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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