ITEMS	PAGE
APPRECIATION	3
OBJECTIVES	4
INTRODUCTION	5
PROCEDURE AND FINDINGS	6-16
Ø FURTHER EXPLORATION	17-18
CONCLUSION	19
REFLECTION	20
REFERENCES	21

Firstly, I would like to give a big thanks to my parents for providing everything, such as money, to buy things that are related to this project work, their advice and support. Then, I want to thank my teacher Mr.Liow Hoi Lee for teaching me Additional Mathematics form 5 and guiding me throughout this project.

Last but not least, my friends who were doing this project with me and sharing our ideas and knowledge. They were helping each other so we can complete our project without any problems and in a short period of time. Thank you so much to everyone!

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The objectives of this project work are:

ü Apply and adapt a variety of problem-solving strategies to solve problems.

ü Develop mathematical knowledge through problem solving in a way that increases students’ interest and confidence.

ü Developing positive attitude towards mathematics in form 5 students.

ü Improve thinking skills and creativity of students.

ü Promote efficiency of mathematical communication which can benefit students in the future.

ü Provide learning environment that stimulates and enhances effective learning so that students can learn effectively.

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We can normally see cakes in grand functions such as birthday parties,wedding anniversaries and may more. Cakes are sweet creamed breads which are baked in ovens. Many people loves to eat cakes. There is a variety of cakes such as chocolate cakes,cheese cakes,butter cakes and various shapes and designs such as round cake,oval cake,multistorey cakes and many more. But most of the peoples did not know how the skills of Mathematics are involved in the cake industry. Every cake baking requires some mathematical skills in order to make them look good and neat.Mathematics are often used to bake and decorate cakes, especially in the following actions:

Ø Measurement of Ingredients

Ø Calculation of Price and Estimated Cost

Ø Estimation of Dimensions

Ø Calculation of Baking Times

Ø Modification of Recipe according to scale

So, in this project work,we will see how the Mathematics skills are applied when we bake a cake.

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Question

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:

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.

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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 3800 , 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

3800 x 5 = (3.142)( )² x 7

(3.142) = ( )²

863.872 = ( )²

= 29.392

d = 58.784 cm

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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.

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

=

= d²

d =

Example:

a) h= 1.0 cm , d=

=155.53 cm

b) h= 2.0 cm , d=

=109.98 cm

c) h=3.0 cm , d=

=89.80 cm

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The other diameters for their respective heights are calculated using the same formula as in example above and tabulated in the table below.

Height,h (cm)	Diameter,d(cm)
1.0	155.53
2.0	109.98
3.0	89.80
4.0	77.77
5.0	68.56
6.0	63.49
7.0	58.78
8.0	54.99
9.0	51.84
10.0	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 answers.

Answer:

h < 7cm is NOT suitable, because the resulting diameter produced is too large to fit into the oven(since the diameter of the oven is 80.0 cm). 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.

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(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)( )²h

19000/(3.142)h =

= d²

d =

d =

log d =

log d = 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 x Height of cream

= (3.142)( )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:

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1 – Rectangle-shaped base (cuboid)

19000 = base area x height

base area =

length x width = 2375

By trial and improvement, 2375 = 50 x 47.5 (length = 50, width = 47.5, height = 8)

Therefore, volume of cream

= 2(Area of left/right side surface)(Height of cream) + 2(Area of front/back side surface)(Height of cream) + Vol. of top surface

= 2(8 x 50)(1) + 2(8 x 47.5)(1) + 2375 = 3935 cm³

2 – Triangle-shaped base

19000 = base area x height

base area = 2375

x length x width = 2375

length x width = 4750

By trial and improvement, 4750 = 95 x 50 (length = 95, width = 50)

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) + Vol. of top surface

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= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68 cm³

3 – Pentagon-shaped base

19000 = base area x height

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 = 25, width = 19)

Therefore, amount of cream

= 5(area of one rectangular side surface)(height of cream) + vol. of top surface

= 5(8 x 19) + 2375 = 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 your answers.

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 = → (3)

Sub. (3) into (2):

V = (3.142)r² + 2(3.142)r( )

V = (3.142)r² + ( )

V = (3.142)r² + 38000r^-1

( ) = 2(3.142)r – ( )

0 = 2(3.142)r – ( ) -->> minimum value, therefore = 0

= 2(3.142)r

= r³

6047.104 = r³

r = 18.22

Sub. r = 18.22 into (3):

h =

h = 18.22

therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cm

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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 + )² – ( )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

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 easily.

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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 is10% 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 2^nd cake = 10% smaller than 1^st cake

radius of 3^rd cake = 10% smaller than 2^nd cake

31, 27.9, 25.11, 22.599…

a = 31, r =

V = (3.142)r²h

Radius of 1^st cake = 31, volume of 1^st cake = (3.142)(31)²(6) = 18116.772

Radius of 2^nd cake = 27.9, vol. of 2^nd cake = 14674.585

Radius of 3^rd cake = 25.11, vol. of 3^rd cake = 11886.414

Radius of 4^th cake = 22.599, vol. of 4^th cake = 9627.995

18116.772, 14674.585, 11886.414, 9627.995, …

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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:

S_n =

S_n = 57000, a = 18116.772 and r = 0.81

57000 =

1 – 0.81ⁿ = 0.59779

0.40221 = 0.81ⁿ

og_0.81 0.40221 = n

n =

n = 4.322

therefore, n ≈ 4

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· 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

· Geometry is the study of angles and triangles, perimeter, area and volume. It differs from algebra in that one develops a logical structure where mathematical relationships are proved and applied.

· Differentiation is essentially the process of finding an equation which will give you the gradient (slope, "rise over run", etc.) at any point along the curve. Say you have y = x^2.The equation y' = 2x will give you the gradient of y at any point along that curve.

· 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

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After completing this project work, I found myself becoming quite interested in Additional Mathematics. So, I would like to dedicate this beautiful song to my beloved subject Additional Mathematics.

Lyrics to The Math Song :

(Jack) Math is a wonderful thing
Math is a really cool thing
So get off your "ath"
Let's do some math
Math(x5)

(Jack) Three minus Four is?
(Summer) Negative one
(Jack) That's right

(Jack) 6 times a billoin is?
(Carrot Top) 6 billion
(Jack) Nailed it

And 54 is 45 more then what is the answer Marta?
(Marta) 9
(Jack) No it's 8
(Marta) No it's 9
(Jack) Yes I was testing you it's 9 and that's a magic number

By: Jack Black lyrics

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