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Multivariable Optimization Problem

  1. Aug 10, 2004 #1
    I have two questions.

    A) Show the parallelipided with fixed surface area and maximum volume is a cube.

    I've already proven that we can narrow down the proof to a box. So, basically, I'm really lost on how do prove that a cube is the box with a fixed surface area and maximum volume.

    B) We might not have covered how to do part B yet, so i'll create a new topic if I still don't understand after tomorrow's lecture.
  2. jcsd
  3. Aug 10, 2004 #2


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    How is a "box" different from a parallelpiped?

    Call the lengths of the sides of your parallelpiped x, y, and z.

    The volume is V= xyz.

    The surface area is 2xy+ 2xz+ 2yz= A (a constant).

    Now use the "Lagrange multiplier" method.

    In order that V= xyz be a minimum (or maximum!) on the surface U=2xy+2xz+ 2yz- A=0, the two gradient vectors, grad V= <yz, xz, xy> and grad U= <2y+ 2z,2x+ 2z, 2x+ 2y> must be parallel. That is we must have <yz, xz, xy>= some multiple of <2y+2z, 2x+ 2z, 2x+ 2y> so that yz= &lambda;(2y+ 2z), xz= &lambda;(2x+ 2z), and
    xy= &lambda;<2x+ 2y>. Eliminate &lambda; from tose equations and see what happens.
  4. Aug 11, 2004 #3


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    Dearly Missed

    A box is characterized with right angles, whereas a parallellepiped need not be subject to this constraint.
  5. Aug 11, 2004 #4


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    Ah, right. Thanks.
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