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Creating a box and finding volume

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data

    There is a sheet of metal 30 x 12 inches long and you want to make a box by cutting x by x corners out at each corner of the sheet metal. (x will then be your height) The box will have no top to maximize volume. What is the maximum volume and what are the side lengths?

    2. Relevant equations

    n/a

    3. The attempt at a solution

    cPmkgLD.png

    I don't really remember the question too much, but that is the image I drew associated with it. Basically, I am trying to find the maximum volume, and side lengths of the box.

    I made the equation f(x)=x(30-2x)(12-2x)
    Now, how do I find the maximum volume and side lengths?
     
    Last edited: Apr 17, 2013
  2. jcsd
  3. Apr 17, 2013 #2
    You will want to use the derivative of the volume function to find the max volume. Remember that a function has a (local) max where its derivative changes from positive to negative. Technically you also have to check the endpoints for x (x=0 and x=6 or 10, see below but it is obvious that neither of these will give you a max as they both give a volume of zero).
    (Also, you have a mistake in either the question or your diagram. The question says the second side length is 20; your diagram shows a 12).
     
  4. Apr 17, 2013 #3
    Oh ok, thanks it is a 12 not a 20, also the graph looks like this:

    EQsDStL.png

    Would the red point(local maximum) be the maximum volume? and if that is the maximum volume, would that x value be the sides lengths? so you would have to plug in x for height, length and width?
     
  5. Apr 17, 2013 #4
    Yes, that red point would represent your max. Its y-coordinate would be the volume. You know the three dimensions of your box in terms of x, so you are done once you find x. Were you expected to find it by graphical methods or by calculating it?
     
    Last edited: Apr 17, 2013
  6. Apr 17, 2013 #5
    Calculating it but we get to use a graphing calculator, so how would you do it by calculating it?

    Just to be clear, I would take the x-value and plug it into the three equations for height, length and width?
     
  7. Apr 17, 2013 #6
    If you know calculus you could use derivatives. If not, there is an algebraic way using arithmetic mean-geometric mean inequality, but it would be really sloppy for three factors.

    Yes, you would plug it back into the three equations. For example, if x was 1, your box would have dimensions of 28x10x2.
     
  8. Apr 17, 2013 #7
    Alright thanks, we have not yet learned derivatives yet so I think I will just saying, "according to the graph..."

    You have been a GREAT help, really appreciate it
     
  9. Apr 17, 2013 #8

    Mark44

    Staff: Mentor

    This graph is not accurate for this problem, as there's a domain to consider.

    Your volume function is V(x) = x(30 - 2x)(12 - 2x). Since you are making a physical box, none of the dimensions can be negative, and if you want to be picky, they can't be zero, either. This means that x ≥ 0, 30 - 2x ≥ 0, and 12 - 2x ≥ 0. Since both of these inequalities must be true, we have 0 ≤ x ≤ 6. This interval corresponds to the first arch in your graph.
     
  10. Apr 17, 2013 #9
    But the question asks for a distinct answer maybe corresponding to the max volume but I understand what you are saying though
     
  11. Apr 17, 2013 #10

    Mark44

    Staff: Mentor

    The high point in the first arch represents the maximum volume. The y coordinate at that point is the maximum volume, and the x coordinate indicates how deeply you can cut each corner. With a graphing calculator, you can zoom in on the high point and read the coordinates fairly accurately.

    My point is that you can ignore everything on the graph for which x < 0 or for x > 6.
     
  12. Apr 17, 2013 #11
    Thanks I understand as the sides need to follow basic logic of being greater than zero but also staying with positive volume within the first arch of the graph
     
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