KEDAH ADDITIONAL MATHEMATICS PROJECT WORK 2015

PART II

Wawasan Bakery shop received an order from your school to bake a 5 kg of round cake as

shown in Diagram 1 for the Merdeka Day celebration.

Diagram 1

1) If a kilogram of cake has a volume of 38000cm3, and the height of the cake is to be 7.0 cm,

the diameter of the baking tray to be used to fit the 5 kg cake ordered by your school 3800 is

Volume of 5kg cake = Base area of cake x Height of cake

3800 x 5 = (3.142)(

) x 7

(3.142) = (

)

863.872 = (

)

= 29.392

d = 58.784 cm

2) The inner dimensions of oven: 80cm length, 60cm width, 45cm height a) The formula that formed for d in terms of h by using the formula for volume of cake, V = 19000 is:

19000 = (3.142)(

)h

=

= d

d =

Table 1 b) i) 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) The most suitable dimensions (h and d) for the cake is h = 8cm, d = 54.99cm, because it can fit into the oven, and the size is suitable for easy handling.

c) i) The same formula in 2(a) is used, that is 19000 = (3.142)(

)h. The same process is also

used, that is, make d the subject. An equation which is suitable and relevant for the graph:

19000 = (3.142)(

)h

=

= d

Height,h Diameter,d

1.0 155.53

2.0 109.98

3.0 89.79

4.0 77.76

5.0 69.55

6.0 63.49

7.0 58.78

8.0 54.99

9.0 51.84

10.0 49.18

d =

d =

log d =

log d =

log h + log 155.53

Table of log d =

log h + log 155.53

Table 2

Height,h Diameter,d Log h Log d

1.0 155.53 0.00 2.19

2.0 109.98 0.30 2.04

3.0 89.79 0.48 1.95

4.0 77.76 0.60 1.89

5.0 69.55 0.70 1.84

6.0 63.49 0.78 1.80

7.0 58.78 0.85 1.77

8.0 54.99 0.90 1.74

9.0 51.84 0.95 1.71

10.0 49.18 1.0 1.69

Graph of log d against log h

ii) Based on the graph: 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 3) The cake with fresh cream, with uniform thickness 1cm is decorated a) The amount of fresh cream needed to decorate the cake, using the dimensions I've

suggested in 2(b)(ii)

From 2(b)(ii) h = 8cm, d = 54.99cm

Amount of fresh cream = volume of fresh cream needed (area x height)

Amount of fresh cream = volume of cream at the top surface + volume 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

Volume of cream at the top surface

= Area of top surface x Height of cream

= (3.142)(

) x 1

= 2375 cm

Volume 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)(

)(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 which

is depends on the 2(b)(ii) and volume 19000cm.

The volume of top surface is always the same for all shapes (since height is same),

My answer (with h = 8cm, and volume of cream on top surface =

= 2375 cm):

1 Rectangle-shaped base (cuboid)

height

width

length

19000 = base area x height

base area =

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 and right side surface)(Height of cream) + 2(Area of front and back side

surface)(Height of cream) + volume of top surface

= 2(50 x 8)(1) + 2(47.5 x 8)(1) + 2375

= 3935 cm

2 Triangle-shaped base width

slant height

19000 = base area x height

base area =

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 slant rectangular left/right

side surface)(Height of cream) + Volume of top surface

= (50 x 8)(1) + 2(98.23 x 8)(1) + 2375 = 4346.68 cm

3 Pentagon-shaped base

width 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(19 x 8) + 2375 = 3135 cm

c) Based on the values above, the shape that requires the least amount of fresh cream to be

used is:

Pentagon-shaped cake, since it requires only 3135 cm of cream to be used.

PART III (1) When there's minimum or maximum, well, there's differentiation and quadratic

functions. The minimum height, h and its corresponding minimum diameter, d is calculated by using the differentiation and function.

Method 1: Differentiation Two equations for this method: the formula for volume of cake (as in 2(a)), and the formula for amount (volume) of cream to be used for the round cake (as in 3(a)).

19000 = (3.142)rh (1)

V = (3.142)r + 2(3.142)rh (2)

From (1): h =

(3)

Sub. (3) into (2):

V = (3.142)r + 2(3.142)r(

)

V = (3.142)r + (

)

V = (3.142)r + 38000r-1

(

) = 2(3.142)r (

)

0 = 2(3.142)r (

) -->> minimum value, therefore

= 0

= 2(3.142)r

= r

6047.104 = r

r = 18.22

Sub. r = 18.22 into (3):

h =

h = 18.22

therefore, h = 18.22cm, d = 2r = 2(18.22) = 36.44cm

Method 2: Quadratic Functions 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)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 +

) (

) ] -->> completing square, with a = (3.142), b = 2h and c = 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 value of

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

(2) I would not choose to bake a cake of such dimensions due to the height of the cake which is

far more extreme than a usual cake. This will make the cake more likely to become instable and hard to handle. Apart from that, the cake will look slightly messy with such height.

Diagram 2 Wawasan Bakery received an order to bake a multi-storey cake for Merdeka Day celebration, as

shown in Diagram 2.

The height of each cake is 6.0 cm and the radius of the largest cake is 31.0 cm. The radius of the

second cake is 10% less than the radius of the first cake, the radius of the third cake is 10% less

than the radius of the second cake and so on.

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

31, 27.9, 25.11, 22.599,

a = 31, r =

V = (3.142)rh,

a) By using the formula for volume V = (3.142)rh, with h = 6 to get the volume of cakes.

Volume of 1st, 2nd, 3rd, and 4th cakes:

Radius of 1st cake = 31, volume of 1st cake = (3.142)(31)(6) = 18116.772

Radius of 2nd cake = 27.9, volume of 2nd cake = (3.142)(27.9)(6) 14674.585

Radius of 3rd cake = 25.11, volume of 3rd cake = (3.142)(25.11)(6) 11886.414

Radius of 4th cake = 22.599, volume of 4th cake = (3.142)(22.599)(6) 9627.995

The volumes form number pattern:

18116.772, 14674.585, 11886.414, 9627.995,

(it is a geometric progression with first term, a = 18116.772 and ratio, r = T2/T1 = T3 /T2 = =

0.81)

b) The total mass of all the cakes should not exceed 15 kg ( total mass < 15 kg, change to

volume: total volume < 57000 cm), so the maximum number of cakes that needs to be baked is

Sn =

Sn = 57000, a = 18116.772 and r = 0.81

57000 =

1 0.81n = 0.59779 0.40221 = 0.81n og0.81 0.40221 = n

n =

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 suitable)

When n = 4:

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