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Math Help - minimum Dimensions of an open box

  1. #1
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    minimum Dimensions of an open box

    1. The problem statement, all variables and given/known data
    Determine the dimensions of a rectangular box, open at the top, having volume
    4 m3, and requiring the least amount of material for its construction. Use the
    second partials test. (Hint: Take advantage of the symmetry of the problem.)



    2. Relevant equations



    3. The attempt at a solution

    Volume, V= 4m^3
    let x = length
    y = width
    z = height

    4m^3 = xyz
    x = 4/yz

    because it is an open at the top rectangular box,
    the Surface, S = 2xz + 2yz + xy

    substitute x=4/yz inside the surface equation..
    S = 8/y + 2yz + 4/z

    to find the critical points, take the partial with respect to y and z.. the equal it to zero..

    S'(y)= 2z - 8/y^2
    S'(z)= 2y - 4/z^2

    solve the equations, i get,
    when y = 0, z = 0..
    y = 8, z = 1/16

    the problem is, what should i do next?
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  2. #2
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    Quote Originally Posted by nameck View Post
    1. The problem statement, all variables and given/known data
    Determine the dimensions of a rectangular box, open at the top, having volume
    4 m3, and requiring the least amount of material for its construction. Use the
    second partials test. (Hint: Take advantage of the symmetry of the problem.)



    2. Relevant equations



    3. The attempt at a solution

    Volume, V= 4m^3
    let x = length
    y = width
    z = height

    4m^3 = xyz
    x = 4/yz

    because it is an open at the top rectangular box,
    the Surface, S = 2xz + 2yz + xy

    substitute x=4/yz inside the surface equation..
    S = 8/y + 2yz + 4/z

    to find the critical points, take the partial with respect to y and z.. the equal it to zero..

    S'(y)= 2z - 8/y^2
    S'(z)= 2y - 4/z^2

    solve the equations, i get,
    when y = 0, z = 0..
    y = 8, z = 1/16

    the problem is, what should i do next?

    The solutions are y=2\,,\,z=1\,\Longrightarrow x=2 and you're done (just check why you got y = 8...??)

    Perhaps you oversaw the fact that you're positioning your box at the origin in the 3-dimensional space, with sides parallel to the three axis. Then, x = 2 is the length of the box's side along the x-axis, y = 2 the length along the y-axis and etc.

    If you already studied Lagrange's Multipliers then you can double-check with them that the answer indeed is the above one.

    Tonio
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  3. #3
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    ok.. i repaired my calculation..
    got problem with the simultaneous equation..
    i got ur answer where y=2 and z=1..
    the i substitute the value into xyz=4..
    i got x=2..
    however, is this my final answer?
    should i use second partial test?
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  4. #4
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    Quote Originally Posted by nameck View Post
    ok.. i repaired my calculation..
    got problem with the simultaneous equation..
    i got ur answer where y=2 and z=1..
    the i substitute the value into xyz=4..
    i got x=2..
    however, is this my final answer?
    should i use second partial test?


    Well...yes, in order to be formal. You'll see that the Hessian is a positive definite matrix and thus you got a minimum (or f_{yy}>0\,,\,\,f_{yy}f_{zz}-(f_{yz})^2>0) when evaluated at (2,2,1)

    Tonio
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  5. #5
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    nice! i got it now tonio!!
    u r really helpful!!
    thanks =)
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