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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 1

ADDITIONAL MATHEMATICS PROJECT WORK 2/

2011

NAME : ROSNI NISHA MYDIN PITCHAI

I/C NUMBER : 941019-10-6572

TEACHER NAME : PN AZNIZAH BT MUHAMAAD

SCHOOL NAME :SMK SEKSYEN 4 KOTA DAMANSARA


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CONTENT PAGE


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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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PART 1

USAGE OF MATHEMATICS

WHEN MAKING A CAKE


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PART I

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.

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

in a \\\\\\\\\\\\\\\\\\\\\\\


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MATHEMATICS IN CAKE BAKING AND CAKE DECORATING

y 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

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 long time 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.

y 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

produce


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y 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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Part 2

PART 2

MATHEMATICS SOLVE THE

PROBLEMS OF BAKER


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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 cake3800 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 of cake,V = 19000 is:

19000 = (3.142)(

)h

=

= d

d =

Height,h Diameter,d


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Table 1

b) i) h < 7cm is NOT suitable, because the resulting diameter produced is too large to fitinto the oven. Furthermore, the cake would be too short and too wide, making it less

attractive.

b) ii) The most suitable dimensions (h and d) for the cake is h = 8cm, d = 54.99cm, becauseit 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

=

= d

d =

2

1

53.155

! hd

log d =

log d =

log h + log 155.53

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 d =

log h + log 155.53

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 d o a ainst o 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 = 42cm

d = 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 of cream at the top surface

= Area of top surface x Height of cream

= (3.142)(

) x 1

= 2375 cm

Volume 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)(

)(8) x 1

= 1382.23 cm

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


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2 Triangle-shaped base

width

slant

height

19000 = base area x height

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 = )2595(22

= 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) + 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

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(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 itrequires only 3135 cm of cream to be used.


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PART 3

CALCULAS IN BAKING &

DECORATING CAKE !


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Part 3

When there's minimum or maximum, well, there's differentiation and quadraticfunctions. The minimum height, h and its corresponding minimum diameter, d is calculated

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 formulafor 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

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

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)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

I would choose notto bake a cake with such dimensions because its dimensions arenot suitable (the height is too high) andtherefore less attractive. Furthermore, such

cakes are difficultto handle easily.


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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,

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

Volume of 1st, 2

nd, 3

rd, and 4

thcakes:

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) 14674.585Radius 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,


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(it is a geometric progression with first term, a = 18116.772 and ratio, r= T2/T1 = T3 /T2 =

= 0.81)

FUTHER EXPLORATION

EXPLORE MATHS WHEN BAKE A

CAKE


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Further Exploration

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

log0.81 0.40221 = n

n =

n = 4.322

therefore, n 4

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)


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REFLECTION

When we work hard,

we can archivewhatever we want.

We should help peoplewho needs our help.

As a human, we shouldkeep on explore and

learn new things to

keep ourself up-to-

date.


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Self confident is

important to make a

person success.

Work in team is

better than work in

one .

Being patience

when doing a work

is a good thing .


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ReferenceWikipedia

www.one-school net

additional mathematics textbook form 4 and form 5

Additional Mathematics Project Work Auto Saved)

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