57208997 Add Maths Project

8/6/2019 57208997 Add Maths Project

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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


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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


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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


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= 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)


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1 Rectangle-shaped base (cuboid)


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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


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2 Triangle-shaped base


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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


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3 Trapezium-shaped base


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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).


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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


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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.


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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)


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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)


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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%


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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


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57208997 Add Maths Project

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