ADDITIONAL MATHEMATICS PROJECT WORK

FORM 52011

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Curriculum Development Centre Ministry Of Education

CONTENT

No. Content Page1.0 INTRODUCTION2.0 PART I3.0 PART II

Question 1 Question 2 (a) Question 2 (b) Question 2 (c) Question 3 (a) Question 3 (b) Question 3 (c)

4.0 PART III5.0 FURTHER EXPLORATION6.0 REFLECTION7.0 REFERENCE

1.0 INTRODUCTION

Mathematics is important in almost all field in world. The mathematics of cooking often

goes unnoticed, but in reality, there is a large quantity of math skills involved in cooking

especially in baking and decoration of cake. Math has a lot to do with baking. Baking is

focused around different types of math. Baking can teach math and they will not even

realize it

Baking a cake required two main techniques, which is cooking technique and of course

measuring technique. Measuring technique for example measuring the amount sugar,

flour, amount of water, shape of cake, temperature needed for the cake to bake, setting the

oven timer, provide basic math practice for young children.

Besides, for decorative purposes, most cake will be covered with cream. Decorating with

cream needs mathematics. Well cake bakers if they're making a symmetrical cake need to

get all the angles of the cream decorations right so they don’t want one side to have a 30

degree angle and the other to have a s0 degree angle of cream. That would make cakes to

become less attractive.

The math involved in baking is a much simple mathematical system. A person only needs a

basic knowledge of math in order to succeed. This is because even though there are many

math calculations, it is all simple measurements and most can be done with a calculator if

needed. Always double-check ones’ math in order to not ruin the dish that is being

prepared. 

2.0 – PART I

There are several uses of Mathematics in cake baking and decorating such as measurement of ingredients, calculation of price and estimated cost, estimation of dimensions, cake decorating, calculation of baking time, and modification of recipe according to scale

Amount of ingredients used should be accurate to produce a perfect cake. Measuring the amount of ingredients such as, baking soda, flour, sugar, should be in proper way to maintain the taste of the cake

Before a cake is being produce, a rough calculation is needed to be calculate such as the cost needed to buy ingredients, cost to buy other materials such as aluminium foil, and of course the amount of electricity needed to bake a cake.

Before baking cake, size of cake should be estimated. These included the volume, height, diameter and shape of cake. Cakes may come different heights and shape such as triangle, round, trapezium and, rectangle to meet the requirement of customers.

Cake decorating is one of the frosting and other edible decorative elements to make otherwise plain visually interesting. Alternatively, cakes can be mould into three-dimensional persons, places and things. In many areas of the world, decorated cakes are often a focal point of a special celebration such as a bridal shower, wedding, or anniversary.

Time needed to bake a cake is needed to estimate before baking. This is because, size of cake affect the baking time. Therefore estimated time should be calculated to make sure cake is not over baked

Modification of recipe according to scale require mathematics skills. Most recipes give guidelines as to how much a single batch will produce.  But what if you want more?  It seems too time consuming to mix up another batch.  What if the recipe makes only small rectangular size cakes and you need to double the size?  Clearly, double is two times more than original size, so we can multiply all the ingredients by two to make a larger batch

3.0 PART II

3.1 – Question 1

1 kg = 3800 cm3

h = 7 cm5 kg = 3800 x 5 = 19000 cm3

V=πr2 h19000 = 3.142 x r2 x 7

r2=1900

3.142 Χ 7

r2 = 863.872 cmr = 29.392 cmd = 2rd = 58.783 cm

3.2 – Question 2

The cake is in round shape. Therefore the maximum value ofd = 60.0 cmh = 45.0 cm

a) Different height from 1.0 cm to 45.0 cm is used to calculate the diameter. All the values is tabulated below

h/cm d/cm1.0 155.52625192.0 109.97366743.0 89.793123394.0 77.763125945.0 69.55345436.0 63.493326457.0 58.783397838.0 54.986833689.0 51.8420839610.0 49.1817191911.0 46.8929293212.0 44.89656169

13.0 43.1352212214.0 41.5661392315.0 40.15670556

16.0 38.8815625717.0 37.7206567118.0 36.6578891219.0 35.6801692120.0 34.7767271521.0 33.9386105622.0 33.1583083123.0 32.4294652824.0 31.7466632325.0 31.1052503726.0 30.5012074327.0 29.9310411328.0 29.3916989129.0 28.8804999430.0 28.39507881

31.0 27.9333394432.0 27.4934168433.0 27.0736453734.0 26.6725321535.0 26.288734736.0 25.9210419537.0 25.5683583138.0 25.229689639.0 24.9041315840.0 24.5908595941.0 24.2591198342.0 23.9982216743.0 23.7175310644.0 23.4464646645.0 23.18448477

b) (i) The range of height that is not suitable :

h < 7 and h > 45

As i stated in (a), the maximum value of diameter should be 60 cm and

maximum value of height is 45 cm. Any heights lower than 7 cm will result in the

diameter of the cake being too big to fit into the baking oven while any heights

higher than 45 cm will cause the cake being too tall to fit into the baking oven.

