ADDITIONAL MATHEMATICS

PROJECT WORK 2/2011

MATH IN CAKE BAKING

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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

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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.

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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.

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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.

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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.

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Numerous shapes of cakes by using Geometry skill:

Half Sphere-shaped

Other shapes :

Multi-storey cakes design by using the Progressions method:

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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.

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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,

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(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

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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

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(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

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= (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

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= 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)

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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.

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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

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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

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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

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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:

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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.

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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.

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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,

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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...

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