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AcknowledgeI want to express our gratitude to all the people who have given their heart

whelming full support in making this compilation a magnificent experience.

First of all, grace be upon ALLAH, the almighty, with blessing,this Additional

Mathematics project work finally have been done. I would like to say Alhamdulillah,

for giving me the 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, computers and all that. In giving us not just

financial, but morally and spiritually. 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, Puan Hajah Mastura bt Yunus for guiding

me and my friends throughout this project. We had some difficulties in doing this

task, but she taught us patiently until we knew what to do. She 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:

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

y To improve thinking skills

y To promote effective mathematical communication

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

increases students interest and confidence

y To use the language of mathematics to express mathematical ideas precisely

y To provide learning environment that stimulates and enhances effective

learning

y To develop positive attitude towards mathematics


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(p a r t o n e)


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INTRODUCTION

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

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 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.


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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.

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 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 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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( p a r t t w o)


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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 1

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

d = 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 ofcake,

V = 19000 is:


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19000 = (3.142)(

)h

=

= d

d =

Table 1

b) i) h < 7cm is NOT suitable, because the resulting diameter produced is too large tofit

into the oven. Furthermore, the cake would be too short and too wide, making itless attractive.

ii) The most suitable dimensions (h and d) for the cake is h = 8cm, d = 54.99cm,because it can 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)(

)h. 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)(

)h

=

Height,h Diameter,d

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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0

= d

d =

d =

log d =

log d =

log h + log 155.53

Table of log d =

log h + log 155.53


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Table 2

Height,h Diameter,d Log h Log d

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 d against log h


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ii) Based on the graph:

a) d when h = 10.5cmh = 10.5cm, log h = 1.021, log d = 1.680, d = 47.86cm

b) h when d = 42cmd = 42cm, log d = 1.623, log h = 1.140, h = 13.80cm

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

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

I've

suggested in 2(b)(ii)

My answer in 2(b)(ii) ==> h = 8cm, d = 54.99cm

Amount of fresh cream = volume of fresh cream needed (area x 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 we're decorating the visible

part of the cake only (top and sides). Obviously, we don't decorate the bottom part of

the cake

Volume ofcream at the top surface

= Area of top surface x Height of cream

= (3.142)(

) x 1

= 2375 cm

Volume ofcream 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)(

)(8) x 1

= 1382.23 cm

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


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b) Three other shapes (the shape of the base of the cake) for the cake with same

height

which is depends on the 2(b)(ii) and volume 19000cm.

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

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

= 2375 cm):

1 Rectangle-shaped base (cuboid)

height

width

length

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 ofcream

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

= 3935 cm


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2 Triangle-shaped base

width

slantheight


base area =

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 = (95 + 25)= 98.23

Therefore, amount ofcream

= 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 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68 cm


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3 Pentagon-shaped base

width


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 ofcream

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

= 5(19 x 8) + 2375 = 3135 cm

c) Based on the values above, the shape that require the least amount of fresh cream

to be used is:

Pentagon-shaped cake, since it requires only 3135 cm ofcream to be used.


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(P a r t T h r e e)


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When there's minimum or maximum, well, there's differentiation and quadraticfunctions. The minimum height, h and its corresponding minimum diameter, d iscalculated by using the differentiation 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)).

19000 = (3.142)rh (1)

V = (3.142)r + 2(3.142)rh (2)

From (1): h =

(3)

Sub. (3) into (2):

V = (3.142)r + 2(3.142)r(

)

V = (3.142)r + (

)

V = (3.142)r + 38000r-1

(

) = 2(3.142)r (

)

0 = 2(3.142)r (

) -->> minimum value, therefore

= 0

= 2(3.142)r

= r

6047.104 = r

r = 18.22


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Sub. r = 18.22 into (3):

h =

h = 18.22

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

Method 2: Quadratic Functions

Two same equations as in Method 1, but only the formula for amount of cream is themain equation used as the quadratic function.

Let f(r) = volume of cream, r = radius of round cake:

19000 = (3.142)rh (1)

f(r) = (3.142)r + 2(3.142)hr (2)

From (2):

f(r) = (3.142)(r + 2hr) -->> factorize (3.142)

= (3.142)[ (r +

) (

) ] -->> completing square, with a = (3.142), b = 2h and c = 0

= (3.142)[ (r + h) h ]

= (3.142)(r + h) (3.142)h

(a = (3.142) (positive indicates min. value), min. value = f(r) = (3.142)h,

corresponding value of x = r = --h)

Sub. r = --h into (1):

19000 = (3.142)(--h)h

h = 6047.104

h = 18.22

Sub. h = 18.22 into (1):

19000 = (3.142)r(18.22)

r = 331.894

r = 18.22

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


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I would choose not to bake a cake with such dimensions because its dimensionsare not suitable (the height is too high) and therefore less attractive.Furthermore, such cakes are difficult to handle easily.

F u r t h e r

e x p l o r a t I o n


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Diagram 2

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.

Given: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 =

V = (3.142)rh,


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a)By using the formula for volume V = (3.142)rh, with h = 6 to get the volume of

cakes.

Volume of 1st, 2nd, 3rd, and 4th cakes:

Radius of 1st cake = 31, Volume of 1st cake = (3.142)(31)(6) = 18116.772

Radius of 2nd cake = 27.9, Volume of 2nd cake =

(3.142)(27.9)(6) 14674.585

Radius of 3rd cake = 25.11, Volume of 3rd cake =

(3.142)(25.11)(6) 11886.414

Radius of 4th cake = 22.599, Volume of 4th cake = (3.142)(22.599)(6) 9627.995

The volumes form number pattern:

18116.772, 14674.585, 11886.414, 9627.995,

(it is a geometric progression with first term, a = 18116.772and ratio, r = T2/T1 =

T3 /T2 = = 0.81)

b) The total mass of all the cakes should not exceed 15 kg ( total mass < 15 kg,

change to volume: total volume < 57000 cm), so the maximum number of

cakes that needs to be baked is

Sn =

Sn = 57000, a = 18116.772 and r = 0.81

57000 =

1 0.81n = 0.59779

0.40221 = 0.81n

og0.81 0.40221 = n

n =

n = 4.322

Therefore, n 4


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Verifying the answer:

When n = 5:

S5 = (18116.772(1 (0.81)5)) / (1 0.81) = 62104.443 > 57000 (Sn > 57000, n = 5 is not

suitable)

When n = 4:

S4 = (18116.772(1 (0.81)4)) / (1 0.81) = 54305.767 < 57000 (Sn < 57000, n = 4 is

suitable)

r e f l e c t I o n


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TEAMWORKISIMPORTANT BEHELPFUL

ALWAYSREADY TOLEARNNEW THINGS BE AHARDWORKINGSTUDENT

BEPATIENT ALWAYSCONFIDENT


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Conclusiony 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.

y 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

y 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

y Differentiation is essentially the process of finding an equation which will giveyou 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.


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Referencey Wikipedia

y www.one-school.net

y Additional mathematics textbook form 4

y Additional mathematics textbook form 5

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