additional ma thematic 1
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AdditionalMathematics
Project Work 2
Written by: ALVIN SOO CHUN KITI/C Num :
Angka Giliran
:School :Date
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Alvin Soo Chun Kit
Additional
Mathematics Project
Work 2 2011Method 2: Using
differentiationAssuming that the
surface area of the
cake is
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proportionate to the
amount of freshcreamneeded to
decorate thecake.*Formula for
surface area= r2
+ 2 rhh = 19000 /
3.142r2
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Surface area in
contact with cream=
r
2
+ 2 r(19000 /
3.142r2
)= r
2
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+ (38000/r)The
values, when plottedinto a graph will
from a minimumvalue that can be
obtainedthroughdifferentiation.dy =
0dxdy = 2 r ±
(38000/r2
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)dx0 = 2 r ±
(38000/r2
)0 = 6.284r3
± 3800038000 =
6.284r3
6047.104 = r3
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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
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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
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not be too big to
bake in aconventional oven.*
The aboveconjecture is proven
by thefollowingWhen r =
10,~ the total
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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
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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
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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.
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Alvin Soo Chun KitAdditional
Mathematics ProjectWork 2
2011FURTHER
EXPLORATION
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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
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x 6= 18116.772 cm3
= 3.142 x (27.9)2
x 6= 14676.585 cm3
Volume of cake 3Volume of cake 4=
r2
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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
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x 6 = 3.142 x
(22.599)2
x 6= 11886.414 cm3
= 9627.995 cm
3The values
118116.772,
14676.585,
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11886.414,
9627.995 form anumber pattern.The
pattern formed is ageometrical
progression.This isproven by the fact
that there is a
common ratio
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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
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n
) = 18116.772 ( 1-0.8n
)1-r 1-0.815 kg =57000 cm
357000 >
18116.772(1-0.8n
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)0.211400 >
18116.772(1-0.8n
)0.629 > 1-0.8n
-0.371 > - 0.8
n0.371 < 0.8
n
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log 0.371 < n log
0.8log 0.371 < nlog0.84.444 < nn = 4
Alvin Soo Chun Kit
Additional
Mathematics Project
Work 2 2011
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Verification of
answer If n = 4Totalvolume of 4 cakes=
18116.772 cm3
+ 14676.585 cm3
+ 11886.414 cm3
+ 9627.995 cm3
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= 54307.766 cm3
Total mass of
cakes= 14.29 kgIf n= 5Total volume of 5
cakes= 18116.772
cm3
+ 14676.585 cm3
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+ 11886.414 cm3
+ 9627.995 cm3
+ 7798.676 cm3
= 62106.442 cm3
Total mass of
cakes= 16.34kgTotal mass of
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cakes must not
exceed 15kg.Therefore,
maximum numberof cakes needed to
be made =