curriculum development division ministry of education malaysia

ADDITIONAL MATHEMATICS PROJECT WORK 2/2011

7/18/2011

SMK RAJA MUDA MUSA/5 SCIENCE 1

SAFIYYAH NAJWA BINTI SHAMSUL ANUAR

940901-14-5072

MADAM HOW YANN PYNG

2 ADDITIONAL MATHEMATICS PROJECT WORK 2/2011

Numb. Title of the content Pages

1. TITLE 2

2. CONTENT 3

3. COMMENDATION 4

4. OBJECTIVES 5

5. INTRODUCTION

DEFINITON

HISTORY

6

7

8 –11

6. PART I

CONSTRUCTING

APPLICATIONPF

MATHEMATICAL

PRICIPLES

12

13 – 15

16 – 17

7. PART II

QUESTION 1

QUESTION 2

PART (a)

PART (b)

PART (c)[i/ii]

GRAPH PART (b)

QUESTION 3

PART (a)

PART (b)

PART (c)

18

19

20

21

21 – 22

23

24

25 – 27

27

8. PART III 28 – 32

9. FURTHER EXPLORATION

PART (a)

PART (b)

33

34 – 35

36 – 37

10. REFLECTION 38

11. CONCLUSION 39

12. ATTACHMENT 40 – 44


First of all, I wish to express gratitude to Allah SWT for His guidance and also giving me the

strength and health to do this project work. Finally, the Additional Mathematics Project Work 2 was

successfully completed with the inspiration, the gift and the instructions were given by Him.

Not forgotten to my parents who always make sure that I have been educating and nurturing of

small through to this day in devotion to God Almighty. They also had providing everything, such as

money, to buy anything that are related to this project work, their advice, which is the most needed for

this project and facilities such as Internet, books, computers and all that. They also supported me and

encouraged me to complete this task so that I will not procrastinate in doing it.

Then, infinity of appreciation and gratitude goes to my teacher, Madam How Yann Pyng for

guiding me throughout this project. Even I had some difficulties in doing this task, but she taught me

patiently until we knew what to do. She tried and tried to teach me until I understand what I’m

supposed to do with the project work.

Besides that, my friends who were doing this project with me. Even this project is individually

but we are cooperated doing this project especially in discussion and sharing ideas to ensure our task

will finish completely.

Last but not least, any party which involved either directly or indirect in completing this project

work. May Allah SWT give a great reward for all His servants that are always helpful in upholding His

religion. Thank you everyone.


The aims of carrying out this project work are to enable students to:

1. Apply mathematics to everyday situations and appreciate the importance and the beauty of

mathematics in everyday lives.

2. Improve problem-solving skills, thinking skills, reasoning and mathematical communication.

3. Develop positive attitude and personalities and instrinsic mathematical values such as accuracy,

confidence anf systematic reasoning.

4. Stimulate learning environment that enhances effective learning inquiry-base and teamwork.

5. Develop mathematical knowledge in a way which increase student’s interest and confidence.


There are a lot of things around us related to circles or parts of circle. A circle is a simple shape

of Euclidean geometry cinsisting of those points in a plane which is the same distance from a given point

called the centre. The common distance of the points of a circle from its centre is called radius.

Circles are simple closed curves which divide the plane into two regions, an interior and

exterior. In everyday use, the term ‘circle’ may be used interchangeably to refer to either the boundary

of the figure (known as the perimeter) or the whole figure including its interior. However, in strict

technical usage, ‘circle’ refers to the perimeter while the interior of the circle is called a disk. The

circumference of a circle is the perimeter of the circle (especially when referring to its length).

A circle is a special ellipse in which the two foci are coincident. Circles are conic sections attained

whe a right circular cone is intersected with a plane perpendicular to the axis of the cone.

The circle has been known since before the beginning of recorded history. It is the basis for the

wheel, which, with related inventions such as gears, makes such of modern civilization possible. In

mathematics, the study of the circle has helped inspire the development of geometry and calculus.

Circles had been used in daily lives to help people in their living.

