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ADDITIONAL MATHEMATICS PROJECT WORK 

2011

  TITLE: CAKE

NAME: YAASHINIE K.KANNAN

I/C : 940209-01-5474

CLASS: 5 AL-BIRUNI

TEACHER: PN UMI KALSOM MOHD YUSOF

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FIRST OF ALL, I WOULD LIKE TO THANK GOD THAT I HAD

SUCEEDED IN FINISHING THIS PROJECT WORK.

I WOULD ALSO LIKE TO EXPRESS MY GRATITUDE TO MY

ADD MATHS TEACHER, PN UMI KALSOM MOHD YUSOF FOR HER

HELPING HAND IN COMPLETING THIS PROJECT WORK WITH

SUCCESS. I APPRECIATE ALL THE HARD WORK THAT SHE PUT IN

HELPING ME DURING THE PROCESS OF COMPLETING THIS PROJECT 

WORK.

I WOULD ALSO THANK BOTH MY PARENTS IN THEIR FULL

SUPPORT DURING THURING THE PROCESS IN COMPLETING THIS

PROJECT WORK. I SINCERELY APPRECIATE THEIR COMMITMENT IN

SUPPLYING THE THINGS I NEEDED THROUGHOUT THIS PROCESS.

MY FELLOW PEERS HAD ALSO HELPED ME A LOT IN

FINISHING THIS PROJECT WORK. I WOULD LIKE TO THANK EACH

AND EVERYONE WHO HELPED ME TO COMPLETE THIS PROJECT.

THANK YOU SO MUCH FOR SPENDING TIME IN EXPLAINING AND

DISCUSSING ABOUT THE COURSE WORK WITH ME.

FINALLY, I WOULD LIKE TO EXPRESS MY GRATITUDE TO

ALL THOSE WHO GAVE ME THE POSSIBILITY TO COMPLETE THIS

PROJECT WORK SUCCESSFULLY.

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TO APPLY AND ADAPT A VARIETY OF PROBLEM

SOLVING SRATEGIC TO SOME PROBLEMS.

TO IMPROVE THINKING SKILLS

TO PROMOTE EFFECTIVE MATHEMATICAL

COMMUNICATION

TO DEVELOP MATHEMATICAL KNOWLEDGE

THROUGH PROBLEM SOLVING IN A WAY THAT 

INCREASES STUDENT’S 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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HISTORY OF CAKE

Cakes are made from various combinations of refined flour, some form of 

shortening, sweetening, eggs, milk, leavening agent, and flavoring. There areliterally thousands of cakes recipes (some are bread-like and some rich and

elaborate) and many are centuries old. Cake making is no longer a complicatedprocedure.

Baking utensils and directions have been so perfected and simplified that even

the amateur cook may easily become and expert baker. There are five basictypes of cake, depending on the substance used for leavening.

The most primitive peoples in the world began making cakes shortly after they

discovered flour. In medieval England, the cakes that were described in writingswere not cakes in the conventional sense. They were described as flour-basedsweet foods as opposed to the description of breads, which were just flour-based

foods without sweetening.

Bread and cake were somewhat interchangeable words with the term "cake"

being used for smaller breads. The earliest examples were found among the

remains of Neolithic villages where archaeologists discovered simple cakes madefrom crushed grains, moistened, compacted and probably cooked on a hot stone.

Today's version of this early cake would be oatcakes, though now we think of 

them more as a biscuit or cookie.

Cakes were called "plakous" by the Greeks, from the word for "flat." These cakes

were usually combinations of nuts and honey. They also had a cake called"satura," which was a flat heavy cake.

During the Roman period, the name for cake (derived from the Greek term)became "placenta." They were also called "libum" by the Romans, and were

primarily used as an offering to their gods. Placenta was more like a cheesecake,baked on a pastry base, or sometimes inside a pastry case.

The terms "bread" and "cake" became interchangeable as years went by. Thewords themselves are of Anglo Saxon origin, and it's probable that the term cake

was used for the smaller breads. Cakes were usually baked for special occasions

because they were made with the finest and most expensive ingredients available

to the cook. The wealthier you were, the more likely you might consume cake ona more frequent basis.

By the middle of the 18th century, yeast had fallen into disuse as a raising agentfor cakes in favor of beaten eggs. Once as much air as possible had been beaten

in, the mixture would be poured into molds, often very elaborate creations, butsometimes as simple as two tin hoops, set on parchment paper on a cookie

sheet. It is from these cake hoops that our modern cake pans developed.

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PART1

 

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Geometry ± To determine suitable dimensions forthe cake, to assist in designing anddecorating cakesthat comes in many attractive shapes and designs, toestimate volume of cake to be produced. 

Calculus (differentiation) ± To determine minimum ormaximum amount of ingredients for cake-baking, toestimate min. or max. amount of cream needed fordecorating, to estimate min. or max. size of cake

produced.

Progressions ± To determine total weight/volume of multi-storey cakes with proportionaldimensions, toestimate total ingredients needed for cake-baking, toestimate total amountof cream for decoration.

