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Max Box problem

  1. Jan 25, 2006 #1
    For those who don't know the max box problem here is a quick recap...

    A lidless box is to be constructed from a square sheet of paper measuring 15cm by 15cm, by cutting out squares from the corners and then folding up the sides. Let x cm be the size of one of the sides of the small corner squares.

    Question: prove the following "the maximum volume will always be of a height which is one sixth of the length of one side of the square piece of paper used". If S is the length of one side of the large square and x as before, write down the volume in terms of S and x and use calculus to prove that indeed this is right?

    Here is my work so far....

    Volume is x (S-2x)^2
    = x (S-2x)(S-2x)
    = x (S^2 - 4Sx + 4x^2)
    = 4x^3 - 4Sx^2 + xS^2

    Can anyone help with the rest please?
     
  2. jcsd
  3. Jan 25, 2006 #2

    StatusX

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    So... maximize that.
     
  4. Jan 25, 2006 #3
    can't do it... don't know what you mean by maximize sorry :frown:
     
  5. Jan 25, 2006 #4

    StatusX

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    I don't understand why you were given this problem if you don't know how to maximize something. Just think of the graph as a function of f. What will be true at the point where the volume is a max (hint: what will the derivative of f be)?
     
  6. Jan 25, 2006 #5
    if y= 4x^3 -4 Sx^2 + xS^2

    then dy/dx = 12 x^2 + 2 xS- 8 Sx right?

    min/max when dy/dx = 0 so....

    x = 0.73
     
  7. Jan 25, 2006 #6

    StatusX

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    I think you're missing an S^2, but besides that it looks ok.
     
  8. Jan 27, 2006 #7
    I don't get this :-(. This is my work so far...

    Length Width height Volume
    13 13 1 169
    11 11 2 242
    10.8 10.8 2.1 245
    10.6 10.6 2.2 247
    10.4 10.4 2.3 249
    10.2 10.2 2.4 250
    10 10 2.5 250
    9.8 9.8 2.6 250
    9.6 9.6 2.7 249
    9.4 9.4 2.8 247
    9.2 9.2 2.9 245

    Now for x = 2.5 cm we have the highest volume which is V = 250 cm^2 (sorry forgot to precise I was working in cm, obviously as it's a piece of paper or cardboard)... anyway the point is ow can "the maximum volume will always be of a height which is one sixth of the length of one side of the square piece of paper used" because this would mean in my example above I get for a height of 2.5 cm gives 10 cm (length) for this box. How can this be 1/6???

    Also how can I do it in terms of S being the length and x as before i.e the size of one of the corners of this sheet of paper (basically the height)???

    Please help :frown:
     
    Last edited: Jan 27, 2006
  9. Jan 27, 2006 #8

    HallsofIvy

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    Okay, so you know how find a derivative- maybe you missed the class in which you were told that at a max or min, the derivative is 0.

    y= 4x^3 -4 Sx^2 + xS^2 so dy/dx= 12 x^2 - 8 Sx + S^2
    (you had "2 xS". Did you forget that S is a constant?)

    Solve the quadratic equation 12x^2- 8Sx+ S^2= 0.
    (Of course, in your first problem S= 15. Leave S in to see that x is always 1/6 S.)
     
  10. Jan 27, 2006 #9
    I must be really thick but I get x = 1/2 and x = 0.167 but I can't see the x is 1/6 S? Can someone just show it to me please :uhh:
     
  11. Jan 27, 2006 #10

    TD

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    Have you tried solving that quadratic equation which HallsofIvy gave you?
    You can use the quadratic formula for example.
     
  12. Jan 27, 2006 #11
    I have used it to get the two x values of above
     
  13. Jan 27, 2006 #12

    TD

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    I mean this one: "Solve the quadratic equation 12x^2- 8Sx+ S^2= 0"

    You couldn't have found those x-values from this equation since it still contains S!
     
  14. Jan 27, 2006 #13
    I would love the answer please... :cry:
     
  15. Jan 27, 2006 #14

    TD

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    It shouldn't be too hard now (although I didn't follow the problem from the start). Just start with the quadratic equation [itex]12x^2-8Sx+S^2=0[/itex] and solve it with the quadratic formula, treating S as an unknown constant. So use:

    [tex]x_{1,2} = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}[/tex]
     
  16. Jan 27, 2006 #15
    Well this gives me x1 = 0.5 and x2 = 0.167 (which is 1/6)
     
  17. Jan 27, 2006 #16

    TD

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    It can't, where did S go? Remember that you're not supposed to fill in a value of S, just leave it as an unknown constant. It can't just disappear... In the formula: a is now 12, b is -8S and c is .
     
  18. Jan 27, 2006 #17
    ahhhhhhhhhhhhhhhhhhhhhhhh! I was using a = 12, b = -8 and c = 1
     
  19. Jan 27, 2006 #18

    TD

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    That's only in the case when S would be 1, but for now we're trying to solve it for any S so you shouldn't leave that out :wink:

    Well I'm logging off in a minute, I assume you'll be able to handle it now?
     
  20. Jan 27, 2006 #19
    wait another 2 mins please I get 8S +or - sqrt 64/24S^2 - 2S^2
     
  21. Jan 27, 2006 #20

    TD

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    Well I'm getting

    [tex]x_{1,2} = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} = \frac{{8s \pm \sqrt {64s^2 - 4 \cdot 12 \cdot s^2 } }}{{24}}[/tex]

    The expression under the square root simplifies nicely to a square...
     
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