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Additional Mathematics Project Work 2 Written By : Nurul Hazira Syaza Abas I/C : 940602-01-6676 Angka Giliran : School : SMK Kangkar Pulai Copyright 2011 ©. Hazira Syaza, All Right Reserve Numb| Title| Page| 1| Acknowledge| 1| 2| Objective| 2| 3| Introduction Part I| 3| 4| Mathematics In Cake Baking And Cake Decorating| 4 5| 5| Part II| 6 14| 6| Part III| 15 17| 7| Further Exploration| 18 21| 8| Reflection| 22 23| 9| Conclusion| 24| 10| Reference| 25| Table. of. Content Copyright 2011 ©. Hazira Syaza, All Right Reserve Acknowledge

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First of all, I would like to say Alhamdulillah, for giving me the strength and health to do this project work. Not forgotten my parents for providing everything, such as money, to buy anything that are related to this project work and their advise, which is the most needed for this project. Internet, books, computers and all that. They also supported me and encouraged me to complete this task so that I will not procrastinate in doing it. Then I would like to thank my teacher, Puan Andek for guiding me and my friends throughout this project.

Copyright 2011 ©. Hazira Syaza, All Right Reserve 3 MATHEMATICS IN CAKE BAKING AND CAKE DECORATING 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 When making a batch of cake batter, you end up with a certain volume, determined by the recipe. The baker must then choose the appropriate size and shape of pan to achieve the desired result. If the pan is too big, the cake becomes too short. If the pan is too small, the cake becomes too tall. This leads into the next situation.

The ratio of the surface area to the volume determines how much crust a baked good will have. The more surface area there is, compared to the volume, the faster the item will bake, and the less “inside” there will be. For a very large, thick item, it will take a long time for the heat to penetrate to the center. To avoid having a rock-hard outside in this case, the baker will have to lower the temperature a little bit and bake for a longer time. We mix ingredients in round bowls because cubes would have corners where unmixed ingredients would accumulate, and we would have a hard time scraping them into the batter. 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. Copyright 2011 ©. Hazira Syaza, All Right Reserve 4 PROGRESSION 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. For example when we make a cake with many layers, we must fix the difference of diameter of the two layers. So we can say that it used arithmetic progression. When the diameter of the first layer of the cake is 8? nd the diameter of second layer of the cake is 6? , then the diameter of the third layer should be 4?. In this case, we use arithmetic progression where the difference of the diameter is constant that is 2. When the diameter decreases, the weight also decreases. That is the way how the cake is balance to prevent it from smooch. We can also use ratio, because when we prepare the ingredient for each layer of the cake, we need to decrease its ratio from lower layer to upper layer. When we cut the cake, we can use fraction to devide the cake according to the total people that will eat the cake.

Copyright 2011 ©. Hazira Syaza, All Right Reserve 5 Part 11 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. 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)( d/2)? x 7 1900/7 (3. 142) = ( d/2)? 863. 872 = (d/2 )? d/2 = 29. 392 d d = 58. 784 cm Copyright 2011 ©. Hazira Syaza, All Right Reserve 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)(d/2)(d/2)  ? h 1900/(3. 142)2 = d  ? /4 24188. 415/h = d  ? d = 155. 53/ h Copyright 2011 ©. Hazira Syaza, All Right Reserve 7 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| 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: Copyright 2011 ©. Hazira Syaza, All Right Reserve 8 1900= (3. 1420(d/2)  ? h 119000/(3. 142)h = d  ? /4 24188. 415/h = d  ? d = 155. 53/ vh d = 155. 53h(1-/? ) log d = log 155. 3h(-1/? ) log d = (-1/? )log h + log 155. 53 Table of log d = (-1/? )log h + log 155. 53 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| Table 2 Copyright 2011 ©. Hazira Syaza, All Right Reserve 9 Graph of log d against log h Copyright 2011 ©. Hazira Syaza, All Right Reserve 10 ii) Based on the graph: a) d when h = 10. 5cm h = 10. 5cm, log h = 1. 21, 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) My answer in 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)(54. 99/2) ? )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)( 54. 99/2)(8) x 1 = 1382. 23 cm? Therefore, amount of fresh cream = 2375 + 1382. 23 = 3757. 23 cm? Copyright 2011 ©. Hazira Syaza, All Right Reserve 11 c) 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?