Part 2Q1) Volume of 5kg cake = Base area of cake xHeight of cake3800 x 5 = (3.142)(d/2)² x 7

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19000/7(3.142)= (d/2)²863.872 = (d/2)²d/2 = 29.392d = 58.784 cm 

Q2)(a)19000 = (3.142)(d/2)²h19000/(3.142)h= d²/424188.415/h = d²d = 155.53/¥h 

HEIGHT,h(cm) DIAMETER,d(cm)

1.0 155.53

2.0 109.983.0 89.80

4.0 77.76

5.0 69.56

6.0 63.49

7.0 58.78

8.0 54.99

9.0 51.84

10.0 49.18

(b)(i)

h < 7cm is NOT suitable, because the resultingdiameter produced is too large to fit into the oven.Furthermore, the cake would be too short and toowide, making it less attractive.

(b)(ii)h = 8cm, d = 54.99cm, because it can fit into theoven, and the size is suitablefor easy handling.

(c)(i)

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19000 = (3.142)(d/2)²h19000/(3.142)h = d²/424188.415/h = d²d = 155.53/¥hd = 155.53h-1/2

log d = log 155.53h-1/2log d = -1/2 log h + log 155.53

log h log d0.00 2.190.30 2.040.48 1.950.60 1.89

0.70 1.840.78 1.800.85 1.770.90 1.740.95 1.711.00 1.69

 ii) Use the graph you've drawn to determine:a) d

when h = 10.5cm h = 10.5cm, log h = 1.021, log d =1.680,d = 47.86cm  b) h when d = 42cm d = 42cm, log d = 1.623, log h= 1.140,h = 13.80cm 

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Q3)

Decorate the cake with fresh cream, with uniformthickness 1cm. a) Estimate the amount of freshcream needed to decorate the cake, using the

dimensions you'vesuggested in Q2/b/ii My answer inQ2/b/ii ==> h = 8cm, d = 54.99cm Amount of freshcream = VOLUME of fresh cream needed (area xheight)Amount of fresh cream = Vol. of cream at thetop surface + Vol. of cream at the side surface(The bottom surface area of cake is NOT COUNTED,because we're decorating the visiblepart of the cakeonly (top and sides). Obviously, we don't decorate

the bottom part of thecake, right?)Vol. of cream at the top surface= Area of top surface

x Height of cream= (3.142)(54.99/2)² x 1=2375 cm³Vol. of cream at the side surface= Area of side

surface x Height of cream= (Circumference of cake xHeight of cake) x Height of cream= 2(3.142)(54.99/2)(8) x 1=

1382.23 cm³Therefore, amount of fresh cream = 2375 + 1382.23=3757.23 cm³  b) Suggest THREE other shapes (the shape of thebase of the cake) for the cake with sameheight(depends on the Q2/b/ii) and volume(19000cm³). Estimate the amount of fresh cream

(thevolume) to be used for each of those cakes. 

NOTE: Circles may NOT be accepted, because it's

already part of the questions earlier, so it'ssafe to

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provide three different shapes for thisquestion. Depends on your choice of shapes, butusually the volume of top surface is always the samefor all shapes (since height is same), therefore your

 job is to first find out the lengths and widths of the

base shape, then find vol. of side surfaces only. Thisprocess can be quite difficult, especiallyif you chooseshapes that has more than 4 sides, such as pentagonor hexagon. By the way, drawthe shapes that youchoose for each question, so that you'll betterunderstand the area involvedfor cake-decorating. Myanswer (with h = 8cm, and volume of cream on topsurface = 19000/8 = 

2375 cm³): 1 ± Rectangle-shaped base (cuboid) (draw cuboid) 19000 = base area x height

 

base area = 19000/8length x width = 2375By trialand 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)(Heightof cream) + Vol. of top surface= 2(8 x 50)(1) + 2(8 x47.5)(1) + 2375 =3935 cm³ 2 ± Triangle-shaped base (draw an isosceles triangle, then make it a 3D shape

by drawing vertical lines on the vertices.Then join thevertices to form the top surface of the cake) 19000 =base area x height base area = 2375½ x length xwidth = 2375length x width = 4750By trial andimprovement,4750 = 95 x 50 (length = 95, width =50)Slant length of triangle = ¥(95² + 25²)=

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98.23Therefore, amount of cream= Area of rectangular front side surface(Height of cream) +2(Area of slant rectangular left/rightside surface)(Height of cream) + Vol. of top surface= (50 x 8)(1)+ 2(98.23 x 8)(1) + 2375 =

4346.68 cm³ 3 ± Pentagon-shaped base (draw a regular pentagon, then draw vertical lineson its vertices, then join the vertices to formthe topsurface of cake) 19000 = base area x height basearea = 2375 = area of 5 similar isosceles triangles ina pentagontherefore:2375 = 5(length x width)475 =

length x widthBy trial and improvement, 475 = 25 x19 (length = 25, width = 19)Therefore, amount of cream= 5(area of one rectangular side surface)(height of cream) + vol. of top surface= 5(8 x 19) +2375 =3135 cm³ (all the answers above may differ from yours,depending on how do you do the calculations) c)