(ii) The dimensions that i think most suitable for the cake is 29 cm in height and

diameter of 28.88049994 cm which is approximately 29 cm. This dimension is

most suitable because it is more symmetrical, balanced and easier to be

decorated.

c) (i) V=πr2 h

V=19000 cm3

d=2r

r=d2

19000cm3=3 . 142 Χd2h

19000cm3=3 . 142 Χ ( d2 )2

h

19000cm3=3 . 142 Χd2

4h

d2=19000cm3 Χ 43 . 142 Χh

d2=76000cm3

3 .142 Χh

Value of h, 1/h and d2 is calculated and tabulated below based on the above equation.

h 1h

d2 h 1h

d2 h 1h

d2

1 1.0000 24188.415 19 0.0526 1273.074 37 0.0270 653.741

2 0.5000 12094.208 20 0.0500 1209.421 38 0.0263 636.537

3 0.3333 8062.805 21 0.0476 1151.829 39 0.0256 620.216

4 0.2500 6047.104 22 0.0455 1099.473 40 0.0250 604.710

5 0.2000 4837.683 23 0.0435 1051.670 41 0.0244 539.961

6 0.1667 4031.403 24 0.0417 1007.851 42 0.0238 575.915

7 0.1429 3455.488 25 0.0400 967.537 43 0.0233 562.521

8 0.1250 3023.552 26 0.0385 930.324 44 0.0227 549.737

9 0.1111 2687.602 27 0.0370 895.867 45 0.0222 537.520

10 0.1000 2418.842 28 0.0357 863.872 46 0.0217 525.835

11 0.0909 2198.947 29 0.0345 834.083 47 0.0213 514.647

12 0.0833 2015.701 30 0.0333 806.281 48 0.0208 503.925

13 0.0769 1860.647 31 0.0323 780.271 49 0.0204 493.641

14 0.0714 1727.744 32 0.0313 755.888 50 0.0200 483.768

15 0.0667 1612.561 33 0.0303 732.982 51 0.0196 474.283

16 0.0625 1511.776 34 0.0294 711.424 52 0.0192 465.162

17 0.0588 1422.848 35 0.0286 691.098 53 0.0189 456.385

18 0.0556 1343.801 36 0.0278 671.900 54 0.0185 447.934

0 0.030.060.090.120.150.180.210.240.27 0.3 0.330.360.390.420.450.480.510.540.57 0.6 0.630.660.690.720.750.780.810.840.87 0.9 0.930.960.990

80016002400320040004800560064007200800088009600

10400112001200012800136001440015200160001680017600184001920020000208002160022400232002400024800

d2

1h

(ii) a) Based on the graph,

When h = 10.5 cm,

1h = 0.095

From the graph (red line), when

1h = 0.095 , d2 = 2300

When d2 = 2300, d = √2300 , d = 47.95cm Therefore the value of d is approximately 48 cm

b) Based on the graph, When d = 42 cm , d2 = 1764

From the graph (blue line), when d2 = 1764,

1h = 0.075

h =

10 .0075 = 13.333cm

Therefore the value of h is approximately 13.3cm

3.3 – Question 3a) From the dimensions that i have suggested in 2(b)(ii), the height of the cake is 29 cm,

the diameter is 28.88049994 cm and the radius is

28 .88049994 2 = 14.44 cm

The volume of cake without cream :h = 29.0 cm , r = 14.44 cm

Therefore volume of cake without cream is 19000 cm3

14.44 cm

29.0 cm

The volume of cake with cream :

The volume of cake with cream, V = 3.142 x (15.44)2 x (30) = 22471 cm3

Therefore the amount cream used is :Vcream = 22471 cm3 – 19000 cm3

= 3471 cm3

b) (i) Triangle Shaped Cake Cake without cream :

Volume of cake without cream : V = 19000 cm3

1 cm

36.2 cm36.2 cm

29 cm

30 cm

1 cm

15.44 cm

Cake with cream :

(2.5 cm is obtained by scale drawing) Front view of cake with cream

Plan view of cake with cream Volume of cake with cream :

V =

12 x 39.7 cm x 39.7 cm x 30 cm

= 23641.4 cm3

Therefore the amount of cream needed for triangular shape cake is Vcream = 23641.4 cm3 – 19000 cm3

= 4641.4 cm3

(ii) Trapezium Shaped Cake Cake without cream :

Volume of cake without cream V = 19000 cm3

Cake with cream :

(Value 2 cm for each side is obtained by scale drawing) Plan view of cake Volume of cake with cream :

V =

12 x (33.6 cm + 16.8 cm) x 31.9 cm x 29 cm

= 23312.5 cm3 Therefore the amount of cream needed for triangular shape cake is Vcream = 23312.5 cm3 – 19000 cm3

= 4312 cm3

(ii) Rectangle Shaped Cake Cake without cream :

Volume of cake without cream : V = 19000 cm3

Cake with cream :

Plan View Front ViewVolume of cake with cream :V = 27.6 cm x 27.6 cm x 30 cm = 22852.8 cm3 Therefore the amount of cream needed for triangular shape cake is Vcream = 22852.8 cm3 – 19000 cm3

