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Maximise the volume of a rectangular prism with 2 constraints

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Maximise the volume of a rectangular prism with the following constraints: the surface area must equal 2m^2 and the total edge length must be 12m.

    2. Relevant equations

    Using Lagrange multipliers, we construct the function we want to optimise with

    [itex]h(x,y,z, λ_{1}, λ_{2}) = f(x,y,z) + λ_{1}g_{1}(x,y,z) + λ_{2}g_{2}(x,y,z)[/itex]

    3. The attempt at a solution

    In this case our f(x,y,z) is the volume of the rectangular prism, so [itex]f(x,y,z)=xyz[/itex], where we take x to be the length, y the widge and z the depth. [itex]g_{1}(x,y,z)=λ_{1}(xy + yz + zx - 1)[/itex], the surface area constraint and [itex]g_{2}(x,y,z)=λ_{2}(x+y+z-3)[/itex] (the edge length constraint).

    The function to be optimised is then [itex]h(x,y,z, λ_{1}, λ_{2}) = xyz + λ_{1}(xy + yz + zx - 1) + λ_{2}(x+y+z-3)[/itex]

    Obtain all the partial derivatives:

    1. [itex]\frac{∂h}{∂x} = yz + λ_{1}(y+z) + λ_{2} = 0[/itex]

    2. [itex]\frac{∂h}{∂y} = xz + λ_{1}(x+z) + λ_{2} = 0[/itex]

    3. [itex]\frac{∂h}{∂z} = yx + λ_{1}(y+z) + λ_{2} = 0[/itex]

    4. [itex]\frac{∂h}{∂λ_{1}} = xy + yx + zy -1 = 0 [/itex]

    5. [itex]\frac{∂h}{∂λ_{2}} = x + y + z -3 = 0 [/itex]

    Add 1, 2 and 3. Then sub in 4 and 5 to obtain the result [itex]λ_{1}=-(3λ_{2}+1)/6[/itex]

    But from here I'm totally lost. I can re-arrange 1,2 and 3 to get λ2 by itself, then equate them all to get

    [itex]\frac{\frac{1}{6}(y+z) - yz}{1-\frac{1}{2}(y+z)}=\frac{\frac{1}{6}(y+x) - yx}{1-\frac{1}{2}(y+x)}=\frac{\frac{1}{6}(z+x) - zx}{1-\frac{1}{2}(z+x)}[/itex]

    But doesn't this imply that x=y=z? This cannot satisfy both 4 *and* 5.
     
  2. jcsd
  3. Aug 12, 2014 #2

    HallsofIvy

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    What that tells you is that the constraints are impossible- you cannot have a rectangular prism with surface area 2 square meters and total edge length 12 meters.
     
  4. Aug 12, 2014 #3

    Ray Vickson

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    No, you can have such a prism. In fact, over the physically allowable region ##x,y,z \geq 0, x+y+z = 3## the half-area ##A = xy + xz + yz## ranges from ##A = 0## at the points ##(x,y,z) = (3,0,0), (0,3,0), (0,0, 3)## to ##A = 3## at the point ##(x,y,z) = (1,1,1)##. Any value of ##A## from 0 to 3 is attainable.

    However, the OP made a number of errors (perhaps typographical, perhaps not).

    I found it much easier to use one of the constraints to eliminate one of the variables. For example, from ##g_2 = 3## we have ##z = 3 - x - y##, and putting that in ##g_1## yields ##G_1(x,y) = g_1(x,y,3-x-y) = 3x+3y-x^2-y^2-xy-1## Now the problem is to maximize ##F = x y (3-x-y)##, subject to ##G_1(x,y) = 0##. This can be attacked using a Lagrange multiplier method; there are solutions.
     
    Last edited: Aug 12, 2014
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