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ACKNOWLEDGE

First of all, I would like to say Alhamdulillah, for giving me strength and health to do this project

work.

Not forgotten my parents for providing everything, such as money, to buy anything that are

related to this project work and their advise, which is the most needed for this project.

Internet, books, computer and all that. They also supported me and encouraged me to

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

Then I would like to thank my teacher, Cik Siti Nurazlini binti Md Yusoff for guiding me and my

friend throughout this project. We had some difficulties in doing this task, but she taught uspatiently until we knew what to do. He tried and tried to teach us until we understand what we

supposed to do with the project work.

Last but not least, my friends who were doing this project with me and sharing our ideas. They

were helpful that when we combined and discussed together, we had this task done.

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OBJECTIVE

The aims of carrying out this project work are:

To apply and adapt a variety of problem solving strategies to solve problems

To improve thinking skills

To promote effective mathematical communication

To develop mathematical knowledge through problem solving in a way that increase

students interest and confidence

To use the language of mathematical ideas precisely

To provide learning environment that stimulates and enhances effective learning

To develop positive attitude towards mathematics

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Part One

INTRODUCTION

Cakes come in a variety of forms and flavours and among favourite deserts served during

special occasions such as birthday parties, Hari Raya, wedding and others. Cakes are treasured

not only because of their wonderful taste but also in the art of cake baking and cake decorating.

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 timer, provide basic math practise 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.

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MATHEMATICS IN CAKE BAKING AND CAKE DECORATING

GEOMETRY

To determine suitable dimension 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.

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

become too tall.This leads into the next situations.

The ratio of the surface area to the volumes determine how much crust a baked good will

have.The more surface area there is, compared to the volume,the faster the item will bake,andthe less inside there will be.For a very large,thick item, it will take a long time for the heat to

penetrate to the centre.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

ingredient would accumulate, and we would have a hard time scraping them into the batter.

CALCULUS (DIFFERENTATION)

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

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

maximum size of cake produced.

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PROGRESSION

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

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 is 2. 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 devide the cake according to the

total people that will eat the cake.

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Part TwoBest Bakery shop received an order from your school to bake a 5kg of round cake as shown in

Diagram 1 for the Teachers Day celebration.

1) If a kilogram of cake has a volume of 38000cm3, and the height of the cake is to be 7.0 cm,

the diameter of the baking tray to be used to fit the 5 kg cake ordered by your school 3800 is

Volume of 5kg cake = Base area of cake x Height of cake

3800 x 5 = (3.142)() x 7

(3.142) = (

)

863.872 = ()

= 29.392

= 58.784 cm

2) The inner dimensions of oven: 80cm length, 60cm width, 45cm height

a)The formula that formed for d in terms of h by using the formula for volume of cake,V = 19000 is:

19000 = (3.142)()h

=

=

=

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Table 1

b) i) < 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.

b) ii) The most suitable dimensions ( and d) for the cake is = 8cm, = 54.99cm, because itcan fit into the oven, and the size is suitable for easy handling.

c) i) The same formula in 2(a) is used, that is 19000 = (3.142)(

). The same process is also

used, that is, make d the subject. An equation which is suitable and relevant for the graph:

19000 = (3.142)()

=

=

=

=

log =

log =

log + log 155.53

Height, Diameter,1.0 155.53

2.0 109.98

3.0 89.79

4.0 77.76

5.0 69.55

6.0 63.49

7.0 58.78

8.0 54.99

9.0 51.84

10.0 49.18

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Table of log = log + log 155.53

Table 2

Height, Diameter, Log Log 1.0 155.53 0.00 2.19

2.0 109.98 0.30 2.04

3.0 89.79 0.48 1.95

4.0 77.76 0.60 1.89

5.0 69.55 0.70 1.84

6.0 63.49 0.78 1.80

7.0 58.78 0.85 1.77

8.0 54.99 0.90 1.74

9.0 51.84 0.95 1.71

10.0 49.18 1.0 1.69

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Graph of log against log

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ii) Based from the graph:

a) when = 10.5 cm = 10.5 cm, log = 1.021, log = 1.680, = 47.86 cm

b) when = 42 cm = 42 cm, log = 1.623, log = 1.140, = 13.80 cm

3) The cake with fresh cream, with uniform thickness 1cm is decorated

a) The amount of fresh cream needed to decorated the cake, using the dimensions Ive

suggested in 2(b)(ii)

My answer in 2(b)(ii) = 8cm, = 54.99cmAmount of fresh cream = volume of fresh cream needed (area height)Amount of fresh cream = volume of cream at the top surface + volume of cream at the side

surface

The bottom surface area of cake is not counted, because were decorating the visible part of

the cake only (top and sides). Obviously, we dont decorate the bottom part of the cake.

