
1 – 0.81n = 0.59779
0.40221 = 0.81n
lg0.81 0.40221 = n
n = lg0.40221
lg0.81
n = 4.322
Therefore, n ≈ 4
Verifying the answer:
19
When n = 5,
S5 = (18116.772(1– (0.81)5))
(1−0.81)
S5 = 11799.8442
0.19
S5 = 62104.443 cm3
Given 1kg = 3800 cm3
62104.443 = 62104.443 ÷ 3800
= 16.3433 kg
Therefore, n=5 is not suitable as the mass has exceeded 15kg.
When n = 4,
S4 = (18116.772 (1– (0.81)4))
(1−0.81)
S4 = 10318.0957
0.19
S4 = 54305.7669 cm3
Given 1kg = 3800 cm3
54305.7669 = 54305.7669 ÷ 3800
= 14.2910 kg
Therefore, n=4 is suitable as the mass is less than 15kg.
20
CONCLUSION
In conclusion, cake baking required a lot of mathematical knowledge including geometry,
calculus, ratios and progressions in order to produce a cake in a systematic way. Those skills are
important especially for pastry industry so that the cake produced can be unified with the same
quality. Without it, the outlook of the cakes will be less attractive. So, I should be thankful of the
people who contribute in the idea of geometry, calculus, ratios and progressions. Those
mathematical theories make daily problems easier to be solved.
From this project I have learned the importance of perseverance as time will be invested to
ensure the completion and excellence of this project. Similarly, I have learned the virtue of working
together as I have help and receive help from my fellow peers in the production of this project as
sharing knowledge is vital achieving a single goal. Also, I have learned to be thankful and
appreciative. This is because I have been able to apply my mathematical knowledge in daily life and
appreciate the beauty of mathematics.
21
REFLECTION
During the period of completing the project work, I had discovered that additional
mathematics have multi-uses especially in pastry industry. The inventions of different mathematics
theorem are to solve daily problems and even make our works more efficient. Furthermore, this
project also encourages knowledge transfer as students practice what they had learnt and found out
the information they don’t know. I also had experienced team work, patience and creative thinking while
conducting the project. Personally, Calculus used to be a very confusing and difficult topic for me. But after
applying it in this project, I have built a deeper understanding toward Differentiation and Integration.
The poem below is what I feel about Additional Mathematics :-
ADD MATH,
A well known taught subject.
Perhaps there’s a concept,
That I wouldn’t able to accept.
Addition, Subtraction, Multiplication and Division,
Miraculously transform to Differentiation,
Then followed by Integration.
Yet, I never fail to figure out the solution,
22
But it always ends up with correction!
To my surprise,
Cake baking requires geometry to apply,
In determine its shape and size.
If I have been wise,
And listen to Calculus’s advice,
Baking will never fail once or twice...
Add Math used to be my pain,
But if I do practice again and again,
Works will never be in vain...
23