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Finding the max volume of a box with an open lid

  1. Nov 22, 2009 #1
    I need help with this problem please.
    Finding the max volume of a box with an open lid.
    Using a 20 in by 20 inch sheet of cardboard and cutting out the corners.

    Equations i been using:
    Volume: Length*Width*height
    Area: L*W
    Perimeter: 2L+2W

    Can someone help me get started
     
  2. jcsd
  3. Nov 22, 2009 #2
    Re: Optimizations

    you have the volume equation correct, but your constraint equation should be (surface area) = 20in * 20in = 400 in^2. so what is the surface area of a topless box?

    figure that out, then its a straightforward application of the lagrange multiplier technique.
     
  4. Nov 22, 2009 #3
    Re: Optimizations

    ?? ok im confused wouldnt that be Area if 20in * 20in?, so the surface area of the sheet of cardboard is 400 in^2?.

    Then the surface area of the box with no lid would be: SA= 2(hw)+2(hl)+(lw)?
     
  5. Nov 22, 2009 #4
    Re: Optimizations

    This may be irelevant but i went and set the surface area of the sheet to the surface area of the box with no lid.

    400=2hw+2hl+lw => 400-wl=h(2w+2l) => (400-wl)/(2w+2l)=h
     
  6. Nov 22, 2009 #5
    Re: Optimizations

    yep, my interpretation of the problem is that you have 400in^2 of cardboard to make as large of a topless box as you can.

    So you want to maximize V(l,h,w) = l*h*w given that S(l,h,w) = 2(hw)+2(hl)+(lw) = 400.

    If you know the Lagrange multiplier technique, use that. If not, you will need to use the S equation to solve for one of the variables l,h,w (it doesn't matter which) in terms of the other variables, substitute that into the V equation, and then maximize the resultant function of two variables. Make sense?
     
  7. Nov 22, 2009 #6
    Re: Optimizations

    Kinda. So i got to use S(l,h,w)= 2(hw)+2(hl)+(lw)=400 and V=lwh?
     
  8. Nov 22, 2009 #7
    Re: Optimizations

    Indeed. Using what you wrote above, we have:

    400=2hw+2hl+lw => 400-wl=h(2w+2l) => (400-wl)/(2w+2l)=h


    Hence V = lwh = lw*(400-wl)/(2w+2l).

    Now what do you do to find the maximum of V as a function of w and l?
     
  9. Nov 22, 2009 #8
    Re: Optimizations

    Ok so say we want to maximize w from that equation:
    we would have to use the chain rule as well as the quotient rule?
     
  10. Nov 22, 2009 #9

    Office_Shredder

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    Gold Member

    Re: Optimizations

    I don't believe this is set up correctly. When you cut the corners off of a piece of paper/cardboard you can fold up the sides to make a box. You will lose surface area depending on how much stuff you have cut away

    Here's a picture I found on google that demonstrates the idea

    http://i.ehow.com/images/GlobalPhoto/Articles/4855626/143703-main_Full.jpg [Broken]


    You have to make your cuts all squares of the same size in order to make the box actually line up
     
    Last edited by a moderator: May 4, 2017
  11. Nov 22, 2009 #10
    Re: Optimizations

    Ok i understand that you will loose Surface area when you make cuts. Still confused though
     
  12. Nov 22, 2009 #11
    Re: Optimizations

    You're right. The way I set up the problem is incorrect. Well, you're kind of right. I don't think the problem is asking for foldable tabs as in the picture.

    They are probably asking for something like this: http://imgur.com/EEJy4" [Broken] (sorry for the crappy paint drawing), where you are to cut out the shaded corners. Sorry for any confusion!

    Anyway, you still apply the same technique I outlined above: figure out the constraint equation, solve for one of the variables, substitute that into the volume equation, and maximize the volume.
     
    Last edited by a moderator: May 4, 2017
  13. Nov 22, 2009 #12
    Re: Optimizations

    ok thanks for the help
     
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