Examples of circle


Pi, π , has the value of 3.14159265. In Euclidean plane geometry, π is defined as the ration of a

circle’s circumference, C, to its diameter, d .

π=Cd

The ration Cd

is constant, regardless of a circle’s size. For example, if a circle has twice the

diameter of another circle it will also have twice the circumference, C, preserving the ratio Cd

.

Alternatively π can be also defined as the ratio of a circle’s area (A) to the area of a square

whose side is equal to radius.

π= Ar


History of Pi

Pi or π is a mathematical constant whose value is the ratio of any circle’s circumference to its

diameter in Euclidean space; this is the same as the ratio of a circle’s area to the square of its radius. It is

approximately equal to 3.14159 in the usual decimal notation. π is one the most important

mathematical and physical constants; many formulae from mathematics, science and engineering

involve π .

π is an irrational number, which means that its value cannot be expressed exactly as a fraction

xy

, where x and y are integers. Consequently, its decimal representation never ends or repeats. It also a

transcendental number, which means that no finite sequence of algebraic operations on integers

(powers, roots, sums, etc) can be equal to its value; proving this was a late achievement in mathematical

history and a significant resuth on 19th century German mathematics. Throughout the history of

mathematics, there has been much effort to determine π more accurately and to understand its nature;

fascination with the number has even carried over into non-mathematical culture.

The Greek letter, π , often spelled outpi in text, was adopted for the number from the Greek word

forperimeter ‘ ’, first by William Jones in 1707 and popularized by Leonhard Eular in 1737.περίμετρος

The constant is occasionally also referred to as the circular constant, Archimede’s constant (not to be

confused with an Archimede’s number) or Ludolph’s number (from a German mathematician whose

efforts to calculate more of its digits became famous).

The name of the Greek letter π is pi and this spelling is commonly used in typographical

contexts when the Greek letter is not available or its usage could be problematic. It is not normally

capitalized ( ) even at the beginning of a sentences. When referring to this constant, the symbol Π π is

always pronounced like ‘pie’ in English, which is the conventional English pronounciation of the Greek

letter. In Greek, the letter is pronounced as ‘pi’.

The constant is named π because π is the first letter of the Greek words περιφέρεια

(periphery) and (perimeter), probably referring to its use in the formula to find theπερίμετρος

circumference or perimeter of a circle. π is Unicode character U+03C0 (‘Greek small letter pi’).


History of geometry

Geometry (Greek) γεωμετρία; geo=earth, metria=measure) arose as the field of knowledge

dealing with spatial relationships. Geometry was one of the two fields of pre-modern mathematics, the

other being the study of numbers (arithmetic).

Classic geometry was focused in compass and straight-edge constructions. Geometry was

revolutionized by Euclid, who introduced mathematical rigor and the axiomatic method still in use

today. His book, The Elements is widely considered the most influential textbook of all time, and was

known to all educated people in the West until the middle of the 20th century.

In modern times, geometric concepts have been generalized to a high level of abstraction and

complexity, and have been subjected to the methods of calculus and abstract algebra, so that many

modern branches of the field are barely recognizable as the descendants of early geometry.

History of Arithmetic progression

In Mathematics, an arithmetic progression (AP) or arithmetic sequence is

a sequence of numbers such that the difference of any two successive members of the sequence is a

constant. For instance, the sequence 3, 5, 7, 9, 11, 13, … is an arithmetic progression with common

difference 2. If the initial term of an arithmetic progression is a1 and the common difference of

successive members is d, then the nth term of the sequence is given by:

And in general :

A finite portion of an arithmetic progression is called a finite arithmetic progression and

sometimes just called an arithmetic progression. The behavior of the arithmetic progression depends on

the common difference d. If the common difference is:

Positive, the members (terms) will grow towards positive infinity.

Negative, the members (terms) will grow towards negative infinity.


History of Geometric progression

In Mathematics, a geometric progression, also known as a geometric sequence, is

a sequence of numbers where each term after the first is found by multiplying the previous one by a

fixed non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric

progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common

ratio 12

. The sum of the terms of a geometric progression is known as a geometric series.

