Friday, June 17, 2011

addmath project work (work 2)

Question
Part I
Cakes come in a variety of forms and flavors and are among favourite desserts served during special occasions such as birthday parties, Hari Raya, weddings and etc. Cakes are treasured not only because of their wonderful taste but also in the art of cake baking and cake decorating. Find out how mathematics is used in cake baking and cake decorating and write about your findings.
Answer :
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.
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.
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.












Part II
Best Bakery shop received an order from your school to bake a 5 kg of round cake as shown in Diagram 1 for the Teachers’ Day celebration. (Diagram 11)

1)If a kilogram of cake has a volume of 3800 cm3, and the height of the cake is to be 7.0cm, calculate the diameter of the baking tray to be used to fit the 5 kg cake ordered by your school.

[ use π = 3.142 ]

Answer:

Volume of 5kg cake = Base area of cake x Height of cake
3800 x 5 = (3.142)( d/2 )2 ×7
19000/7(3.142) = ( d/2 )2
863.872 = ( d/2 )2
d/2 = 29.392
d = 58.784 cm
2)The cake will be baked in an oven with inner dimensions of 80.0 cm in length, 60.0 cm in width and 45.0 cm in height.
a) If the volume of cake remains the same, explore by using different values of heights ,h cm , and the corresponding values of diameters of the baking tray to be used ,d cm. Tabulate your answers.




Answer:
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 = d^2/4
24188.415/h = d2
d = 155.53/(√h)
Height,h (cm) Diameter,d (cm)
1.0 155.53
2.0 109.93
3.0 89.80
4.0 77.77
5.0 68.56
6.0 63.49
7.0 58.78
8.0 54.99
9.0 51.84
10.0 49.18

(b)Based on the values in your table,
i)state the range of heights that is NOT suitable for the cakes and explain your answers.

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.


(ii) suggest the dimensions that you think most suitable for the cake. Give reasons for your answer.

Answer:

h = 8cm, d = 54.99cm, because it can fit into the oven, and the size is suitable for easy handling

(c)(i) Form an equation to represent the linear relation between hand d. Hence, plot a suitablegraph based on the equation that you have formed. [You may draw your graph with the idea of computer software.]


19000 = (3.142) ( d^2/2)h
19000/ (3.142)h = d^2/4
24188.415/h = d2
d = 155.53/(√h)
d = 155.53 (-1)/2
log d = 155.53 (-1)/2
log d = (-1)/2 log h + log 155.53
Log h 0 1 2 3 4
Log d 2.19 1.69 1.19 0.69 0.19










(ii)(a) If Best Bakery received an order to bake a cake where the height of the cake is 10.5cm, use your graph to determine the diameter of the round cake pan required.
Answer:
h = 10.5cm, log h = 1.021, log d = 1.680, d = 47.86cm

(b) If Best Bakery used a 42 cm diameter round cake tray, use your graph to estimate the height of the cake obtained.

Answer:

d = 42cm, log d = 1.623, log h = 1.140,h = 13.80cm

3)Best Bakery has been requested to decorate the cake with fresh cream. The thickness of the cream is normally set to a uniform layer of about1cm.

(a)Estimate the amount of fresh cream required to decorate the cake using the dimensions that you have suggested in 2(b) (ii).

Answer:

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 + Vol. of cream at the side surface

Vol. of cream at the top surface
= Area of top surface x Height of cream
= (3.142) (54.99/2)2×1
=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
=1382.23cm3

Therefore, amount of fresh cream = 2375 + 1382.23 =3757.23 cm3
(b)Suggest three other shapes for cake, that will have the same height and volume as those suggested in 2(b)(ii). Estimate the amount of fresh cream to be used on each of the cakes.



Answer:

1 – Rectangle – shaped base (cuboid)








19000 = base area x height
base area = 19000/2


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 based







19000 = base area x heightbase area
Base area = 2375
1/2 length × width = 2375
length × 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 based





19000 = base area x height
base area = 2375 = area of 5 similar isosceles triangles in a pentagon
therefore :
2375 = 5(length × 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


(c)Based on the values that you have found which shape requires the least amount of fresh cream to be used?

Answer:

Pentagon-shaped cake, since it requires only 3135cm3 of cream to be used.

Part III
Find the dimension of a 5 kg round cake that requires the minimum amount of fresh cream to decorate. Use at least two different methods including Calculus .State whether you would choose to bake a cake of such dimensions. Give reasons for your answers.




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) r2 h → (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 + 38000r2

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