57266259 kerja kursus matematik tambahan 2011

8/6/2019 57266259 Kerja Kursus Matematik Tambahan 2011

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ACKNOWLEDGEFirst of all, 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. They also supported me and encouraged me to

this task so that I will not procrastinate in doing it.

Then I would like to thank my teacher, Encik Mohamed bin Masri for guiding me

and my friends throughout this project. We had some difficulties in doing this task, but he taught

us patiently until we knew what to do. He 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:

to apply and adapt a variety of problem-solving strategies to solve problems

to improve thinking skills

to promote effective mathematical communication

to develop mathematical knowledge through problem solving in a way that increases

students interest and confidence

to use the language of mathematics to express mathematical ideas precisely

to provide learning environment that stimulates and enhances effective learning

to develop positive attitude towards mathematics


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INTRODUCTION

Cakes come in a variety of forms and flavours and are among favourite desserts served duringspecial 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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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 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.

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.


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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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Best Bakery shop received an order from your school to bake a 5 kg of round cake as shown inDiagram 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 of cake,V = 19000 is:

19000 = (3.142)()h

=

= d


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d =

Table 1

b) i) h < 7cm is NOT suitable, because the resulting diameter produced is too large to fit into theoven. 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 itcan 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 =

Height,h Diameter,d

1.0 155.53

2.0 109.983.0 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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log d =

log h + log 155.53

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 log d against log h


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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

widthlength

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 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


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

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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When there's minimum or maximum, well, there's differentiation and quadratic functions.The minimum height, h and its corresponding minimum diameter, d is calculated by using thedifferentiation and function.

Method 1: Differentiation

Two equations for this method: the formula for volume of cake (as in 2(a)), and the formula foramount (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 mainequation 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 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.


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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 thesecond cake is 10% less than the radius of the first cake, the radius of the third cake is 10% lessthan the radius of the second cake and so on.

Given:height, h of each cake = 6cmradius of largest cake = 31cm

radius of 2nd

cake = 10% smaller than 1st

cakeradius 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, 2nd, 3rd, and 4th cakes:

Radius of 1st cake = 31, volume of 1st cake = (3.142)(31)(6) = 18116.772Radius 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) 11886.414Radius of 4th cake = 22.599, volume of 4th cake = (3.142)(22.599)(6) 9627.995

The volumes form number pattern:


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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 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

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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TEAM WORK IS IMPORTANT BE HELPFUL

ALWAYS READY TO LEARN NEW THINGS BE A HARDWORKING STUDENT

BE PATIENT ALWAYS CONFIDENT


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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 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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ReferenceWikipedia

www.one-school net

additional mathematics textbook form 4 and form 5

57266259 kerja kursus matematik tambahan 2011

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