Thus, the general form of a geometric sequence is:

and that of a geometric series is:

where r ≠ 0 is the common ratio and a is a scale factor, equal to the sequence's start value.


Differentiation (History of Differentiation)

The concept of a derivative in the sense of a tangent line is a very old one, familiar

to Greek geometers such as Euclid (c. 300 BC), Archimedes (c. 287–212 BC) and Apollonius of Perga (c.

262–190 BC).Archimedes also introduced the use of infinitesimals, although these were primarily used

to study areas and volumes rather than derivatives and tangents.

The use of infinitesimals to study rates of change can be found in Indian mathematics, perhaps

as early as 500 AD, when the astronomer and mathematician Aryabhata (476–550) used infinitesimals

to study the motion of the moon. The use of infinitesimals to compute rates of change was developed

significantly by Bhāskara II (1114-1185); indeed, it has been argued that many of the key notions of

differential calculus can be found in his work, such as "Rolle's theorem". ThePersian

mathematician, Sharaf al-Dīn al-Tūsī (1135-1213), was the first to discover the derivative of cubic

polynomials, an important result in differential calculus; his Treatise on Equations developed concepts

related to differential calculus, such as the derivative function and the maxima and minima of curves, in

order to solve cubic equations which may not have positive solutions.

Since the 17th century many mathematicians have contributed to the theory of differentiation.

In the 19th century, calculus was put on a much more rigorous footing by mathematicians such

as Augustin Louis Cauchy (1789 – 1857), Bernhard Riemann (1826 – 1866), and Karl Weierstrass (1815

– 1897). It was also during this period that the differentiation was generalized to Euclidean space and

the complex plane.

Isaac Newton (1643 – 1727) and Gottfried Leibniz (1646 – 1716), who provided

independent and unified approaches to differentiation and derivatives. The key insight, however, that

earned them this credit, was the fundamental theorem of calculus relating differentiation and

integration: this rendered obsolete most previous methods for computing areas and volumes, which had

not been significantly extended since the time of Ibn al-Haytham (Alhazen). For their ideas on

derivatives, both Newton and Leibniz built on significant earlier work by mathematicians such as Isaac

Barrow (1630 – 1677), René Descartes (1596 – 1650), Christiaan Huygens (1629 – 1695), Blaise

Pascal (1623 – 1662) and John Wallis (1616 – 1703). Isaac Barrow is generallly given credit for the early

development of the derivative. Nevertheless, Newton and Leibniz remain key figures in the history of

differentiation, not least because Newton was the first to apply differentiation to theoretical physics,

while Leibniz systematically developed much of the notation still used today.


Question: 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

finding.

Constructing the structure of a cake

These cakes are made by using different sizes of circular pans, then stacking the baked cake

sections on top of each other.

You are to plan for a cake that will serve between 200 and 250 people.

The wedding cake must feed between 200 and 250 people.

You have four different sizes of pans of you can use (all pans have the same height).

r = 10 cm r = 15 cm r = 20 cm r = 25 cm

Each layer of the cake must remain a cylinder.

You can stack layers. Each layer can be then be separated and cut individually.

Each layer of cake will be cut into sectors that have a top area of exactly 50 cm2

You may have some left-over cake from a layer.

Example of 50 cm2 top area of sector

One sector feeds one person.

Your final ingredients list must be proportional to the ingredients list provided for you.


By using the theory of arithmetic and geometric progression in Chapter One Form Five, the concept can

be used to:

Decide on how many layers of each size of cake you will need for your cake.

Show how many can you cut the layers of the cake into equivalent sectors having a top area of

50 cm2 each, in order to feed between 200 and 250 people.

Complete the ingredients list by identifying the quantities needed for each ingredient in cake.

