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Constrained optimization troubles

  1. Jul 2, 2008 #1
    I'm trying to find the regular parallelepiped with sides parallel to the coordinate axis inscribed in the ellipsoid x[2]/a[2] + y[2]/b[2] + z[2]/c[2] = 1 that has the largest volume. I've been trying the Lagrangian method: minimize f = (x)(y)(z), subject to the constraint (x[2]/a[2] + y[2]/b[2] + z[2]/c[2] - 1 = 0.

    My new function F = (x)(y)(z) + L(x[2]/a[2] + y[2]/b[2] + z[2]/c[2] - 1) however leads to zeroes after differentiation and the linear system is solved. I'm stuck . . .
     
  2. jcsd
  3. Jul 2, 2008 #2

    HallsofIvy

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    The linear system is solved? What's bad about that? If you mean x= 0, y= 0, z= 0 are solutions, that's a solution but is not the only solution.
    What you get, using the LaGrange multiplier method, is:
    [itex]yz=2\lambda x/a[/itex]
    [itex]xz= 2\lamda y/b[/itex]
    [itex]xy= 2\lambda z/c[/itex]

    Since any one of those being 0 will give volume 0, that's obviously the minimum. Assuming none are 0, we can eliminate [itex]\lambda[/itex] by dividing one equation by another. For example, dividing the second equation by the first gives x/y= ay/bx which, in turn gives [itex]y^2= b^2x/a^2[/itex] so (we can assume x and y are positive here) y= bx/a. Similarly, dividing the second equation by the third results in z= cx/a. replace y and z in [itex]x^2/a^2+ y^2/b^2+ z^2/c^2= 1[/itex] and solve for x.
     
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