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Maximize the volume without using Lagrange multipliers

  1. Feb 28, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    When a rectangular box is sent through the mail, the post office demands that the length of the box plus twice the sum of its height and width be no more than 250 centimeters. Find the dimensions of the box satisfying this requirement that encloses the largest possible volume. (Solve this problem without using Lagrange multipliers.)


    2. Relevant equations
    Partial differentiation and equations of constraint for each variable.


    3. The attempt at a solution
    My attempt at a solution is attached however, given that I got length = l = 0 (even though I get a nonzero width and height) which gives a volume of 0, I'd say I did something wrong and I don't have the solutions or answer for this particular problem so I can't check what's wrong.
     

    Attached Files:

  2. jcsd
  3. Feb 28, 2012 #2

    lanedance

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    did you just equate the volume with zero? shouldn't you differentiate to maximise the volume and find where the derivative is zero?
     
    Last edited: Feb 28, 2012
  4. Feb 28, 2012 #3

    SammyS

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    attachment.php?attachmentid=44490&d=1330441256.jpg

    (Your solution is correct for the minimum volume.)

    Once you have w = h, put that back into the volume formula so V is only a function of w or h .

    Maximize that.
     
  5. Mar 1, 2012 #4

    s3a

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    lanedance, no that's not what I did. I did differentiate (partially) hence the subscripts. ;)

    SammyS, it makes sense that I got the minimum :) (thanks for mentioning it though because I initially thought I was doing something redundant rather than getting a minimum). What you said sounds like what I did though. Could you please be a bit more descriptive algorithmically?
     
  6. Mar 1, 2012 #5

    SammyS

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    [itex]\displaystyle V=\ell hw[/itex]

    If h = w , then = 250 - 4h ,

    and [itex]\displaystyle V=(250 - 4h)h^2\,.[/itex]

    Maximize that.
     
  7. Mar 1, 2012 #6

    s3a

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    In the attachment above, I already have

    V = 250w^2 - 4w^3

    which is basically what you said with w instead of h which is okay since w = h.

    As for maximizing that, do you mean taking dV/dw = 0 and solving for w?
     
  8. Mar 1, 2012 #7

    SammyS

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    That's the usual way.

    Try it.
     
  9. Mar 1, 2012 #8

    s3a

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    Is this what you meant?

    (By the way, I choose to reject w = 0 since it yields a minimum.)
     

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