Work and calculations to determine the ingredients of the cake

Baking a cake offers a tasty way to practice math skills, such as fractions and ratios, in a real-

world context. Many steps of baking a cake, such as counting ingredients and setting the oven time,

provide basic math practice for young children. Older children and teenagers can use more sophisticated

math to solve baking dilemmas, such as how to make a cake recipe larger or smaller or how to

determine what size slices you should cut. Practicing math while baking not only improves your math

skills, it helps you become a more flexible and resourceful baker.

Calculate the proportions of different ingredients. For example, a frosting recipe that calls for 2

cups cream cheese, 2 cups confectioners' sugar and 1/2 cup butter has a cream cheese, sugar

and butter ratio of 4:4:1. Identifying ratios can also help you make recipes larger or smaller.

Use as few measuring cups as possible. For example, instead of using a 3/4 cup, use a 1/4 cup

three times. This requires you to work with fractions.

Determine what time it will be when the oven timer goes off. For example, if your cake has to

bake for 30 minutes and you set the timer at 3:40, the timer will go off at 4:10.

Calculate the surface area of the part of the cake that needs frosting. For example, a sheet cake in

a pan only needs the top frosted, while a sheet cake on a tray needs the top and four sides

frosted. A round layer cake requires frosting on the top, on each layer and on the sides.

Determine how large each slice should be if you want to serve a certain amount of people. For

example, an 18 by 13 inch sheet cake designed to serve 25 people should be cut into slices that

measure approximately 3 by 3 inches.

Add up the cost of your ingredients to find the cost of your cake. Estimate the cost of partially

used ingredients, such as flour, by determining the fraction of the container used and

multiplying that by the cost of the entire container


Initial draft of the cake

r = 10 cm

r = 15 cm

r = 20 cm h = 20 cm

r = 25 cm


Application of Mathematical Principles to Answer The Questions:

Geometry

To determine suitable dimensions for the cake, to assist in designing and decorating

cakes that comes in many attractive shapes and designs, to estimate volume of cake to

be produced

Calculus (differentiation)

To determine minimum or maximum amount of ingredients for cake-baking, to estimate

min. or max. amount of cream needed for decorating, to estimate min. or max. size of

cake produced.

Progressions

To determine total weight/volume of multi-storey cakes with proportional dimensions,

to estimate total ingredients needed for cake-baking, to estimate total amount of cream

for decoration.

Geometry

Geometry is needed to determine suitable dimensions for the cake, to assist in designing and

decorating cakes that comes in many attractive shapes and designs, and also to estimate volume of cake

to be produced. When making a batch of cake batter, you end up with a certain volume, determined by

the recipe. The baker must then choose the appropriate size and shape of pan to achieve the desired

result. If the pan is too big, the cake becomes too short. If the pan is too small, the cake becomes too tall.

This leads into the next situation. The ratio of the surface area to the volume determines how much

crust a baked good will have. The more surface area there is ,compared to the volume, the faster the item

will bake, and the less "inside" there will be. For a very large, thick item, it will take a longtime for the

heat to penetrate to the center. To avoid having a rock-hard outside in this case, the baker will have to

lower the temperature a little bit and bake for a longer time. We mix ingredients in round bowls because

cubes would have corners where unmixed ingredients would accumulate, and we would have a hard

time scraping them into the batter.


Calculus (Differentiation)

Calculus is used to determine minimum or maximum amount of ingredients for cake baking, to

estimate minimum or maximum amount of cream needed for decorating, and to estimate minimum or

maximum size of cake produced.

Progressions

Progressions are functioned to determine total weight or volume of multi-storey cakes with

proportional dimensions, to estimate total ingredients needed for cake-baking, and to estimate total

amount of cream for decoration. For example when we make a cake with many layers, we must fix the

difference of diameter of the two layers. So we can say that it used arithmetic progression. When the

diameter of the first layer of the cake is 8´ and the diameter of second layer of the cake is 6´, then the

diameter of the third layer should be 4´.In this case, we use arithmetic progression where the difference

of the diameter is constant that is2. When the diameter decreases, the weight also decreases. That is the

way how the cake is balance to prevent it from smooch. We can also use ratio, because when we prepare

the ingredient for each layer of the cake, we need to decrease its ratio from lower layer to upper layer.

