Question Part I Cakes come in a variety of forms and flavours 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 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)( d2)² x 7 190007(3.142) = ( d2)²
863.872 = ( d2)² d2 = 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 cmin 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 = d²4 24188.415h = d²
d = 155.53h
Height,h (cm)
Diameter,d(cm)
1.0
155.53
2.0
109.98
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 h and d. Hence, plot a suitable graph based on the equation that you have formed. [You may draw your graph with the aid of computer software.] Answer: 19000 = (3.142)( d2)²h 19000/(3.142)h = d²4 24188.415h = d²
d = 155.53√h d = 155.53h-12 log d = log 155.53h-12 log d = -12 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 10.5 cm, 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 about 1cm (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 = Vol. 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)( 54.992)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 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 = 190002 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)(Heig s urface)(Height ht of cream) + 2(Area of front/back 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
19000 = base area x height base area = 2375 12 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
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 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 3135 cm³ 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. 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)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 + ( 38000r ) V = (3.142)r² + 38000r -1 (dVdr) = 2(3.142)r – ( 38000r²) 0 = 2(3.142)r – ( 38000r²) -->> minimum value, therefore dVdr = 0 38000r² = 2(3.142)r 380002(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 + 2h2)² – ( 2h2)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. FURTHER EXPLORATION
Best 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 is10% less than the radius of the second cake and so on.(a) Find the volume of the first, the second, the third and the fourth cakes. By comparing all these values, determine whether the volumes of the cakes form a number pattern? Explain and elaborate on the number patterns. Answer: height, h of each cake = 6cm radius of largest cake = 31cm radius of 2 nd cake = 10% smaller than 1 st cake radius of 3 rd cake = 10% smaller than 2 nd cake 31, 27.9, 25.11, 22.599… a = 31, r = 910 V = (3.142)r²h Radius of 1 st cake = 31, volume of 1 st cake = (3.142)(31)²(6) = 18116.772 Radius of 2 nd cake = 27.9, vol. of 2 nd cake = 14674.585 Radius of 3 rd cake = 25.11, vol. of 3 rd cake = 11886.414 Radius of 4 th cake = 22.599, vol. of 4 th cake = 9627.995
18116.772, 14674.585, 11886.414, 9627.995, … a = 18116.772, ratio, r = T 2 /T /T1 = T3 /T2 = … = 0.81
(b) If the total mass of all the cakes should not exceed 15 kg, calculate the maximum number of cakes that the bakery needs to bake. Verify your answer using other methods. Answer: Sn = (a(1 - rn)) (1 - r) Sn = 57000, a = 18116.772 and r = 0.81 57000 = (18116.772(1 – (0.81)n))(1 - 0.81) 1 – 0.81 n = 0.59779 0.40221 = 0.81 n og0.81 0.40221 = n n = log 0.40221 log 0.81
n = 4.322 therefore, n ≈ 4
REFLECTION In the making of this project, I have spent countless hours doing this project.I realized that this subject is a compulsory to me. Without it, I can’t fulfill my big dreams and wishes…. I used to hate Additional Mathematics… It always makes me wonder why this subject is so difficult… I always tried to love every part of it… It always an absolute obstacle for me… Throughout day and night… I sacrificed my precious time to have fun… From.. Monday,Tuesday,Wednesday,Thursday,Friday And even the weekend that I always looking forward to…
1 28ve 980 ADDITIONAL MATHEMATICS
(Cover the top part of the phrase “1 28ve 980” From now on, I will do my best on every second that I will learn Additional Mathematics.
