Project Work 22011

8/4/2019 Project Work 22011

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Part I-Question

Cakes come in a variety of forms and flavours and are amongfavourite desserts servedduring special occasions such asbirthday parties, Hari Raya, weddings and etc. Cakes aretreasurednot only because of their wonderful taste but also inthe art of cake baking and cake decorating.

Baking a cake offers a tasty way to practice math skills, such as fractions and ratios, in areal-

world context. Many steps of baking a cake, such as counting ingredients and setting theoven

timer, provide basic math practice for young children. Older children and teenagers can usemore

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 notonly

improves your math skills, it helps you become a more flexible and resourceful baker.

Answer:

Geometry To determine suitable dimensions for the cake, to assist in designing and decoratingcakes that comes in many attractive shapes and designs, to estimate volume of cake to be produced

Calculus (differentiation) To determine minimum or maximum amount of ingredientsfor cake-baking, to estimate min. or max.amount of cream needed for decorating, to estimate min. ormax. size of cake produced.

Progressions To determine total weight/volume of multi-storey cakes with proportionaldimensions, to estimate total ingredients needed for cake-baking, to estimate total amount of creamfor decoration.


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Best Bakery shop received an order from your school to bake a 5 kg of round cake as showninDiagram 1 for the Teachers Daycelebration.

1) If a kilogram of cake has a volume of 3800, and the height of the cake is to be 7.0cm,

calculate the diameter of the baking tray to be used to fit the 5 kg cake ordered by yourschool.

[Use = 3.142]

Answer:

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


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2)The cake will be baked in an oven with inner dimensions of 80.0 cm in length, 60.0 cmin widthand 45.0 cm in height.

a) If the volume of cake remains the same, explore by using different values of heights,hcm,and the corresponding values of diameters of the baking tray to beused,d cm. Tabulate youranswers

Answer:

First, form the formulaford in terms ofh by using the above formula for volume of cake, V =19000, that is:

Volume of cake,V =V = 19000

19000 = (3.142)(d/2)h

=

= d

d =

Height,h (cm) Diameter,d(cm)

1.0 155.53

2.0 109.98

3.0 89.80

4.0 77.77

5.0 68.56

6.0 63.49

7.0 58.78

8.0 54.99

9.0 51.84

10.0 49.18


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(b)Based on the values in your table,

(i)state the range of heights that isnot suitable for the cakes andexplain your answers.

Answer:

h< 7cm is not suitable. This is due to 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.Moreover,it willdecrease the number of cake that can be sell.

(ii)suggest the dimensions that you think most suitable for the cake. Givereasons for your answer.

Answer:

Size of oven=

=

=

h = 8cm, d = 54.99cm, because it can fit into the oven, and the size is suitable for easy handling.

(c)(i) Form an equation to represent the linear relation betweenhand d. Hence, plot a suitable graphbased on the equation that you haveformed. [You may draw your graph with the aid ofcomputersoftware.]

Answer:

19000 = (3.142)(

)h

19000/(3.142)h =

= d

d =

d =

log d =

[make this equation equal to ]


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log d =

logh + log155.53

Logh 0 1 2 3 4

Log d 2.19 1.69 1.19 0.69 0.19

(ii)(a) IfBest Bakery received an order to bake a cake where the height of the cake is 10.5 cm, useyour graph to determine the diameter of the round cake pan required.


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Answer:

h = 10.5cm, logh = 1.021, log d = 1.680, d = 47.86cm

(b) IfBest Bakery used a 42 cm diameter round cake tray, use yourgraph to estimate the height of

the cake obtained.

Answer:

d = 42cm, log d = 1.623, logh = 1.140, h = 13.80cm

3)Best Bakery has been requested to decorate the cake with fresh cream. The thicknessof the creamis normally set to a uniform layer of about1cm

(a)Estimate the amount of fresh cream required to decorate the cake using thedimensions that youhave suggested in 2(b)(ii).

Answer:

h = 8cm, d = 54.99cmAmount of fresh cream = VOLUME of fresh cream needed (area x height)Amount of fresh cream = Vol. of cream at the top surface + Vol. of cream at the side surface

Vol. of cream at the top surface= Area of top surface x Height of cream

= (3.142)(

) x 1

= 2375 cm

Vol. 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)(54.99/2)(8) x 1

= 1382.23 cm

Therefore, amount of fresh cream = 2375 + 1382.23 = 3757.23 cm


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(b)Suggestthreeother shapes for cake, that will have the same height andvolume as those suggestedin 2(b)(ii). Estimate the amount of fresh cream tobe used on each of the cakes.

