Date post: | 07-Apr-2018 |
Category: |
Documents |
Upload: | ezwan-razman |
View: | 230 times |
Download: | 0 times |
of 17
8/6/2019 57208997 Add Maths Project
1/17
Part 1 Cakes And Add-Math? Realationship?
Maybe it use progression that are thought in additional mathematics...for examplewhen we make a cake with many layer, we must fix the difference of diameter ofthe two layer. So we can say that it used arithmetic progression. When thediameter of the first layer of the cake is 8 and the diameter of second layer of the
cake is 6, then the diameter of the third layer should be 4. In this case, we usearithmetic progression where the difference of the diameter is constant that is 2.When the diameter decrease, the weight also decrease. That is the way how thecake is balance to prevent it from smooch. We can also use ratio, because when weprepare the ingredient for each layer of the cake, we need to decrease its ratio fromlower layer to upper layer. When we cut the cake, we can use fraction to devide thecake according to the total people that will eat the cake.
Part 2 (The Best Bakery SMK Mahsuri Teachers Day cake problem )
Q1) Lets say1 kg cake has volume 3800cm, and h is 7cm, so find d.
Volume of 5kg cake = Base area of cake x Height of cake
3800 x 5 = (3.142)(d/2) x 7
19000/7(3.142) = (d/2)
863.872 = (d/2)
d/2 = 29.392
d = 58.784 cm
Q2) Inner Dimension of Oven is-- 80cm length, 60cm width, 45cm height
a) corresponding values of d with different values of h, and tabulate the answers.
V = 19000, that is:
19000 = (3.142)(d/2)h
19000/(3.142)h = d/4
24188.415/h = d
d = 155.53/h
Draw and complete table of 2 columns
8/6/2019 57208997 Add Maths Project
2/17
b) i) State the range of heights that is NOT suitable for the cakes and explain.
H < 10cm is NOT suitable, because the resulting diameter produced is too largepluse there arent many teacher to finish all the cake . Futhermore its will coastmore to the school and the decoration of the cake will be simple as the ingredientgo deco will be expensive
b) ii) Suggest and explain the most suitable dimensions (h and d) for the cake.
h = 6cm, d = 63.493cm, because it is the standart diameter of the cake pluse theprice will be much cheaper. Futhermore the cake is wide enough to treateveryone,the cake will be much decorative and pluse less ingredient to use..
linear equation relating d and h. Hence, plot a suitable (linear, best fit) graph basedon that equation.
19000 = (3.142)(d/2)h
19000/(3.142)h = d/4
24188.415/h = d
d = 155.53/h
d = 155.53h-1/2
log d = log 155.53h-1/2
log d = -1/2 log h + log 155.53
8/6/2019 57208997 Add Maths Project
3/17
ii) Use the graph you've drawn to determine:when h = 10.5cmh = 10.5cm, log h = 1.021, log d = 1.680, d = 47.86cmh when d = 42cm
d = 42cm, log d = 1.623, log h = 1.140, h = 13.80cm
Q3) Decorate the cake with fresh cream, with uniform thickness 1cm.
The amount of fresh cream needed to decorate the cake, using the dimensionsyou've suggested in Q2/b/ii
Amount of fresh cream = VOLUME of fresh cream needed (area x height)
Amount of fresh cream = Vol. of cream at the top surface + Vol. of cream at the sidesurfac
Vol. of cream at the top surface
8/6/2019 57208997 Add Maths Project
4/17
= Area of top surface x Height of cream
= (3.142)(54.99/2) x 1
= 2375 cm
Vol. of cream at the side surface
= Area of side surface x Height of cream
= (Circumference of cake x Height of cake) x Height of cream
= 2(3.142)(54.99/2)(8) x 1
= 1382.23 cm
Therefore, amount of fresh cream = 2375 + 1382.23 = 3757.23 cm
B) THREE other shapes (the shape of the base of the cake) for the cake with same
height (depends on the Q2/b/ii) and volume (19000cm)
8/6/2019 57208997 Add Maths Project
5/17
1 Rectangle-shaped base (cuboid)
8/6/2019 57208997 Add Maths Project
6/17
19000 = base area x height
base area = 19000/8
length x width = 2375
By trial and 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 sidesurface)(Height
of cream) + Vol. of top surface
= 2(8 x 50)(1) + 2(8 x 47.5)(1) + 2375 = 3935 c
8/6/2019 57208997 Add Maths Project
7/17
2 Triangle-shaped base
8/6/2019 57208997 Add Maths Project
8/17
19000 = base area x height
base area = 2375
x length x width = 2375
length x width = 4750
By trial and improvement, 4750 = 95 x 50 (length = 95, width = 50)
Slant length of triangle = (95 + 25)= 98.23
Therefore, amount of cream
= Area of rectangular front side surface(Height of cream) + 2(Area of slantrectangular left/rightside surface)(Height of cream) + Vol. of top surface
= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68 cm
8/6/2019 57208997 Add Maths Project
9/17
3 Trapezium-shaped base
8/6/2019 57208997 Add Maths Project
10/17
19000 = base area x height
base area = 2375 = area of 5 similar isosceles triangles in a pentagon
therefore:
2375 = 5(length x width)
475 = length x width
By trial and improvement, 475 = 25 x 19 (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 cm
c) shape that require the least amount of fresh cream to be used.
Trapezium-shaped cake, since itrequires only 3135 cm of cream to be used.
