Math Project Part 5.doc - Introduction In this part of project we base on the dirt evaluated in Part IV to design a box that Holds the dirt However we

# Math Project Part 5.doc - Introduction In this part of...

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Introduction In this part of project, we base on the dirt evaluated in Part IV to design a box that Holds the dirt. However we are not finding any dimension of the box to hold dirt, but the smallest dimension of box that holds the dirt, that means the length of the box should be as small as possible but can hold the same amount of dirt. For example a box with dimension of 9 cm, 3.8 cm, and 17.5 cm has a surface area of 516.4 cm 2 and holds a volume of 598.5 cm 3 (Volume: 9 • 3.8 • 17.5 = 598.5 cm 3 , area: 2(9 • 3.8) + 2(3.8 • 17.5) + 2(3.8 • 17.5) = 516.4 cm 2 ). However a box with a dimension of 8.43 cm can holds the same amount of dirt but only ahs a surface area of 426.4 cm 2 (Volume: 8.43 • 8.43 • 8.43 = 599 cm 3 , area: 2(8.43 • 8.43) + 2(8.43 • 8.43) + 2(8.43 • 8.43) = 426.4 cm 2 ).
A(x) The total volume of dirt below A(x) is 10462.8852 ft 3 . The dimension of the box that holds the dirt is 21.87176445 ft. Work: XYZ = 10462.8852 ft 3 SA(surface area) = 2xy + 2xy + 2yz SA = 2x (10462.8852/(xy)) + 2xy + 2y(10462.8852/(xy)) SA = 20925.7704y -1 + 2xy + 20925.7704x -1 SA = 2y – 20925.7704x -2 = 0 SA = 2x – 20925.7704y -2 = 0 Since 2y – 20925.7704x -2 = 0, and 2x – 20925.7704y -2 = 0, x = y. 2x – 20925.7704x -2 = 0 2x = 20925.7704/x 2 2x 3 = 20925.7704 x 3 = 10462.8852

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Unformatted text preview: x = 21.87176445 B(x) The total volume of dirt below B(x) is 2884.506427 ft 3 . The dimension of the box that holds the dirt is 14.23499013 ft. Work: XYZ = 2884.506427 ft 3 SA = 2x (2884.506427/(xy)) + 2xy + 2y(2884.506427/(xy)) SA = 5769.012854y-1 + 2xy + 5769.012854x-1 SA = 2y – 5769.012854x-2 = 0 SA = 2x – 5769.012854y-2 = 0 x = y. 2x – 5769.012854x-2 = 0 2 2x = 5769.012854/x 2 2x 3 = 5769.012854 x 3 = 2884.506427 x = 14.23499013 C(x) The total volume of dirt below C(x) is 6768.616573 ft 3 . The dimension of the box that holds the dirt is 18.91617418 ft. Work: XYZ = 6768.616573 ft 3 SA = 2x (6768.616573/(xy)) + 2xy + 2y(6768.616573/(xy)) SA = 13537.23315y-1 + 2xy + 13537.23315x-1 SA = 2y – 13537.23315x-2 = 0 SA = 2x – 13537.23315y-2 = 0 x = y. 2x – 13537.23315x-2 = 0 2x = 13537.23315/x 2 2x 3 = 13537.23315 x 3 = 6768.616573 x = 18.91617418 3 Conclusion The math that we use in this part of project is very important to the real world because by using this formula to find the dimension of the box, we are able to save many space in the world. For example in business world, it is very important for them to create a box that can hold their box but on the other hand saves money. 4...
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• Spring '14
• DanielMcKinney

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