Mathematica How Much Weight Can a Cardboard Boat Hold?

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The discussion centers on determining the maximum weight a cardboard boat made from a single 4x8 foot sheet can hold, with calculations suggesting a potential capacity of 1010.31 pounds. Participants emphasize the importance of shape and structural design, noting that while complex shapes could complicate calculations, simpler box shapes are more manageable for maximizing volume and weight capacity. The approach involves using calculus, specifically the method of Lagrange multipliers, to find the optimal dimensions for the boat. It is highlighted that the top of the box does not need to be covered to maximize volume, which is crucial for weight support. Overall, the challenge lies in effectively cutting and folding the cardboard to achieve the theoretical weight capacity.
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i was given an assignment in my design engineering class...it goes as follows:

What is the maximum amount of weight that can float in a boat made out of 1 sheet of cardboard 4 feet by 8 feet in size...when flat but it can be folded and cut and glued in any shape you choose.



**you also have to be able to prove your solution**

if anyone could help me figure this out i would really appreciate it

thanks---
brooke
 
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if i did my calculation correct...then the maximum weight it could hold is 1010.31 pounds...im jst not sure how i would cut or fold the 8' by 4' sheet of cardboard to be able to prove my solution
 
I vote none since the boat will sink after a very short time after being introduced to the water.
 
If the cardboard is allowed to take any shape, then this is a very complicated problem involving the calculus of variations. If we are confining ourselves to box shapes, the problem becomes much more managable. Call the dimentions x, y, and z. You have to find the box of maximal volume with a given surface area. (This will be the box that holds the most weight since the amount of weight an object can have before sinking is the wieght that water of an equivalent volume has.) The thing to notice is that the top of the box does not need to be covered and shouldn't be in the box of maximal volume since it would be a waste of surface area. If x is the length, y is the hieght, and z is the other dimention, then one of the xz surfaces does not need to be there. So you must maximize
f(x,y,z)=xyz
subject to
2xy + 2yz + xz=32
The standard way of doing this would be the method of Lagrange multipliers.
 
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