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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
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 = (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
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 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