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Friday, June 3, 2011 Solutions for SPM Additional Mathematics Project Work 2/2011 (for all my lovely friends)

Due to strong demand (I mean, cries of help) from buddies for me to help out with their Project Work, well, here are SOME of the guides and solutions for Parts 1, 2 and 3, as well as the Further Exploration part of the Project Work, which was commissioned to us on Thursday, 26th May 2011 and is to be completed by the 1st week of the 2nd school semester (after holidays lor …)

TAKE NOTE that this isn't the entire folio itself (no content page, no intro, no reflection, no conjectures, no conclusions or any other related stuff) because my intention here is to guide all of you with only the mathematical part of the Project Work. For other parts, well, it's on your own, since I only finished the maths part, while the others are still empty … =p

And, these are INTERNET answers, not teacher's, nor resource books nor something similar, therefore it may not be the complete answers, or even wrong answers… so if you have any problems, uncertainty or just for confirmation, it's best to consult your Add Maths teacher or refer to past year's Project Works... and help me correct any problems that occur here, for the benefit of others. =

(value of 'pi' used = 3.142)

Part 1 (Find out how maths is used in cake baking and cake decorating and write about your findings)

(actually I'm not sure how to write this … some websites tat provide last year's sample wrote loooong answers for this part … ended up writing too little about this … but I'll just give you SOME of the points here. If you have any other suitable ones, do write in down too.)

Geometry – To determine suitable dimensions for the cake, to assist in designing and decorating cakes that comes in many attractive shapes and designs, to estimate volume of cake to be produced, … etc.

Calculus (differentiation) – To determine minimum or maximum amount of ingredients for cake-baking, to estimate min. or max. amount of cream needed for decorating, to estimate min. or max. size of cake produced, … etc.

Progressions – To determine total weight/volume of multi-storey cakes with proportional dimensions, to estimate total ingredients needed for cake-baking, to estimate total amount of cream for decoration, … etc.

(if there's more, add it yourself and don't forget to elaborate on the points you provide …)

Part 2 (bake a 5 kg round cake for your school. given the height of cake, h and the diameter of cake, d.)

Q1) Given 1 kg cake has volume 3800cm³, 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)² x 7
19000/7(3.142) = (d/2)²
863.872 = (d/2)²
d/2 = 29.392
d = 58.784 cm

Q2) Given the inner dimensions of oven: 80cm length, 60cm width, 45cm height

a) Find corresponding values of d with different values of h, and tabulate the answers.

First, form the formula for d in terms of h by using the above formula for volume of cake, V = 19000, that is:
19000 = (3.142)(d/2)²h
19000/(3.142)h = d²/4
24188.415/h = d²
d = 155.53/√h

Then, draw and complete table of 2 columns, 10 rows (example), as shown below: (use that formula to find d, for every value of h)

b) i) State the range of heights that is NOT suitable for the cakes and explain.

(INTERNET answer) h < 7cm is NOT suitable, because the resulting diameter produced is too large to fit into the oven. Furthermore, the cake would be too short and too wide, making it less attractive.

b) ii) Suggest and explain the most suitable dimensions (h and d) for the cake.

NOTE: just provide ONE h and its corresponding d only, it asks for “most suitable” (which means the one and only suitable)
(INTERNET answer) h = 8cm, d = 54.99cm, because it can fit into the oven, and the size is suitable for easy handling.

c) i) Form a linear equation relating d and h. Hence, plot a suitable (linear, best fit) graph based on that equation.

The same formula in Q2/a is used, that is 19000 = (3.142)(d/2)²h. The same process is also used, that is, make d the subject. This time, form an equation which is suitable and relevant for the graph:
19000 = (3.142)(d/2)²h
19000/(3.142)h = d²/4
24188.415/h = d²
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 (the final equation for graph-drawing)

Create another table (or add two extra columns to the 1st table in Q2/a), with one column is log h and the other is log d. Then, plot a graph of log d against log h.

NOTE: 2cm on graph represents 0.1 units for both axes. The x-axis must be from 0 till 1.2 or more, in order to answer the next question.