When we cut the cake, we can use fraction to divide the cake according to the total people that will eat

the cake


Best Bakery shop received an order from your school to bake 5 kg of round cake as shown in Diagram 1 for the Teacher’s Day celebration.


h cm

d cm

1) If a kilogram of cake has a volume of 3800 cm3 and the height of the cake is to be 7.0 cm,

calculate the diameter of the baking tray to be used to fit the 5 kg cake ordered by your school.

(Use = 3.142)π

Answer:

1 kg = 3800 cm3

h = 7 cm

5 kg = 3800 x 5

= 19000 cm3

V = π r2h

19000 = 3.142 x r2 x 7

r2 =19000

3.142×7

r 2 = 863.872

r = ±29.392 cm

using r = 29.392 cm

d = 2r

d = 2(29.392)

d = 58.784 cm


2) The cake will be baked in an oven with inner dimensions of 80.0 cm in length, 60 cm in width

and 45.0 cm in height.

a) If the volume of cake remains the same, explore by using different values of heights, h

cm, and the corresponding values of diameter of the baking tray to be used, d cm.

Tabulate your answer.

Answer:

Given the inner dimensions of the oven is 80.0 cm in length, 60.0 cm in width and 45.0 cm in height.

Hence, maximum dimensions of cake:

Maximum diameter, d = 60.0 cm

Maximum height, h = 45.0 cm

First, form a formula for d in terms of h by using the above formula for volume of cake, V = 19000, that’s:

19000 = π ( d2 )2

h

19000πh

= d2

4

6047.1038h

= d2

4

√ 24188.415h

= d

d =155.53

√h

h, (height) d, (diameter)

1 155.53

2 109.98

3 89.80

4 77.77

5 69.56

6 63.49

7 58.78

8 54.99

9 51.84

10 49.18


b) Based on the values in your table,

i. State the range of heights that is NOT suitable for the cakes and explain your

answer.

Answer: h ¿ 7 cm 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: I would suggest the dimensions of the cake to be 10 cm in height and about 49.18 cm in

diameter. This is because a cake with such dimensions is more symmetrical and easier

to decorate. Moreover, it is easy to handle.

c) Based on the graph,

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: Using V=π r2h and V=19000 cm3, an equation is formed:

19000 = π ( d2 )2

h

19000πh

= d2

4

6047.1038h

= d2

4

√ 24188.415h

= d

d =155.53

√h

d = 155.53×h−12


log10d = log10h−12 + log10155.53

∴ log10d=−12

log10h+ log10155.53 → in linear form, Y=mX+C

Hence, a table is drawn to determine the plotting points:

log10 h 0.0 1.0 2.0 3.0 4.0

log10 d 2.19 1.69 1.19 0.69 0.19

A graph of log10d=−12

log10h+ log10155.53 is plotted (on the next page).

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:

When h=10.5 cm, log10h=1.021

According to the graph,

log10d = 1.68

Therefore,d = 47.86 cm

b) If Best Bakery used a 42 cm diameter round cake tray, use your graph to

estimate the height of the cake obtained.

Answer:

When d = 42 cm, log10d=1.62

According to the graph,

log10 h = 1.15

Therefore, h = 14.12 cm


11 cm

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 1 cm.

a) Estimate the amount of fresh cream required to decorate the cake using the dimension

that you have suggested in 2(b)(ii).

Answer:

h = 10 cm

r = 49.18 cm

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

symbolised as the smaller cylinder.


