add mathematics 2011

8/6/2019 Add Mathematics 2011

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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


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


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

add mathematics 2011

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