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Math Help - dimensions of a box w/ fixed volume

  1. #1
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    dimensions of a box w/ fixed volume

    Find the dimensions of a rectangular box, open at the top, with a volume of 32 cubic feet, using the least amount of material.
    I'm not really sure even how to start this problem. Any help?
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  2. #2
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    Are you familiar with Lagrange multipliers?.

    We must minimize S=xy+2xz+2yz

    subject to the constraint xyz=32

    If we let f(x,y,z)=xy+2xz+2yz and g(x,y,z)=xyz

    then, {\nabla}f=(y+2z)i+(x+2z)j+(2x+2y)k

    and

    {\nabla}g=yzi+xzj+xyk

    If follows that {\nabla}g\neq{0} at any point on the surface

    xyz=32, since x, y, and z are all non zero on this surface. So, we must have

    {\nabla}f={\lambda}{\nabla}g

    So, we get the 3 equations:

    y+2z={\lambda}yz

    x+2z={\lambda}xz

    2x+2y={\lambda}xy

    These can be rewritten like:

    \frac{1}{z}+\frac{1}{y}={\lambda}...[1]

    \frac{1}{z}+\frac{2}{z}={\lambda}....[2]

    \frac{2}{y}+\frac{2}{z}-{\lambda}....[3]

    From [1] and [2] we get y=x

    From [2] and [3] we get z=\frac{x}{2}

    Sub these into the constraint and get:

    \frac{x^{3}}{2}=32

    Now, I will leave finding the values up to you. Not much left to do.

    You can also do this with partial derivatives.
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  3. #3
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    Hello, ikickyouindanuts!

    Find the dimensions of an open-top rectangular box with a volume of 32 ft³,
    using the least amount of material.
    You need to be familiar with partial derivatives . . .
    Code:
              *-----------*
             /|          /|
            / |         / |
           /  |        /  |H
          *-----------*   |
          |           |   |
          |           |   *
        H |           |  /
          |           | /W
          |           |/
          *-----------*
                L

    The volume is 32 ft³: . LWH \:=\:32\quad\Rightarrow\quad H \:=\:\frac{32}{LW}\;\;{\color{blue}1]}


    The base has area: .  LW

    The front and back has area: . 2LH

    The sides have area: . 2WH

    The total area is: . A \:=\:LW + 2LH + 2WH\;\;{\color{blue}[2]}


    Substitute [1] into [2]: . A \;=\;LW + 2L\left(\frac{32}{LW}\right) + 2W\left(\frac{32}{LW}\right)

    . . and we have: . A \;=\;LW + 64W^{-1} + 64L^{-1}


    Take partial dereivatives and equate to zero:

    . . \begin{array}{cccccccccc}\frac{\partial A}{\partial L} &=& W - 64L^{-2} &=& 0 & \Rightarrow & L^2W &=& 64 & {\color{blue}[3]}\\ \frac{\partial A}{\partial W} &=& L - 64W^{-2} &=&0 & \Rightarrow & LW^2 &=& 64 & {\color{blue}[4]}\end{array}

    Equate [3] and [4]: . L^2W \:=\:LW^2\quad\Rightarrow\quad L \:=\:W

    Substitute into [3]: . L^3 \:=\:64\quad\Rightarrow\quad L \:=\:4\quad\Rightarrow\quad W \:=\:4

    Substitute into [1]: . H \:=\:\frac{32}{4\cdot4} \;=\;2

    . . Therefore: . (L,W,H) \;=\;(4,4,2)

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  4. #4
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    thanks a lot guys, this really helps. We have covered partial derivatives, and I knew it was possible, but I couldn't figure out where to start.
    I have never heard of Lagrange multipliers.
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