Additional Ma Thematic 1

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AdditionalMathematics

Project Work 2

Written by: ALVIN SOO CHUN KITI/C Num :

Angka Giliran

:School :Date



Alvin Soo Chun Kit

Additional

Mathematics Project

Work 2 2011Method 2: Using

differentiationAssuming that the

surface area of the

cake is



proportionate to the

amount of freshcreamneeded to

decorate thecake.*Formula for

surface area= r2

+ 2 rhh = 19000 /

3.142r2



Surface area in

contact with cream=

r

2

+ 2 r(19000 /

3.142r2

)= r

2



+ (38000/r)The

values, when plottedinto a graph will

from a minimumvalue that can be

obtainedthroughdifferentiation.dy =

0dxdy = 2 r ±

(38000/r2



)dx0 = 2 r ±

(38000/r2

)0 = 6.284r3

± 3800038000 =

6.284r3

6047.104 = r3



18.22 = r When r =

18.22 cm, h = 18.22cmThe dimensions

of the cake thatrequires the

minimum amountof fresh cream to

decorate

isapproximately 18.2



cm in height and

18.2 cm in radius.Iwould bake a cake

of such dimensionsbecause the cake

would not be toolarge for thecutting

or eating of said

cake, and it would



not be too big to

bake in aconventional oven.*

The aboveconjecture is proven

by thefollowingWhen r =

10,~ the total



surface area of the

cake is 4114.2 cm2

~ the amount of fresh cream needed

to decorate the cake

is 4381.2 cm3

~ the ratio of totalsurface area of cake



to amount of fresh

cream needed is0.94When r = 20,~

the total surfacearea of the cake is

3156.8 cm2

~ the amount of

fresh cream needed



to decorate the cake

is 3308.5 cm3

~ the ratio of totalsurface area of cake

to amount of fresh

cream needed is0.94Therefore, the

above conjecture isproven to be true.



Alvin Soo Chun KitAdditional

Mathematics ProjectWork 2

2011FURTHER

EXPLORATION



a) Volume of cake 1

Volume of cake 2=

r

2

h = r2

h= 3.142 x 31 x 31 x6 = 3.142 x (0.9 x

31)2



x 6= 18116.772 cm3

= 3.142 x (27.9)2

x 6= 14676.585 cm3

Volume of cake 3Volume of cake 4=

r2



h = r2

h= 3.142 x (0.9 x 0.9

x 31)2

x 6 = 3.142 x (0.9 x

0.9 x 0.9 x 31)2

x 6= 3.142 x (25.11)2



x 6 = 3.142 x

(22.599)2

x 6= 11886.414 cm3

= 9627.995 cm

3The values

118116.772,

14676.585,



11886.414,

9627.995 form anumber pattern.The

pattern formed is ageometrical

progression.This isproven by the fact

that there is a

common ratio



between subsequent

numbers, r =0.81.14676.585 =

0.81 11886.414 =0.8118116.772

14676.585.

9627.995 =

0.8111886.414 b) Sn

= a(1-r



n

) = 18116.772 ( 1-0.8n

)1-r 1-0.815 kg =57000 cm

357000 >

18116.772(1-0.8n



)0.211400 >

18116.772(1-0.8n

)0.629 > 1-0.8n

-0.371 > - 0.8

n0.371 < 0.8

n



log 0.371 < n log

0.8log 0.371 < nlog0.84.444 < nn = 4

Alvin Soo Chun Kit

Additional

Mathematics Project

Work 2 2011



Verification of

answer If n = 4Totalvolume of 4 cakes=

18116.772 cm3

+ 14676.585 cm3

+ 11886.414 cm3

+ 9627.995 cm3



= 54307.766 cm3

Total mass of

cakes= 14.29 kgIf n= 5Total volume of 5

cakes= 18116.772

cm3

+ 14676.585 cm3



+ 11886.414 cm3

+ 9627.995 cm3

+ 7798.676 cm3

= 62106.442 cm3

Total mass of

cakes= 16.34kgTotal mass of



cakes must not

exceed 15kg.Therefore,

maximum numberof cakes needed to

be made =

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