Add Maths Project 2011,Hantar!!

8/2/2019 Add Maths Project 2011,Hantar!!

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CONTENTS

Num

Question Page1 Part I 2,3

Part II-Question 1 4Question 2(a) 5-Question 2(b) 6

2 -Question 2(c) 6,7,8-Question 3(a) 8,9-Question 3(b) 9,10,11,12-Question 3(c) 123 Part III 13,14

4 FurtherExploration

15,16,175 Reflection 186 Reference 19,20,21,22

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Additional mathematics project work (TASK 1)

Part ICakes come in a variety of forms and flavours and are amongfavourite desserts served during special occasions such as birthdayparties, Hari Raya, weddings and etc. Cakes are treasured not onlybecause of their wonderful taste but also in the art of cake bakingand cake decorating. Find out how mathematics is used in cakebaking and cake decorating and write about your findings.

Answer Part I :

Baking a cake offers a tasty way to practice math skills, such asfractions and ratios, in a real-world context. Many steps of baking acake, such as counting ingredients and setting the oven timer,provide basic math practice for young children. Older children andteenagers can use more sophisticated math to solve bakingdilemmas, such as how to make a cake recipe larger or smaller orhow to determine what size slices you should cut. Practicing mathwhile baking not only improves your math skills, it helps you becomea more flexible and resourceful baker.

When we make a cake with many layer, we must fix the difference ofdiameter of the two layer. So we can say that it used arithmeticprogression. When the diameter of the first layer of the cake is 8 andthe diameter of second layer of the cake is 6, then the diameter ofthe 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 decrease, the weight also decrease. That is theway how the cake is balance to prevent it from smooch. We can also

2
http://www.ehow.com/recipes/http://www.ehow.com/how_11464_make-yellow-cake.htmlhttp://www.ehow.com/recipes/http://www.ehow.com/how_11464_make-yellow-cake.html


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use ratio, because when we prepare the ingredient for each layer ofthe cake, we need to decrease its ratio from lower layer to upperlayer. When we cut the cake, we can use fraction to divide the cakeaccording to the total people that will eat the cake.

On a large scale you need to calculate what ratio of ingredients are

needed. You also need to be more mindful of the ordering of supplies.

We also use math when cooking and baking to estimate the cost of acertain dish. We can understand that cheesecake is more expensiveto make than a batch of cookies, particularly when people buyingredients such as flour, sugar, and butter in bulk and cream cheeseis more expensive. When comparing recipes, it may be beneficial toestimate the cost of each recipe. Mathematical skills are used quitefrequently when baking and cooking. It can be very helpful tounderstand how math affects the quality of culinary in order to makethe most delicious meals and treats.

Geometry is used to determine suitable dimensions for the cake, toassist in designing and decorating cakes that comes in manyattractive shapes and designs, to estimate volume of cake to beproduced, etc.

Differentation is used to determine minimum or maximum amount ofingredients for cake-baking, to estimate min. or max. amount ofcream needed for decorating, to estimate min. or max. size of cakeproduced, etc.

Progressions is used to determine total weight/volume of multi-storeycakes with proportional dimensions, to estimate total ingredientsneeded for cake-baking, to estimate total amount of cream for

decoration, etc.

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Part IIBest Bakery shop received an order from your school to bake a 5 kgof round cake as shown in Diagram 1 for the Teachers Daycelebration.

(1) If a kilogram of cake has a volume of 3800 cm3, and the heightof the cake is to be 7.0 cm, calculate the diameter of thebaking tray to be used to fit the 5 kg cake ordered by your

school.[Use = 3.142]

Solution :

If 1 Kg of cake = 3800 cm3 of volume,

Then 5 Kg of cake = 3800 5= 19 000 cm3

If h = 7 cm, then

Formulae for Volume, hrV 2= 7142.319000 2r=

7142.3

190002

=r

872.8632 =r872.8632 =r

392.29=r

Diameter, rd 2=)392.29(2=d

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784.58=d

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

(a) If the volume of cake remains the same, explore by usingdifferent values of heights, h cm, and the correspondingvalues of diameters of the baking tray to be used, d cm.Tabulate your answers.