Based on the values above, determine the shape thatrequire the least amount of fresh cream to beused. (depends on your answers, but here's myobvious answer) Pentagon-shaped cake, since itrequires only 3135cm³ of cream to be used. Part

3(Find dimensions of 5kg ROUND cake (volume:19000cm³) that require minimumamount of cream todecorate. Use two different methods, includingCalculus

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differentiation/integration). Also, explain whether youwould choose to bake that cake with suchdimensions

and give reasons why. When there's minimum´ ormaximum´, well, there's differentiation and quadraticfunctions.Use both to find the minimum height, h and its corresponding minimum diameter, d. (Note: This question only gives you just ONE infoonly: the mass of cake (which you can changeinto

volume of cake, as shown), so use this ONE info onlyto find the minimum dimensions) Here's my answer: Method 1: DifferentiationUse two equations for this method: the formula for

volume of cake (as in Q2/a), and the formulaforamount (volume) of cream to be used for the roundcake (as in Q3/a).19000 = (3.142)r²h (1)V =(3.142)r² + 2(3.142)rh (2)From (1): h = 19000/

(3.142)r² (3)Sub. (3) into (2):V = (3.142)r² +2(3.142)r(19000/(3.142)r²)V = (3.142)r² + (38000/r)V= (3.142)r² + 38000r-1dV/dr = 2(3.142)r ± (38000/r²)0 = 2(3.142)r ±

(38000/r²) -->> minimum value, therefore dV/dr =038000/r² = 2(3.142)r 38000/2(3.142) = r 6047.104= r

r = 18.22Sub. r = 18.22 into (3):h = 19000/(3.142)(18.22)²h = 18.22therefore,

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

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Method 2: Quadratic FunctionsUse the two same equations as in Method 1, but

only the formula for amount of cream is themainequation used as the quadratic function.Let f(r) =volume of cream, r = radius of round cake:19000 =

(3.142)r²h (1)f(r) = (3.142)r² + 2(3.142)hr (2)From (2):f(r) = (3.142)(r² + 2hr) -->> factorize(3.142)= (3.142)[ (r + 2h/2)² ± (2h/2)² ] -->>completing square, with a = (3.142), b = 2h and c =0= (3.142)[ (r + h)² ± h² ]= (3.142)(r + h)² ±(3.142)h²(a = (3.142) (positive indicates min. value),min. value = f(r) = ±(3.142)h², correspondingvalueof x = r = --h)

Sub. r = --h into (1):19000 = (3.142)(--h)²hh³ =

6047.104h = 18.22Sub. h = 18.22 into (1):19000 = (3.142)r²(18.22)r² =

331.894r = 18.22therefore,

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

I would choose not to bake a cake with suchdimensions because its dimensions are notsuitable(the height is too high) and therefore less attractive.Furthermore, such cakes aredifficult to handle easily. (add in your own explanation.) Further Exploration

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(order to bake multi-storey cake) 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 2ndcake « etc. From question, you'll get: 31, 27.9,

25.11, 22.599, « (GP with a = 31, r = 9/10) a) Findvolume of 1st, 2nd, 3rd, and 4thcakes. Determine whether the volumes formnumber pattern, then explain and elaborate on thenumber patterns. Use the formula for volume V =(3.142)r²h, with h = 6 to get the volume of cakes.

 The values of r can be obtained from the progressionof radius of cakes given in previous question. My

answer:Radius of 1st cake = 31, volume of 1st cake = (3.142)

(31)²(6) =18116.772 Radius of 2nd cake = 27.9, vol.of 2nd cake =14674.585 Radius of 3rd cake = 25.11,vol. of 3rdcake =11886.414 Radius of 4th cake =22.599, vol. of 4th cake =9627.99518116.772, 14674.585, 11886.414, 9627.995, «(it is aGP with first term, a = 18116.772 and ratio, r

= T2/T1= T3/T2= « = 0.81) 

b) Given the total mass of all the cakes should notexceed 15 kg ( total mass < 15 kg, change tovolume:total volume < 57000 cm³), find the maximum

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number of cakes that needs to be baked.Verify theanswer using other methods. Use Sn= (a(1 - rn

)) / (1 - r), with Sn= 57000, a = 18116.772 and r = 0.81 to findn:57000 = (18116.772(1 ± (0.81)n)) / (1 - 0.81)1 ± 0.81n= 0.597790.40221 = 0.81

nlog

0.810.40221 = nn = log 0.40221 / log 0.81n =4.322therefore,n § 4Verifying the answer: When n = 5:

 S5= (18116.772(1 ± (0.81)5)) / (1 ± 0.81) = 62104.443 > 57000 (Sn> 57000, n = 5 is notsuitable)

 When n = 4: S4= (18116.772(1 ± (0.81)

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4)) / (1 ± 0.81) = 54305.767 < 57000 (Sn< 57000, n = 4 is suitable)