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Math Help - Maximizing a volume

  1. #1
    MHF Contributor arbolis's Avatar
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    Maximizing a volume

    I find the following problem very hard!
    A rectangular box has 3 faces in the planes given by x=0, z=0 and y=0. The vertex that is not in the latter planes is in the plane 4x+3y+z=36. Determine the dimensions that maximizes the volume of the box.
    I tried to visualize the box by drawing a sketch and I think I couldn't. So I'm stuck on giving an expression of the volume of the box. I'd like some help for this task.
    Precisely I've drew a rectangular parallelepiped in the xyz plane but I don't see how it could have a maximum volume so I think I drew it wrongly.
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  2. #2
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    We'll assume that the box is suppose to lie in the first octant. This seems like a reasonable starting point.

    If the vertex in the plane 4x+3y+z=36 has coordinates (x,y,z) then the volume of the box is xyz.

    So the problem boils down to maximising the quantity xyz subject to the conditions that x, y, z are positive and 4x+3y+z=36.

    Now try the AM-GM inequality, namely \frac{4x+3y+z}3\geq\root 3\of{4x.3y.z} with equality iff 4x=3y=z.

    You should now be able to deduce that the maximum volume is 144. Happy hunting.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Ok thanks a lot. I'll try it.
    Err... I would have never thought to suppose that the box has to lie in the first octant. I hope the question will be clearer in the exam.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Instead of using the inequality, I used Lagrange's multipliers method. I found x=3, y=4 and z=12, hence V=144 as you pointed out.
    I still have some difficulties to see why we can suppose the vertex to have the coordinates (x,y,z). I don't know if I could solve a similar problem. (Hopefully yes).
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