Part 3
Dimensions of 5kg ROUND cake (volume: 19000cm) that require minimumamount of cream to decorate. Two different methods, including Calculus(differentiation/integration).
8/6/2019 57208997 Add Maths Project
11/17
Method 1: DifferentiationUse two equations for this method: the formula for volume of cake (as in Q2/a), andthe formulafor amount (volume) of cream to be used for the round cake (as in Q3/a).
19000 = (3.142)rh (1)
V = (3.142)r + 2(3.142)rh (2)From (1): h = 19000/(3.142)r (3)
Sub. (3) into (2):
V = (3.142)r + 2(3.142)r(19000/(3.142)r)
V = (3.142)r + (38000/r)
V = (3.142)r + 38000r-1
dV/dr = 2(3.142)r (38000/r)
0 = 2(3.142)r (38000/r) -->> minimum value, therefore dV/dr = 0
38000/r = 2(3.142)r
38000/2(3.142) = r
6047.104 = rr = 18.22
Sub. r = 18.22 into (3):
h = 19000/(3.142)(18.22)
h = 18.22 therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cSo the Dimensions of 5kg round cake by using this method is h=18.22 and diameterwhere r=18.22 And the diameter is 36.44cm
8/6/2019 57208997 Add Maths Project
12/17
Method 2: Quadratic Functions
Use the two same equations as in Method 1, but only the formula for amount ofcream is themain equation used as the quadratic function.Let f(r) = volume of cream, r = radius of round cake:
19000 = (3.142)rh (1)f(r) = (3.142)r + 2(3.142)hr (2)
From (2):
f(r) = (3.142)(r + 2hr) -->> factorize (3.142)
= (3.142)[ (r + 2h/2) (2h/2) ] -->> completing square, with a = (3.142), b = 2h andc = 0
= (3.142)[ (r + h) h ]
= (3.142)(r + h) (3.142)h
(a = (3.142) (positive indicates min. value), min. value = f(r) = (3.142)h,corresponding valueof x = r = --h)
Sub. r = --h into (1):19000 = (3.142)(--h)h
h = 6047.104
h = 18.22
Sub. h = 18.22 into (1):
19000 = (3.142)r(18.22)
r = 331.894
r = 18.22
therefore, h = 18.22 cm, d = 2r = 2(18.22) = 36.44 cm
So the Dimensions of 5kg round cake by using this method is h=18.22 and diameterwhere r=18.22 And the diameter is 36.44cm
I would choose to bake the cake as the shape is rare in market and is a trill to do.Plus the cake is have many space to decorate with colour and many people canenjoy it.
8/6/2019 57208997 Add Maths Project
13/17
Further Exploration (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 2nd cake
a) Find volume of 1st, 2nd, 3rd, and 4th cakes. Determine whether the volumesform number
b) pattern, then explain and elaborate on the number patterns.Radius of 1st cake = 31, volume of 1st cake = (3.142)(31)(6) = 18116.772
Radius of 2nd cake = 27.9, vol. of 2nd cake = 14674.585
Radius of 3rd cake = 25.11, vol. of 3rd cake = 11886.414
Radius of 4th cake = 22.599, vol. of 4th cake = 9627.995
18116.772, 14674.585, 11886.414, 9627.995,
(it is a GP with first term, a = 18116.772 and ratio, r = T2/T1 = T3/T2 = = 0.81)
c) Given the total mass of all the cakes should not exceed 15 kg ( total mass < 15kg, change to volume: total volume < 57000 cm), find the maximum numberof cakes that needs to be baked.Answer using other methods.
Use Sn = (a(1 - rn)) / (1 - r), with Sn = 57000, a = 18116.772 and r = 0.81 to find n:57000 =(18116.772(1 (0.81)n)) / (1 - 0.81)
8/6/2019 57208997 Add Maths Project
14/17
1 0.81n = 0.59779
0.40221 = 0.81n
log0.81 0.40221 = n
n = log 0.40221 / log 0.81
n = 4.322
therefore, n 4
Verifying 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)4)) / (1 0.81) = 54305.767 < 57000 (Sn < 57000, n = 4 issuitable)
8/6/2019 57208997 Add Maths Project
15/17
REFLECTION HAIZMY PRINSIP AKAUN FOLIO
JUST PASSED TO MY P.A TEACHER
JUST THINKING ABOUT RESTING
FOR A WHILE..BUT THENHERE
COME ANOTHER PROJECT SO
LAME.
ARHMUST DO IT >.
HOW TO DO THIS?
WHAT IS THE
FOMULA? ARE YOUDONE IT YET.
REFER YOUR TEXT
BOOKREFER THE NOTE
BOOKYES I DONE IT
=)
HOW TO OP
MICROSOFT
WORD..IM
COMPUTER
NOOB!!!!!!
I DONT KNOW
EITHER..HERM.LET
ME THINK@_@
WORK IN PROGRESSLOADING.10%
WORK IN
PROGRESS .LOADING..30%
8/6/2019 57208997 Add Maths Project
16/17
The end! At last
my addmath
folio*(I hope Igot 100%) ^__^
mORAL VALUE
nEVER sAY nEVER!
wORK HARD!
tEAMWORK IS IMPORTANT
iTS hard TO SUCCESS IN LIFE
aDD-mATH ARE FUN WHEN WE GET THE ANSWER BUT
WE WILL BE DEPRESSED IF WE CANT GET THE
ANSWER
iNTERNET IS THE BEST DICTIONARY IN WHOLE
8/6/2019 57208997 Add Maths Project
17/17