1 cm

25.59 cm

Diagram 1: Cake without Cream

24.59 cm

Diagram 2: Cake with Cream

1 cm

10 cm

V cream=[3.142× (25.59 )2×11 ]−19000

V cream=22637.86−19000

V cream=3637.86

∴V cream=3637.86cm3

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:

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

A. Rectangular shaped cake (cuboid)

A cake without cream

A cake with cream


10 cm

38 cm

50 cm

11 cm

52 cm

40 cm

Top view

52 cm

Side view

Estimated volume of cream used:

V cream=( 40×52×11)−19000

V cream=22880−19000

∴V cream=3880cm3

B. Triangle shaped cake

Cake without cream

Cake with cream


V cream=( 12×52×78×11)−19000


10 cm

50 cm76 cm

52 cm

78 cm

52 cm

11 cm

Side viewTop view

V cream=22308−19000

∴V cream=3308cm3

C. Trapezium shaped cake

Cake without cream

Cake with cream


V cream=[ 12× (33+47 )×52×11]−19000

V cream=22880−19000

V cream=3880

∴V cream=3880cm3


10 cm

45 cm

50 cm

31 cm

11 cm

52 cm 47 cm

52 cm

33 cm

Side view Top view

c) Based on the values that you have found which shape requires the least amount of fresh

cream to be used?

Answer: Based on the values I have obtained, the triangle shaped cake requires the least

amount of fresh cream (3308 cm3).


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 your answers.

Answer:

Method 1: Using differentiation

Assuming that the surface area of the cake is proportional to the amount of fresh cream needed

to decorate the cake.

Use two equations for this method: the formula for volume of cake as in 2(a) and the formula for

the amount (volume) of fresh cream used.

The formula of surface area in contact with cream can also be used instead of amount (volume)

of fresh cream used.

19000 = π r²h→

A = π r² + 2π r h →

From : h =19000

π r2 →

Substitute into :

A=π r2+2 πr ( 19000

π r2 ) A=π r2+( 38000

r ) A=π r2+38000 r−1


The values, when plotted into a graph will form a minimum value (for dydx

=0) that can be

obtained through differentiation.

A = π r2+38000 r−1

dAdr

= 2πr−( 38000

r2 )0 = 2πr−( 38000

r2 )→ minimum value, therefore dAdr

=0

38000

r2 = 2π r

380002π

= r3

6047.1038 = r3

r = 3√6047.1038

r = 18 .2186

Substitute r=18.2186 into :

h =19000

π (18.2186)2

h = 18 .2187

∴ Hence, r=18.2186 and h=18.2187


Therefore,

h = 18.2187

d = 2r

= 2(18.2187)

= 36.44

Method 2: Quadratic Functions

Use the two same equations as in Method 1, but only the formula for the surface area in

contact in cream is used as the quadratic function.

Let y = surface area in contact in cream, r = radius of round cake:

19000 = (3.142)r²h →

y = (3.142)r² + 2(3.142)r h →

Factorise 3.142 in :

y = (3.142)(r ² + 2hr)

¿ (3.142 )[(r+2h2 )

2

−( 2h2 )

2] → completing square, with a = 3.142, b = 2h and c = 0

¿(3.142)[(r+h)² – h ²]

¿(3.142)(r+h)² – (3.142)h ²

Noted that a=3.142, positive indicates minimum value. Hence, minimum value ¿ – (3.142)h ² ,

corresponding value of r=−h.

Substitute r=−h into :


19000 = (3.142)(−h) ²h

h ³ = 6047.104

h = 18 .2186

Substitute h=18.22 into :

19000 = (3.142)r ² (18.2186)

r ² = 331.894

r = ±18 .2186 but using r=18.2186

Therefore,

h = 18.2187

d = 2r

= 2(18.2187)

= 36.44

∴ Thus, both methods have calculated that r and h is approximately 18.22 and 18.22 respectively.

In conclusion, the dimensions of the cake that requires the minimum amount of fresh cream

to decorate is approximately 18.22 cm in height and 18.22 cm in radius.

In conclusion, I would choose not to bake a cake with such dimensions because its

dimensions are not suitable (the height is too high). Besides, it will look less attractive and could be

difficult to handle such a big volume (size).


Best Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, as shown in Diagram 2 below.

