Addmath Project Work

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Part 1

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 minimum or maximum, amount of cream needed for decorating, to estimate minimum or maximum 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 2

1) Given 1 kg of cake has volume 3800 cm³, and h is 7 cm, so find the d. Volume of 5 kg = 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

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

|Height, h (cm) |Diameter, d (cm) | |1.0 |155.53 | |2.0 |109.98 | |3.0 |89.80 | |4.0 |77.77 | |5.0 |69.56 | |6.0 |63.49 | |7.0 |58.78 | |8.0 |54.99 | |9.0 |51.84 | |10.0 |49.18 |

b) i) h < 7 cm 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) h = 8 cm , d = 54.99 cm , because it can fit into he oven, and the size is suitable for easy handling.

c) i) The same formula in Q2(a) is used, that is 19000 = (3.142)(d/2)²h. The same process ia also used, that is make d as 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 = 155.53hˉ¹ ²
log d = log 155.53hˉ¹ ²
log d = -1/2 log h + log 155.53 (the final equation for graph – drawing)

|Height, h (cm) |Diameter, d (cm) |Log h |Log d | |1.0 |155.53 |0.000 |2.192 | |2.0 |109.98 |0.300 |2.041 | |3.0 |89.80 |0.477 |1.953 | |4.0 |77.77 |0.602...
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