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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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 2
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 5
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 7
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 8
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 9
Part 2
PART 2
MATHEMATICS SOLVE THE
PROBLEMS OF BAKER
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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 10
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 12
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 14
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 16
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 18
PART 3
CALCULAS IN BAKING &
DECORATING CAKE !
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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 19
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 21
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 22
(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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 23
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 25
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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ADDITIONAL MATHEMATICS PROJECT WORK 2/2011 ROSNI Page 26
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
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