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Can anyone help me with this mutltivariable optimisation problem?

  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data
    I have 12m2 of cardboard. I must make a rectangular box with no lid, but I must maximize the potential volume of the box. The length, width and height of the box are x,y,z.


    2. Relevant equations
    I know that xyz = V (volume). I know the surface area is 12m2
    ∴ 2xz+2yz+xy=12
    ∴z=(12-xy)/(2x+2y)
    ∴V = xy(12-xy)/(2x+2y)

    I've done the calculations to find the first and second derivatives:

    ∂V/∂x = -y2(x2+2xy-12)/2(x+y)2
    ∂V/∂y = -x2(y2+2xy-12)/2(x+y)2
    3. The attempt at a solution
    I know I must make these = 0 in order to find the critical points. I must then use the Hessian matrix to apply the second derivative test.

    For ∂V/∂x=0 I got the identity x=(12-y2)/2y
    I then subbed this into x2+2xy-12

    The I got this equation: 3y4+24y2-24=0

    I think the y values of this equations represent the y-values of the critical points, but I don't know how to evaluate them. Or maybe I made a mistake somewhere along the way. Can anyone help me with this? I'd really appreciate it.
     
  2. jcsd
  3. Apr 22, 2013 #2

    Mark44

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    Are you sure you're approaching this the right way? These kinds of problems usually start with a rectangular piece of cardboard or metal or whatever, and have squares cut out of the corners, forming four flaps. The flaps are then folded up to make the sides of the box.

    Also, 12 m2 is a BIG piece of cardboard.
     
  4. Apr 22, 2013 #3
    I'm pretty sure. I even looked over some tutorial notes online and the V=xy(12-xy)/(2x+2y) formula was mentioned. I know of the method with cutting out the corners. I'll try that now instead, but my lecturer has, annoyingly, failed to provide the final answer to the problem, so I can't even check if my work is correct.
     
    Last edited: Apr 22, 2013
  5. Apr 23, 2013 #4
    I just went over my notes. It seems I can only use the 'cutting out of corners' method when I have an actual value associated with the width, breath and height. I have no numerical values other the fact that the surface area = 12m2
     
  6. Apr 23, 2013 #5

    Dick

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    I think you are doing everything ok, basically. But I get 3y^4+24*y^2-144=0. That has simpler roots. Check your derivation. You can solve an equation like that by putting u=y^2 and solving the quadratic in u.
     
  7. Apr 23, 2013 #6
    Ah! Sweet, that makes complete sense. Thanks a million!
     
  8. Apr 23, 2013 #7

    Ray Vickson

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    Your solution is incorrect: the solution of the problem
    [tex] \max V = f(x,y) = \frac{xy(12 - xy)}{2(x+y)}[/tex]
    is ##x = y = 2##, ##V = 4##. However, the true solution to the problem is
    [tex] x = y = \frac{4}{\sqrt{3}}, \; z = \frac{1}{\sqrt{3}}, \: V = \frac{16 \sqrt{3}}{9}
    \doteq 3.0792.[/tex]
    Your expression for area is incorrect.
     
  9. Apr 23, 2013 #8

    Dick

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    Why?
     
  10. Apr 23, 2013 #9

    haruspex

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    I think Ray is assuming that the cardboard is provided as a single square, and that the box is to be made of a single piece cut from this. The second assumption would be reasonable if the first were a given, but since it is not the question becomes somewhat unclear.
    Since the shape of the provided cardboard is not given, we are forced to assume that we can cut it and reglue it as necessary, without loss. This leads to the 2x2x1 solution, but calculus is unnecessary if we assume that the most efficient closed rectangular box is a cube.
     
  11. Apr 23, 2013 #10

    Dick

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    The first part of the question responses went through cut out and fold option and rejected it. I guess Ray Vickson skipped that. Since it's an easy question and the question is labelled 'multivariable optimization' , I don't see why you would choose to 'assume that the most efficient closed rectangular box is a cube'. It's easy enough to prove.
     
  12. Apr 24, 2013 #11

    haruspex

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    Let me put it a different way. If you already have the result that the most efficient closed rectangular box is a cube, by whatever means, you don't need calculus to solve the open rectangular box case. It's a two-liner.
     
  13. Apr 24, 2013 #12

    Ray Vickson

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    As far as I can make out the rejection of the cut-and-fold method was illegitimate. The OP said:
    "I just went over my notes. It seems I can only use the 'cutting out of corners' method when I have an actual value associated with the width, breath and height. I have no numerical values other the fact that the surface area = 12m2".

    He seems to be saying that he is not allowed to use the cut and fold method unless he already knows the numerical values. (I assume he does NOT mean that his instructor will only allow the non-cut-and-fold method!)

    Without knowing the dimensions, the cardboard length is x + 2z and the width is y + 2z, so the area is (x+2z)*(y+2z) = 12. One can show that the optimal solution has x = y (for example, using the Lagrange multiplier method), so the problem is to maximize ##V = x^2 z,## subject to ##(x+2z)^2 = 12,## or ##x = \sqrt{12}-2z.## Now
    [tex]V = z (\sqrt{12}-2z)^2[/tex] is pretty easy to maximize.
     
  14. Apr 24, 2013 #13

    Dick

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    I think the main evidence that the intent was to have the final area of the box be 12m^2 is that Pizzerer found V=xy(12-xy)/(2x+2y) in the tutorial notes, which omitted the final answer. Getting a nice whole number solution that way is also suggestive. Actually, I think the OP's main question was how to solve the quartic.
     
  15. Apr 24, 2013 #14

    Ray Vickson

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    OK, but it would have been much easier if the OP noticed that V_x = 0 gives x^2 + 2xy = 12 and V_y = 0 gives y^2 + 2xy = 12. From these it follows that x = y, hence
    [tex]V = x^2 (12-x^2)/(4x) = (1/4) x (12-x^2).[/tex]
    It is much easier to maximize this simple cubic than to deal with a quartic (athough that quartic is a quadratic in disguise).
     
  16. Apr 27, 2013 #15
    Thanks for all the help, guys. I appreciate it.
     
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