Volume of cream at the top surface

=Area of top surface

Height of cream

= (3.142) 1=2375

Volume of cream at the side surface

=Area of side surface Height of cream= (Circumference of cake Height of cake) Height of cream=2(3.142)(

)(8) 1

=1382.23

Therefore, amount of fresh cream = 2375 + 1382.23 = 3757.23

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b) Three other shapes (the shape of the base of the cake) for the cake with sameheight which is depends on the 2(b)(ii) and volume 19000.

The volume of top surface is always the same for all shapes (since height is same),

My answer (with = 8, and volume of cream on top surface = = 237

) :

1 Rectangle-shaped base (cuboid)

height

length width

1900 = base area height

base area =

length width = 2375

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

volume of cream

= 2(Area of left and right side surface)(Height of cream) + 2(Area of front and back side

surface)(Height of cream) + volume of top surface

=2(50 8)(1) + 2(47.5 8)(1) + 2375

=3935

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2 Triangle-shaped base

width

slant height

1900 = base area height

base area =

base area = 2375

length width = 2375

Length width = 4750

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

Slant length of triangle = = 98.23

Amount of cream

=Area of rectangular front side surface(Height of cream) + 2(Area of slant rectangular left/right

side surface)(Height of cream) + Volume of top surface

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

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3 Pentagon-shaped base

width

19000 = base areaheight

base area = 2375 = area of 5 similar isosceles triangles in a pentagon

2375 = 5(length width)

475 = length width

By trial and improvement, 475 = 25 19 (length = 25, width = 19)

amount of cream

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

=5(19 8) + 2375 = 3135

c)Based on the values above, the shape that required the least amount of fresh cream to be used is :

Pentagon-shapedcake,sinceit requiresonly 3135 ofcream to beused.

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Part ThreeWhen theres minimumor maximum, well, theres differentiation and quadratic functions.The minimum height, and its corresponding minimum diameter, is calculated by using thedifferentiation and function.

METHOD 1 : DIFFERENTIATION

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

amount ( volume ) of cream to be used for the round cake ( as in 3(a)).

1900

V 2

From :

Sub. into :

V 2

V

V 38000

0 minimum value, therefore, 0

6047.104

18.22

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sub. 18.22 into:

18.22

when 18.22, 2 2 36.44cm

METHOD 2 : QUADRATIC FUNCTIONS

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 volume of cream, radius of round cake:

1900

From :

factorize

completing the square, with and c 0

, min. value , correspondingvalue of

Sub. into :

19000

18.22

Sub. 18.22 into :

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19000

Therefore, 18.22cm,

I would choosenotto bakea cake withsuchdimensions becauseitsdimensionsarenot

suitable ( theheightistoohigh) andthereforelessattractive. Furthermore,such cakesare

difficulttohandleeasily.

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FurtherExploration

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 cake, the radius of the third cake is 10% less than

the radius of the second and so on.

Given:

height, h of each cake = 6 cm

radius of largest cake = 31 cm

radius of cake = 10% smaller than cakeradius of cake = 10% smaller than cake31, 27.9, 25.11, 22.599,

a = 31, r =

V = (3.142)

a)By using the formula fir volume V = (3.142)

h, with h = 6 to get the volume of cakes.

Volume of, , , and cakes:

Radius of cake = 31, Volume of cake = (3.142)(6) = 18116.772Radius of cake = 27.9, Volume of cake = (3.142)(6) = 14674.585Radius of cake = 25.11, Volume of cake = (3.142)(6) = 11886.414Radius of cake = 22.599, Volume of cake = (3.142)(6) = 9627.995

The volume form number pattern:

18116.772, 14674.585, 11886.414, 9627.995,

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(it is a geometric progression with first term, a = 18116.772 and ratio, r = / = / = =0.81)

a) The total mass of all the cakes should not exceed 15 kg (total mass < 15 kg, change tovolume: total volume < 57000 ), so the maximun number of cakes that needs to bebaked is

=

= 57000,a = 18116.772 and r = 0.81

= 0.59779

=

0.40221 =

n = 4.322 verifying the answer

therefore, n

verifying the answer:

when = 5:

= (18116.772(1-))/(1-0.81) = 62104.443 > 57000 ( > 57000, n = 5 is not suitable)

when n = 4:

= (18116.772(1-))/(1-0.81) = 54305.767 < 57000 ( is suitable)

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Reflection

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Conclusion

Geometry is the study of angles and triangles, perimeter, area and volume.It differsfrom algebra in that one develops a logical structure where mathematical relationships

are proved and applied. An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such

that the different of any two successive members of the sequence is a constant.

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.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 =

.The equation y = 2x will give you the gradient of y at any point along that curve.

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Reference

MWikipediaMwww.one-school.netMadditional mathematics textbook form 4 and form 5