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Math Help - box dimensions

  1. #1
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    box dimensions

    The question is : Find the dimensions of the rectangular box with largest volumes if the total surface area is given as 64cm^2. (it is a closed top box)

    What I have done
    Let x =lenght, y = width, and z =height
    The volume of the box is equal to V = xyz
    Subject to the surface area
    S = 2xy + 2xz + 2yz = 64
    = 2(xy + xz + yz)
    = 2[xy + x(64/xy) + y(64/xy)]
    S(x,y)= 2(xy + 64/y + 64/x)
    Then
    Mx(x, y) = y = 64/x^2
    My(x, y) = x = 64/y^2
    y^2 = 64/x
    (64/x^2)^2 = 64
    4096/x^4 = 64/x
    x^3 = 4096/64
    x^3 = 64
    x = 4
    y = 64/x^2
    y = 4
    z= 64/yx
    z= 64/16
    z = 4

    Therefor the dimensions are cube 4.

    Is this correct or was i suppose to use a different equation?
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  2. #2
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    Smile

    Quote Originally Posted by mellowdano View Post
    The question is : Find the dimensions of the rectangular box with largest volumes if the total surface area is given as 64cm^2. (it is a closed top box)

    What I have done
    Let x =lenght, y = width, and z =height
    The volume of the box is equal to V = xyz
    Subject to the surface area
    S = 2xy + 2xz + 2yz = 64
    = 2(xy + xz + yz)
    = 2[xy + x(64/xy) + y(64/xy)]
    S(x,y)= 2(xy + 64/y + 64/x)
    Then
    Mx(x, y) = y = 64/x^2
    My(x, y) = x = 64/y^2
    y^2 = 64/x
    (64/x^2)^2 = 64
    4096/x^4 = 64/x
    x^3 = 4096/64
    x^3 = 64
    x = 4
    y = 64/x^2
    y = 4
    z= 64/yx
    z= 64/16
    z = 4

    Therefor the dimensions are cube 4.

    Is this correct or was i suppose to use a different equation?
    x = y = z = 4 doesn't satisfy the surface equation S = 2xy + 2xz + 2yz = 64

    why do you substitute z = \frac{64}{xy} to the surface equation?
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  3. #3
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    You appear to be confusing your area with your volume. z= 64/xy only if xyz, the volume, is 64 and that's not true. It is the area that is 64.

    The simplest way to do this is to use "Lagrange multipliers".

    With V(x,y,z)= xyz and A(x,y,z)= 2xy+ 2yz+ 2xz= 64, \nabla V= yz\vec{i}+ xz\vec{j}+ xy\vec{k} and \nabla A= (2y+ 2z)\vec{i}+ (2x+ 2z)\vec{j}+ (2y+ 2x)\vec{k} and so, at an extemum, we must have \nabla V= \lambda \nabla A for some constant \lambda.

    That reduces to yz= \lambda (2y+ 2z), xz= \lambda (2x+ 2z), and xy= \lambda (2x+ 2y). I find that dividing one equation by another, to immediately eliminate \lambda is often a good idea.

    And, of course, don't forget that you still have 2xy+ 2yz+ 2xz= 64.
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  4. #4
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    Ok, so after changinf a few things around, i got my final amswer to be
    x = y = z =3.2659 cm.

    Correct?
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  5. #5
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    Quote Originally Posted by mellowdano View Post
    Ok, so after changinf a few things around, i got my final amswer to be
    x = y = z =3.2659 cm.

    Correct?
    If you mean 4\sqrt{2/3}, yes, that is what I get.

    Actually, you can "shortcut" the entire calculation by using what Rennaissance philosophers called "lack of sufficient reason". By the symmetry of the problem, there is no reason for one side to be larger than any other- the solid with maximum volume is a cube. (If your teacher objects to 'lack of sufficient reason', just say "by symmetry".) That is x= y= z so 6x^2= 64, x^2= \frac{64}{6}= \frac{32}{3}= \frac{16\cdot 2}{3} and x= y= z= 4\sqrt{\frac{2}{3}}.
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  6. #6
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    Thanks very much for the help!!

    I have mentioned that it is a cube but don't think I will be quoting Renaissance philosophers.

    Again Thanks!
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