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OBJECTIVE
The aims of carrying out this project work are:
1. to apply and adapt a variety of problem-solving strategies to solve problems2. to improve thinking skills3. to promote effective mathematical communication4. to develop mathematical knowledge through problem solving in a way that increases
students interest and confidence
5. to use the language of mathematics to express mathematical ideas precisely6. to provide learning environment that stimulates and enhances effective learning7. to develop positive attitude towards mathematics
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Part 1
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 decoratingBaking 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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MATHEMATICS IN CAKE BAKING AND CAKE DECORATING
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 willhave. 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.
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 fordecoration.
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.
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In t i case, we use arit etic progression where the di erence ofthe diameteris constant
thatis 2. When the diameter decreases, the weight also decreases. Thatis the way how the
cake is balance to preventit from smooch. We can also use ratio, because when we prepare
the ingredient for each layer ofthe cake, we need to decrease its ratio from lowerlayerto
upperlayer. When we cutthe cake, we can use fraction to devide the cake according to thetotal people that will eatthe cake.
art 2
BestBakery shop received an order from your schoolto bake a 5 kg of round cake as shown
in Diagram 1 forthe Teachers Day celebration.
1)If a kilogram of cake has a volume of 38000cm3, and the height ofthe cake is to be7.0cm, the diameter ofthe baking tray to be used to fitthe 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
Diagram 1
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2)The inner dimensions of oven: 80cm length, 60cm width, 45cm heighta) 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 =
Table 1
b) i) h < 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 (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
=
= d
d =
d =
log d =
log d =
log h + log 155.53
Table of log d =
log h + log 155.53
Height,h Diameter,d
1.0 155.532.0 109.98
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 51.8410.0 49.18
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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 1.954.0 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 againstlog h
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c) 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 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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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 of cream
= 2(Area of left and right side surface)(Height of cream) + 2(Area of front and back sidesurface)(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
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 = (95 + 25)= 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
a) Based on the values above, the shape that require the leastamount of fresh cream to be used is
Pentagon-shaped cake, since it requires only 3135 cm of cream to be used.
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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
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I would choose not to bake a cake with such dimensions because its dimensions are not
suitable (the height is too high) and therefore less attractive. Furthermore, such cakes
are difficult to handle easily.
Diagram 2
Best Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, asshown 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) 14674.585Radius 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,
(it is a geometric progression with first term, a = 18116.772 and 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 tovolume: total volume < 57000 cm), so the maximum number of cakes that needs to be bakedis
Sn =
Sn = 57000, a = 18116.772 and r = 0.81
57000 =
1 0.81n = 0.597790.40221 = 0.81nog0.81 0.40221 = n
n =
n = 4.322therefore, n 4
Verifying the answer:When n = 5:S5 = (18116.772(1 (0.81)5)) / (1 0.81) = 62104.443 > 57000 (S n > 57000, n = 5 is notsuitable)
When n = 4:S4 = (18116.772(1 (0.81)
4)) / (1 0.81) = 54305.767 < 57000 (S n < 57000, n = 4 is suitable)
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Reflection
TEAM WORK IS IMPORTANT BE HELPFUL
ALWAYS READY TO LEARN NEW THINGS BE A HARDWORKING STUDENT
BE PATIE NT ALWAYSCONFIDENT
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Conclusion
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.
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
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 thegradient (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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Reference