(NOTE: I chose not to use the d = 155.53(1/√h) to draw the graph because when h = 0, then 1/0, which is undefined (graph will not be a straight-line graph). Therefore, I chose to do the safer way of using other forms or similar equations (such as the log equation above) so that the graph will be a continuous straight-line graph. BUT I'm still open for discussions about this issue. =) )

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

Q3) Decorate the cake with fresh cream, with uniform thickness 1cm
a) Estimate the amount of fresh cream needed to decorate the cake, using the dimensions you've suggested in Q2/b/ii

Internet answer in Q2/b/ii ==> h = 8cm, d = 54.99cm
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 side surface
(The bottom surface area of cake is NOT COUNTED, because we're decorating the visible part of the cake only (top and sides). Obviously, we don't decorate the bottom part of the cake, 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 cm³

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 cm³

Therefore, amount of fresh cream = 2375 + 1382.23 = 3757.23 cm³

b) Suggest 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 (19000cm³). Estimate the amount of fresh cream (the volume) 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's safe to provide three different shapes for this question.
Depends on your choice of shapes, but usually the volume of top surface is always the same for 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. This process can be quite difficult, especially if you choose shapes that has more than 4 sides, such as pentagon or hexagon. By the way, draw the shapes that you choose for each question, so that you'll better understand the area involved for cake-decorating.
internet answer (with h = 8cm, and volume of cream on top surface = 19000/8 = 2375 cm³):
1 – Rectangle-shaped base (cuboid)
(draw cuboid)
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 side surface)(Height of cream) + Vol. of top surface
= 2(8 x 50)(1) + 2(8 x 47.5)(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 the vertices to form the top surface of the cake)
19000 = base area x height
base area = 2375
½ 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 = √(95² + 25²)= 98.23
Therefore, amount of cream
= Area of rectangular front side surface(Height of cream) + 2(Area of slant rectangular left/right side 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 lines on its vertices, then join the vertices to form the top surface of cake)
19000 = base area x height
base area = 2375 = area of 5 similar isosceles triangles in a pentagon
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 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 that require the least amount of fresh cream to be used.

(depends on your answers, but here's my obvious answer) Pentagon-shaped cake, since it requires only 3135 cm³ of cream to be used.

Part 3 (Find dimensions of 5kg ROUND cake (volume: 19000cm³) that require minimum amount of cream to decorate. Use two different methods, including Calculus (differentiation/integration). Also, explain whether you would choose to bake that cake with such dimensions and give reasons why.

When there's “minimum” or “maximum”, well, there's differentiation and quadratic functions. Use both to find the minimum height, h and its corresponding minimum diameter, d.
(Note: This question only gives you just ONE info only: the mass of cake (which you can change into volume of cake, as shown), so use this ONE info only to find the minimum dimensions)

Here's my answer:
Method 1: Differentiation
Use two equations for this method: the formula for volume of cake (as in Q2/a), and the formula for amount (volume) of cream to be used for the round cake (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-1

dV/dr = 2(3.142)r – (38000/r²)
0 = 2(3.142)r – (38000/r²) -->> minimum value, therefore dV/dr = 0
38000/r² = 2(3.142)r
38000/2(3.142) = r³
6047.104 = r³
r = 18.22

Sub. r = 18.22 into (3):
h = 19000/(3.142)(18.22)²
h = 18.22
therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cm

Method 2: Quadratic Functions
Use the two same equations as in Method 1, but only the formula for amount of cream is the main equation 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, corresponding value of x = r = --h)

Sub. r = --h into (1):
19000 = (3.142)(--h)²h
h³ = 6047.104
h = 18.22

Sub. h = 18.22 into (1):
19000 = (3.142)r²(18.22)
r² = 331.894
r = 18.22
therefore, h = 18.22 cm, d = 2r = 2(18.22) = 36.44 cm

I would choose not to bake a cake with such dimensions because its dimensions are not suitable (the height is too high) and therefore less attractive. Furthermore, such cakes are difficult to handle easily.

(add in your own explanation.)

Further Exploration (order to bake multi-storey cake)
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
… etc.

From question, you'll get:
31, 27.9, 25.11, 22.599, … (GP with a = 31, r = 9/10)

a) Find volume of 1st, 2nd, 3rd, and 4th cakes. Determine whether the volumes form number pattern, then explain and elaborate on the number 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 progression of 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 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)

b) Given the total mass of all the cakes should not exceed 15 kg ( total mass < 15 kg, change to volume: total volume < 57000 cm³), find the maximum number of cakes that needs to be baked. Verify the 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)
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 not


When n = 4:
S4 = (18116.772(1 – (0.81)4)) / (1 – 0.81) = 54305.767 < 57000 (Sn < 57000, n = 4 is


Reflection(xpress ur love to Add u love add math)

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