Solution :

Given V = 19 000 cm3,

From the formulae, hrV 2=

hd

2

219000

=

419000

2hd=

hd

)4(190002

=

h

d

)4(190002 =

h

d

76000=

The table below constructed using the formulae,

hd

76000= for 451


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663.4933

3 2133.9386

1 3625.9210

4

7 58.7834 2233.1583

1 3725.5683

6

854.9868

3 2332.4294

7 3825.2296

9

951.8420

8 2431.7466

6 3924.9041

3

1049.1817

2 2531.1052

5 4024.5908

6

1146.8929

3 2630.5012

1 4124.2891

2

1244.8965

6 2729.9310

4 4223.9982

2

1343.1352

2 28 29.3917 4323.7175

3

1441.5661

4 29 28.8805 4423.4464

6

15

40.1567

1 30

28.3950

8 45

23.1844

8

(b) Based on the values in your table,(i) state the range of heights that is NOT suitable for

the cakes and explain your answers.

Solution :

h/cm d/cm h/cm d/cm h/cm d/cm

1155.526

3 1638.8815

6 3127.9333

4

2109.973

7 1737.7206

6 3227.4934

2

389.7931

2 1836.6578

9 3327.0736

5

477.7631

3 1935.6801

7 3426.6725

3

569.5534

5 2034.7767

3 3526.2887

3

6

63.4933

3 21

33.9386

1 36

25.9210

4

7 58.7834 2233.1583

1 3725.5683

6

854.9868

3 2332.4294

7 3825.2296

9

951.8420

8 2431.7466

6 3924.9041

3

1049.1817

2 2531.1052

5 4024.5908

6

1146.8929

3 2630.5012

1 4124.2891

2

1244.8965

6 2729.9310

4 4223.9982

2

1343.1352

2 28 29.3917 4323.7175

314 41.5661 29 28.8805 44 23.4464

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4 6

1540.1567

1 3028.3950

8 4523.1844

8

From the table above, h < 7 cm is NOT suitable

because any heights lower than 7 cm will result inthe diameter of the cake being too big to fit intothe baking oven. Furthermore, the cake will be tooshort and too wide, making it less attractive. Otherthan that height more than 45 cm will be too tall tobe fit into the baking oven.

(ii) suggest the dimensions that you think mostsuitable for the cake. Give reasons for your answer.

Solution :

I suggest the the height, h = 8 cm. It will produce acake with diameter, d = 54.99 cm. This cake willmore attractive and can fit into oven very well andthe size is suitable for easy handling.

(c) (i) Form an equation to represent the linear relationbetween h and d. Hence, plot a suitable

graph based on the equation that you have formed.

Solution :

19000 = (3.142)( )h

19000/(3.142)h =

= d

d =

d =

log d =

log d = log h + log 155.53

Log h 0 1 2 3 4

Log d 2.19 1.69 1.19 0.69 0.19

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(ii)(a) If Best Bakery received an order to bake a cake where the height of thecake is 10.5 cm, use your graph to determine the diameter of the round cakepan required.

Solution:

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

(b) If Best Bakery used a 42 cm diameter round cake tray, use your graph toestimate the height of the cake obtained.Answer:d = 42cm, log d = 1.623, log h = 1.140, h = 13.80cm

(3) Best Bakery has been requested to decorate the cake with

fresh cream. The thickness of the cream is normally set to a uniformlayer of about 1cm.

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(a) Estimate the amount of fresh cream required to decoratethe cake using the dimensions that you have suggested in2(b)(ii).

Solution :

Solution :

Volume of cream used= Vol. of cake with cream Vol. of actual cake

= hr2 19000

= 9)5.28)(142.3( 2 19000

= 22969 19000= 3969 cm3

(b) Suggest three other shapes for cake, that will have the sameheight and volume as those suggested in 2(b)(ii).

Estimate the amount of fresh cream to be used on each ofthe cakes.

9

r = 27.5 cm

h = 8 cm

Actual cake

r = 27.5 cm + 1cm = 28.5 cm

h = 8 cm + 1cm = 9cm

Actual cake with cream

Actual cake


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Solution :

(i) Square shaped cake

Volume of square shaped cake = w l h= l2 h (where w =

l)

l2 h = 19000

l2 =8

19000

l = 2375

l = 49 cm

Estimated volume of cream used= Vol. of cake with cream Vol. of actual cake

= l2 h 19000= (512 9) 19000= 23409 19000= 4409 cm3

(i) Triangle shaped cake

10

8 cm

9 cm

Actual size

Cake with cream

51 cm

8 cm


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Volume of triangle shaped cake

=2

1w l h

21 w 70 h = 19000 (assuming l = 70 cm)

w = 2708

19000

w = 68 cm


= 972702

1 19000

= 22680 19000= 3680 cm3

(i) Trapezium shaped cake

11

9 cm

8 cm


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Volume of triangle shaped cake

=2

1(a + b) l h

2

1(a + b) l h =19000 (assuming a =40 cm b

=50 cm)

l = 2908

19000

l = 53 cm


=2

1(42 + 52) 55 9 19000

= 23265 19000= 4265 cm3

(c) Based on the values that you have found which shape requiresthe least amount of fresh cream to be used?

