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anonymous

  • one year ago

optimization problem. Maximum volume of a box? I don't know how to set this up. We're using a paper with dimensions 8(1/2) inches by 11 inches to form a box WITH a lid with maximum volume. Obviously I'm using the formula v=length x width x height but I don't know how to get v(x) :/

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  1. dan815
    • one year ago
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    |dw:1433466000616:dw|

  2. dan815
    • one year ago
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    also ask yourself are you able to cut the paper in any way u want or do u have to fold it

  3. dan815
    • one year ago
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    is it possible to just wrap this paper around some cube?

  4. dan815
    • one year ago
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    any thoughts so far?

  5. dan815
    • one year ago
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    |dw:1433466271247:dw|

  6. anonymous
    • one year ago
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    ok so I'm able to cut identical squares out of the corners and in the middle so I cold then fold it up into the box.

  7. ybarrap
    • one year ago
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    You might want to try using the method of Lagrange: http://en.wikipedia.org/wiki/Lagrange_multiplier $$ \text{Maximize: }l\times w\times h\\ \text{Subject to the constraint: }2\times(hw+lh+wh)=8.5\times 11\\ \text{Hence,}\\ \Lambda(x,y,z,\lambda)=lwh+\lambda\left (hw+lh+wl-\cfrac{8.5\times 11}{2}\right ) $$ Now find the partials: $$ \Lambda_h=wl+\lambda(w+l)=0\\ \Lambda_w=hl+\lambda(h+l)=0\\ \Lambda_l=hw+\lambda(h+w)=0\\ \Lambda_{\lambda}=hw+lh+wl-\cfrac{8.5\times 11}{2}=0\\ $$ Using first equation, we find for \(\lambda\) $$ \lambda=-\cfrac{wl}{w+l} $$ Plugging this into the second and third equation we get that \(w=l\) Using this result, we get that \(\lambda =-\cfrac{w}{2}\) Using these results into the last two equations we find the final value for \(h\). I'll let you do that. Once you have this result, plug back into your constraint to find the actual values. Does this make sense?

  8. anonymous
    • one year ago
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    I had to reread it a few times but I think I got it. Thank you :)

  9. ybarrap
    • one year ago
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    You're welcome

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