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Homework Help: Findin largest volume

  1. Dec 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Johnny needs to make a rectangular box for his physics class project. He has bought P cm of wire and S cm^2 of special paper. He would like to use all the wire (for the 12 edges) and paper (for the 6 sides) to make the box.

    What is the largest volume of the box that Johnny can make?
    One could use any method to solve the above problem.
    I just need help getting started.

    First for the wire.

    I think the perimeter function for the wire is : 4(x+2y) [ check my work please].

    We are given S, the area of the paper. So a function for the paper might be, x^2

    And we have the volume function for the box : V = xyz

    So we have 3 functions :

    Perimeter :z = 4(x+2y)
    Area : y = x^2
    Volume : V = xyz

    From there I guess I can substitute. But I know something is wrong.

    Any help solving this problem. Hints will be fine. Can I do this with double integrals?
    Last edited: Dec 26, 2009
  2. jcsd
  3. Dec 26, 2009 #2


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    Hi tnutty! :smile:
    Yes, the volume is right, but the perimeter and area aren't.

    Try again … what are the perimeter and area of a reactangular box with sides x y and z ?
  4. Dec 26, 2009 #3
    Perimeter : 2(2x + 2y) + 2(2z) = 4x + 4y + 4z
    Area of a rectangle box = 2(xy) + 2(xz) + 2(yz) = 2xy + 2xz + 2yz
    V = xyz

    What do you think?

    Also how do you think we could incorporate the variable P and S ?
    Last edited: Dec 26, 2009
  5. Dec 27, 2009 #4


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    Hi tnutty! :smile:

    (just got up :zzz: …)
    That's correct (though a little long-winded).

    Hint: what is Area times z ? :wink:
  6. Dec 27, 2009 #5
    Area times z is also volume no? And what is meant by long-winded ?

    V = xyz
    V = A * z
    A = 2x(y+z) + 2yz
    P = 4(x+y+z)

    Last edited: Dec 27, 2009
  7. Dec 27, 2009 #6


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    No, V is not Az :confused:

    A = 2(xy + xz + yz), so what is Az in terms of P and V ?

    (and "long-winded" was referring to the intermediate stage with all those brackets)
  8. Dec 28, 2009 #7

    I thought Az was the volume? Since x*y = Area, then V = Az. ?

    I am not sure about the second question.

    A = 2(xy+xz+yz ) then

    A*z = 2z(xy+xz+yz) ?
  9. Dec 28, 2009 #8


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    You're using "area" to mean two different things.

    xy is the area of one face of the box.

    There are three different sizes of face, and A is the sum of the areas of all the faces.
    Yes … and what is that in terms of P and V ? :smile:
  10. Dec 28, 2009 #9
    A*z = 2z(xy+xz+yz)

    V = xyz
    x = V/yz

    P/4 = x+y+z
    P/4 - x - y = z

    A*z = 2z(xy+xz+yz)

    A( P/4 - x - y ) = 2(P/4 - x -y) ( V/z + V/y + yz )

    is that right?

    For a rectangular box, we could compute the area of one size and multiply by its depth
    to get the volume no? I am confused why it isn't. Can you take a look at this picture, http://img64.imageshack.us/img64/415/31659034.png" [Broken].

    The integration of an area over a region is the object's volume right? So
    the way I thought about it is that we compute the area of a strip of the box( the red strip in the picture), This strip is dz. And since the area does not change over dz, we can pull the area out and the integral reduces to Area * Z . Thus volume is Area * Z, or Area*Depth?
    Last edited by a moderator: May 4, 2017
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