12

9 cm


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Solution :

From the table above, the triangle shaped cake requires the leastamount of fresh cream.

Part IIIFind the dimension of a 5 kg round cake that requires the minimumamount of fresh cream to decorate. Use at least two differentmethods including Calculus. State whether you would choose to bakea cake of such dimensions. Give reasons for your answers.

Solution :

METHOD 1: Differentation

Type of cake Round Square Triangle Trapezium

Volume of creamneeded ( cm3 ) 3969 4409 3680 4265

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Assuming that the surface area of the cake is proportionate to theamount of fresh cream needed to decorate the cake.

Hence,Formula for the surface area

= rhr 22 +

h =2

142.3

19000

r

Surface area in contact with cream

=

+

2

2

142.3

190002

rrr

= + rr 380002

The values, when plotted into a graph will form a minimum value thatcan be obtained through differentiation.

0=dx

dy

+=

2

380002

rr

dx

dy

+=

2

3800020r

r

38000284.60 3 = r 3284.638000 r=

3104.6047 r= r=22.18

When 22.18=r , h = 18.22 cm

METHOD 2: Comparing Value of height against volume of thecream

h / cm r/cmVolume of cream used

/ cm3

1 77.76312594 19983.61097

2 54.98683368 10546.03779

3 44.89656169 7474.4213084 38.88156297 5987.368708

5 34.77672715 5130.07372

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6 31.74666323 4585.132889

7 29.39169891 4217.001202

8 27.49341684 3958.195683

9 25.92104198 3771.409389

10 24.59085959 3634.380579

11 23.44646466 3533.027734

12 22.44828085 3458.02429213 21.56761061 3402.958572

14 20.78306961 3363.284999

15 20.07835278 3335.696569

16 19.44078148 3317.733804

17 18.86032835 3307.53252

18 18.32894456 3303.65622

19 17.84008461 3304.981833

20 17.38836357 3310.620011

21 16.96930528 3319.858421

22 16.57915416 3332.120672

23 16.21473264 3346.936074

24 15.87333161 3363.91706325 15.55262519 3382.742114

From the table above, the minimum value of cream used is 3303.66cm3 when h = 18 cm.

When h = 18 cm, r = 18.3 cm.

I would not choose to bake a cake with such dimension becauseits dimensions are not suitable. For example the height is too high.

Furthermore such cakes are difficult to handle easily.

Further exploration

Best Bakery received an order to bake a multi-storey cake forMerdeka Day celebration, as shown in Diagram 2.

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The height of each cake is 6.0 cm and the radius of the largest cake is31.0 cm. The radius of the second cake is 10% less than the radius ofthe first cake, the radius of the third cake is 10% less than the radiusof the second cake and so on.(a) Find the volume of the first, the second, the third and the fourth

cakes. By comparing all these values, determine whether thevolumes of the cakes form a number pattern? Explain and

elaborate on the number patterns.

Solution :

Volume of cake 1

= hr2= 63131142.3 = 18116.8 cm3

Volume of cake 2

= hr

2

= 6)319.0(142.3 2 = 14676.6 cm3

Volume of cake 3

= hr2

= 6)319.09.0(142.3 2 = 11886.4 cm3

Volume of cake 4

= hr2

= 6)319.09.09.0(142.3 2 = 9628 cm3

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The values 18116.8, 14676.6, 11886.4 and 9628 form anumber pattern.

The pattern formed is a geometrical progression. This is proven that there is a common ratio between

subsequent numbers, r = 0.81

81.08.18116

6.14676=

81.06.14676

4.11886=

81.04.11886

9628

=

(b) If the total mass of all the cakes should not exceed 15 kg,calculate the maximum number of cakes that the bakery needs tobake. Verify your answer using other methods.

Solution :

8.01

)8.01(772.18116

1

)1(

=

=nn

nr

raS

15 kg = 57 000 cm3

2.0

)8.01(772.1811657000

n>

)8.01(772.1811611400 n> )8.01(629.0 n>

n8.0371.0 > n8.0371.0

Add Maths Project 2011,Hantar!!

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