Answer:

1 Rectangle-shaped base (cuboid)

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/right side surface)(Height of cream) + 2(Area of front/back side surface)(Height ofcream) + Vol. of top surface

= 2(8 x 50)(1) + 2(8 x 47.5)(1) + 2375 = 3935cm


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2 Triangle-shaped base


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 sidesurface)(Height of cream) + Vol. of top surface

= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68cm

3 Pentagon-shaped base


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5-Octagon-shape base


base area = 2375 = area of 8similar isosceles triangles in a pentagon

therefore:

2375 = 8(length x width)

297 = length x width

By trial and improvement, 297 = 27 x 11(length = 27, width = 11)

Therefore, amount of cream

= 8(area of one rectangular side surface)(height of cream) + vol. of top surface

= 8(8 x 11) + 2375 =3079cm

(c)Based on the values that you have found which shape requires the leastamount of fresh cream tobe used?

Answer:

Hexagon shaped cake because since it requires only 2999 cm of cream to be used compare to

others.


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Find the dimension of a 5 kg round cake that requires the minimum amount of fresh creamtodecorate. Use at least two different methods including Calculus.State whether you would chooseto bake a cake of such dimensions. Give reasons for youranswers.

Answer:

Method 1: DifferentiationUse two equations for this method: the formula for volume of cake (as in Question 2/a), andthe formula for amount (volume) of cream to be used for the round cake (as in Question3/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):


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h =

h = 18.22

therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cm

Method 2: Quadratic Functions

Use the 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)

Sububstitu r = --h into (1):

19000 = (3.142)(--h)h

h = 6047.104

h = 18.22


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Substitute 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 asthe height is too much and it will also make it looks less attractive. Plus, such cakes like that aredifficult to handle.


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FURTHER EXPLORATION

Best Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, asshown inDiagram 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 is10% less thanthe radius of the second cake and so on.

(a) Find the volume of the first, the second, the third and the fourth cakes. By comparingall thesevalues, determine whether the volumes of the cakes form a number pattern?Explain and elaborate onthe number patterns.

Answer:

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

radius of 2nd cake =

radius of 3rd cake =

= 3.1 = 2.79

radius of 2nd cake = radius of 3rd cake =

=27.9 = 25.11

radius of 4th cake =

radius of 4th cake =

= 2.511 = 22.599

31, 27.9, 25.11, 22.599

a = 31, r =

V = (3.142)rh

Radius of 1st cake = 31, volume of 1st cake = (3.142)(31)(6) = 18116.772

Radius of 2nd cake = 27.9, vol. of 2nd cake =(3.142)(27.9)(6) = 14674.585

Radius of 3rd cake = 25.11, vol. of 3rd cake = (3.142)(25.11)(6) = 11886.414

Radius of 4th cake = 22.599, vol. of 4th cake =(3.142)(22.599)(6) = 9627.995


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18116.772, 14674.585, 11886.414, 9627.995,

a = 18116.772, ratio, r = = =

=

(b) If the total mass of all the cakes should not exceed 15 kg, calculate the maximumnumber ofcakes that the bakery needs to bake. Verify your answer using othermethods.

Answer:

Sn =

Sn = 57000, a = 18116.772 and r = 0.81

57000 =

1 0.81n = 0.59779

0.40221 = 0.81n

og0.810.40221 = n

n =

n = 4.322

therefore, n 4


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REFLECTION

In publishing this project,I have spent a few days of my holiday to finish this AdditionalMathematics project workof2011.I realize that this project work can help me in my trail SPMwhich for me quite difficult.But,although it is quite difficult,I still like this subject so much andwithout it,I cant fulfill my big dreams and wishes.

I loveAdditional Mathematics since first time we met as I also love mathemathic. It always makesme wonder and also make me happy. The thing that make me feel gay is when I can get through ahard question which there is no one can answer it except my teacher.It always an absolute obstacle

for me but I still thinking that I can go fo it. I enjoy sacrificing my precious time to have fun bydoing addmath questions. From Monday till Sunday, I always looking forward to make myAdditional Mathematics get into improvement so that my teacher show true satisfaction.

Plus, I also learn about true meaning of teamwork. My friends and I always share if one of us getuseful information in making this paperwork. This also encourage me to be more hardworking as itneed a lot of research.

Project Work 22011

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