= 3852.8 cm3

*(All the values are only estimated based on the assumption that the layer of cream isuniformly thick at 1 cm )

c) Based on the values of cream that i obtained from four shapes of cakes, round cake requires the least amount of fresh cream to be used which is 3471 cm3

4.0 – PART III

Method 1Differentiation

Volume of Cake without creamV = πr2h19000 = 3.142r2h -------------------①

Volume of Cream = Surface area of cylinderV = πr2 + 2 πrhV = 3.142r2 + ( 2 x 3.142 x rh)V = 3.142r2 + 6.284rh -------------------②

From equation ①h = 19000 -------------------③ 3.142r2

Substitute Equation ③ into ②

V = 3.142r2 + 6.284r 19000 -------------------③ 3.142r2

V = 3.142r2 + 38000 rV = 3.142r2 + 38000r-1

Differentiate V respect to rdV = 2(3.142)r – 38000 ------------------- Minimum value turning point, dV = 0 dr r2 dr

0 = 2(3.142)r – 38000 r2

38000 = 6.284r r2

38000 = 6.284r3

r = 18.22 cm

Therefore, substitute the value or r = 18.22 into Equation ③ = __19000__ 3.142(18.22)2

= 18.22cmThe dimension are r = 18.22cm , h= 18.22cmMethod 2Quadratic Function

Let f(r) = Volume of cream

r=3√380006 . 284

r = radius of round cake

Use same equation as above methodVolume of Cake without creamf(r) = πr2h19000 = 3.142r2h -------------------①

Volume of Cream = Surface area of cylinderf(r) = πr2 + 2 πrhf(r) = 3.142r2 + ( 2 x 3.142 x rh)-------------------②Factorize Equation ②f(r) = 3.142(r2 + 2rh)Do completing the square for the above equationf(r) = 3.142 r + 2h 2 – 2h 2 2 2

f(r) = 3.142(r + h)2 – 3.142h2

Since the value of a = 3.142, thus The minimum value = -3.142h2

When r = -hTherefore substitute r into Equation ①19000 = 3.142(-h)2h19000 = 3.142h3

h=3√190003 .142

h = 18.22 cm

So, to find the value of r, substitute the value of h = 18.22 into Equation ①19000 = 3.142r2(18.22)

r=√190003 .142 Χ 18 .22

r = 18.22cm

The dimension are r = 18.22cm , h= 18.22cm

I will not choose to bake a cake of such dimensions because the height is 18.22cm which is quite tall. This makes the cake less attractive. Furthermore, the diameter of the cake is 2 x 18.22 = 36.44 cm which is quite small to withstand such a tall cake and makes the cake to become unstable. Since the cake is too tall, it is too hard to handle the cake.

5.0 FURTHER EXPLORATION

a) From the question, it is stated that :

The height, h of each cake is 6cmThe radius of largest cake is 31cmThe radius of 2nd cake is 10% smaller than 1st cakeThe radius of 3rd cake is 10% smaller than 2nd cake

Therefore,The radius of 2nd cake is 90% X 31 cm = 27.9cmThe radius of 3rd cake is 90% X 27.90 cm = 25.11The radius of 4th cake is 90% X 25.11 cm = 22.60

The progression will be

31, 27.9, 25.11, 22.60…

a = 31, r = 9

10

V = 3.142r²h

Volume of 1st cake = (3.142)(31)²(6) = 18116.772Volume of 2nd cake =(3.142)(27.9)²(6) =  14674.585Volume of 3rd cake =(3.142)(25.11)²(6) =  11886.414Volume of 4th cake =(3.142)(22.60)²(6) =  9627.995

18116.772, 14674.585, 11886.414, 9627.995, …

a = 18116.772, r = 0.81

The volume of cake shows a geometric progression

b) From Part I, the question stated that “a kilogram of cake has a volume of 3800cm3”This means that 15kg of cake will have : 15 kg X 3800cm3 = 57000cm3

Therefore Sn=

a (1−rn)1−r , where Sn = 57000 ; a = 18116.772 ; r = 0.81

57000=18116 .772 (1−0 .81n)

1−0 .811083018116. 772

=1−0 .81n

0 .81n=1−0 .597790 .81n=0 . 40221nlog10 0.81 = log10 0.40221n = log10 0.40221 log10 0.81 n = 4.322n 4

So, the maximum number of cakes that the bakery needs to bake is approximately 4 cakes

6.0 REFLECTION

Additional Mathematics project work taught me billions of creative work. Before conducting this project, I found that cakes are delicious in taste, but after completing this project work, I found that baking cakes involves lots of calculation, baking cakes involves mathematics, baking cakes involve creativeness, and baking cakes need hardworking. Round cakes looks normal, but how about a Mickey Mouse cake, or superman cake or maybe mathematical symbol cake. Looks creative but involve mathematics. Everything is mathematics. Thank to my entire additional mathematics teacher because I can apply all the formulas and technique that they taught me in class. Most important thing here is, this project taught me how to design a cake with different shape and I’ve try 3 types, Triangle, rectangle and trapezium. I almost spend more than 2days to design those 3 cakes and its works out, making me good in drawing.

In future, maybe I will design a cake something like this using the knowledge that I got from this project work.