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:

Volume of cake 1 Volume of cake 2

= π r2h =π r2h

= 3.142 × 31 × 31 × 6 = 3.142× (0.9×31 )2×6

¿18 116.772 cm3 ¿3.142× (27.9 )2×6

¿14 674.585 cm3

Volume of cake 3 Volume of cake 4

= π r2h = π r2h

¿3.142× (0.9×0.9×31 )2×6 ¿3.142× (0.9×0.9×0.9×31 )2×6

¿ 3.142 × (25.11)2 × 6 ¿ 3.142 × (22.599)2 × 6

¿ 11 886.414 cm3 ¿ 9 627.995 cm3

The values 18 116.772, 14 676.585, 11 886.414, 9 627.995 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.

a=18116.772, ratio, r=T 2

T 1

=T 3

T 2

=…=0.81

14 676.58518116 .772

=0.8111889.41414 676.585

=0.81 9627.995

11886.414=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: By using the formula of the sum of the first n terms, Sn, of a geometric progression:

Sn =a (1−r n )

1−r

= 18116.772 ( 1−0.8n )1−0.8

15 kg = 57 000 cm3

57 000 ¿ 18116.772(1−0.8n)0.2

1 1400 ¿ 18 116.772(1−0.8n)

0.629 ¿ 1−0.8n

−0.371 ¿ −0.8n

0.371 ¿ 0.8n

log10 0.371 ¿ n log10 0.8

log10 0.371

log100.8¿ n

n ¿ 4.444

n ≈ 4


Verification answer

Sn = a(1- rn )1-r

Sn = 57 000, a = 18 116.772 and r = 0.81

If n = 4, total volume of 4 cakes:

= 18 116.772(1- 0.814 )1-0.81

= 54 305.767

Hence, Sn < 57 000, n = 4 is suitable.

If n = 5, total volume of 5 cakes:

= 18 116.772(1- 0.815 )1-0.81

= 62 104.443

Hence, Sn > 57 000, n = 5 is not suitable.

∴ Total mass of cakes must not exceed 15 kg.

Therefore, maximum number of cakes needed to be made is 4.


As the famous saying goes ‘’thing ventured, nothing gained’’. 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 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.

This project 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.

Besides that, I would like to show my gratitude towards my teacher, Madam How Yann Pyng

who has given me guidance throughout this project. So from now on, I will do my best in learning

Additional Mathematics. Nevertheless, it is not only this subject that I will give my best but for every

subject as well.


After doing research, answering questions, drawing graphs and some problem solving, I saw

that the usage of geometry, calculus (differentiation) and progressions are important in daily life.

Geometry is the study of angles and triangles, perimeter, area ang volume. It differs from algebra

in that one develops a logical structure where mathematical relationships are proved and applied.

An arithmetic progression (A.P) is a sequence of numbers such that the difference of any two

successive member of the sequence is constant. Meanwhile, a geometric progression (G.P) is a sequence

of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero

number called the common ratio, r.

Differentiation is essentially the process of finding an equation which will give you the gradient

at any point along the curve. Say you have y=x2. The equation is y=2x will give you the gradient of y

at any point along the curve.

Pi (π) is a very useful mathematics related to circle in which it helps the mankind to solve many

problems easily involving circle. We are able to know how we can use this unit to solve various

problems involving objects that are circular in shape of even part of circle shape.

Besides, in this project we need to use a lot of mathematical concept in order to get the answer.

This makes me understand more about other mathematical concept besides Pi (π). So, after doing this

project, I am quite impressed with the usage of circle and its way to help us in solving problems although

there are some errors occur. Besides that, I also learnt many things for this which I can never find them

in the textbook or reference book or even in our school syllabus.

Furthermore, I am able to interpret carefully when handling such mind twisting problem that is

in Part 3. This experience that I gain from this project work can make me apply to other subjects so that

it will make me more be careful when handling such question mentioned. I am really appreciating the

government as they gave us this opportunity to do this project in this process of understanding and


learning deeply into circles. I would like to thank my Additional Mathematics teacher as without her

help, I would not be able to accomplice this project.


curriculum development division ministry of education malaysia

Documents

examples of circle

term circle

parts of circle

beauty of mathematics

circles circumference

circles area

circles size

mathematical constant