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Maximum volume of a hexahedron

  1. Sep 4, 2008 #1
    Hello everyone, may I ask a question? I've been struggle with it for a few days.

    There's a cuboid (rectangular prism) R, with dimensions of x, y, z respectively. A point P is outside of R. The shortest distance between P and R is d. What is the maximum possible volume of the new hexahedron formed by P and R?

    Is something like d*sqrt((x*y)^2 + (y*z)^2 + (x*z)^2) + x*y*z close?
     
  2. jcsd
  3. Sep 5, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hello huahaiy! Welcome to PF! :smile:

    I don't understand … what do you mean by the hexahedron (six-sided solid) formed by P and R? :confused:
     
  4. Sep 5, 2008 #3
    Re: Welcome to PF!

    Thanks for reply. Sorry I didn't make it clear.

    Basically, I mean the vertex of R that is closest to P is now replaced by P, this gives a new hexahedron which is bigger. Depending on where P is, the volume of this new hexahedron will be different. I would like to know what is the maximum volume. Is this clearer?

     
  5. Sep 5, 2008 #4

    tiny-tim

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    ah!

    ah!

    Hint: if the vertex A is closest to P, and A is connected to vertices B C and D, then the volume of the new hexahedron depends on the volume of tetrahedron PBCD (the rest is fixed).

    And if you have two tetrahedra with the same base (BCD), how do you compare their volumes? :wink:
     
  6. Sep 5, 2008 #5
    Re: ah!

    Look at the height? Thanks for the hint. But the rest doesn't seem to be fixed. My question was in fact wrong. The problem is more complicated than that: if A is replaced by P, to keep the resulting solid convex, a hexahedron won't do. We will get a solid with 9 sides instead of 6:

    Say vertex A is closest to P, A is connected to vertices B, C and D. Both B and C are connected to vertex E. After P replaced A, the rectangle ABCE will be turned into two triangles PBE and PCE, which are not on the same plane! So we end up with a 9 sided solid.

    If this is the case, the part that is fixed is only the tetrahedron to the opposite side of A.

     
  7. Sep 5, 2008 #6
    Re: ah!

    I think I got it. The maximum volume seems to be 2/3*d*S + x*y*z, where S is the area of the triangle BCD. Is that right?

    Thanks a lot for the hint.

     
  8. Sep 6, 2008 #7

    tiny-tim

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    Hi huahaiy! :smile:
    I don't follow that :confused: … if the cuboid is made of twelve pieces of elastic joining eight vertices, then pulling one vertex out of position will still give eight faces …
    E + F = V + 2. :smile:
    Looks good! (except … isn't it 1/3 ?) :biggrin:
     
  9. Sep 6, 2008 #8
    Oh, Euler characteristic, I forget to use it. :shy: V + F = E + 2. For a cuboid, it's 8+6=12+2. But when a vertex is pulled out of place, there will be 3 extra edges, so it become 9 faces. Right?

     
  10. Sep 6, 2008 #9

    tiny-tim

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    But the number of vertices is the same, so you've removed one vertex and replaced it by another.

    And the removed vertex had three edges, so they've gone also.

    So three edges have gone, and three new ones have arrived.

    That's still 12 edges and 6 faces, isn't it? :smile:
     
  11. Sep 6, 2008 #10
    On top of three edges gone and coming, three extra edges have arrived. They were the diagonal lines of the three rectangles facing P, now have been pulled out of the planes to become edges of their own :smile:

     
  12. Sep 6, 2008 #11

    tiny-tim

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    ping!

    ah … got it! … :biggrin:

    (i can't think in 3D! :redface:)

    yes, you're right … the four vertices of each of the three new "faces" aren't in a plane, so each "face" is really two faces, and there are three new edges along the base of our pyramid …

    so the volume calculation is still correct, but it isn't a hexahedron! :smile:
     
  13. Sep 6, 2008 #12
    Re: ping!

    I can't too. I carefully drew a picture with a ruler to figure it out :smile:

     
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