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Maximizing volume of a box without lagrange multipliers

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the largest rectangular box having a fixed surface area must be a cube.

    2. Relevant equations

    ##V(x,y,z) = xyz##
    ##\sigma(x,y,z) = 2(xy + yz + zx) = C \in \mathbf{R}##

    3. The attempt at a solution

    As of this assignment, we haven't yet learned Lagrange multipliers, so that's out of the question. I solved ##\sigma## for ##z##, giving $$z(x,y) = \frac{C - 2xy}{2x + 2y}$$ Substituting in for ##z## in ##V##, I get $$V(x,y) = \frac{Cxy - 2x^2 y^2}{2x + 2y}$$ Taking the Jacobian gives a row matrix whose elements are ##\frac{Cy - 4xy^2}{2x + 2y} + \frac{4x^2y^y - 2Cxy}{(2x + 2y)^2}## and ##\frac{Cy - 4x^2y}{2x + 2y} + \frac{4x^2y^y - 2Cxy}{(2x + 2y)^2}## If I were to proceed, I would set these equal to zero, find the critical points, and use the Hessian, but I feel as if I'm proceeding incorrectly due to the complexity of the derivatives and the presence of ##C##. Any advice?
     
  2. jcsd
  3. Oct 30, 2013 #2

    LCKurtz

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    I don't think there is any shortcut. In your equation for ##\sigma## you could divide both sides by ##2## using a new constant. Also you can combine the fractions in your derivatives. I presume those ##y^y## factors are typos. I didn't work it all out but my advice is to plug and chug.

    If nothing else, it will give you an appreciation for LaGrange multipliers. Could be that is the point of the exercise at this stage of your studies.
     
  4. Oct 30, 2013 #3
    Yeah, those should have been ##y^2## in both cases. I can definitely see that, given how much they stressed that we not use Lagrange multipliers here. Thank you for the confirmation!
     
  5. Oct 30, 2013 #4
    Just as an update, after two and a half pages of scratch work, plugging and chugging did in fact work out.

    Thank god for Lagrange multipliers.
     
  6. Oct 30, 2013 #5

    Ray Vickson

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    Just as a matter of interest: it is much, much simpler to solve the problem of minimizing surface area, subject to a given value of volume. In the fixed-volume problem you have z = V/(x*y), for example, and putting that into σ(x,y,z) yields s quite simple minimization problem. You might then think about whether the fixed-volume, minimum area and fixed-area, maximum volume problems are essentially the same.
     
  7. Oct 31, 2013 #6

    ehild

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    z is function of x and y: ##z(x,y)=\frac{C-2xy}{2x+2y}##

    So the derivatives of V=xyz can be written as

    ##\frac{\partial V}{\partial x}=yz+xy\frac{\partial z}{\partial x}##

    ##\frac{\partial V}{\partial y}=xz+xy\frac{\partial z}{\partial y}##

    with

    ##\frac{\partial z}{\partial x}=-\frac{y^2+C/2}{(x+y)^2}##
    ##\frac{\partial z}{\partial y}=-\frac{x^2+C/2}{(x+y)^2}##

    and both partials of V should be zero:


    ##yz=-xy\frac{\partial z}{\partial x}## *
    ##xz=-xy\frac{\partial z}{\partial y}## **

    divide eq *- with eq ** and simplify.

    ehild
     
    Last edited: Oct 31, 2013
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