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Homework Help: Help with Maximize

  1. May 20, 2006 #1

    tony873004

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    Maximize P=xyz with x+y=30, y+z=30, and x,y, and z >= 0.
    [tex]
    \begin{array}{l}
    {
    \\
    y = 30 - x \\
    \\
    z = 30 - y\,\, \Rightarrow \,\,z = 30 - 30 - x = - x \\
    \end{array}
    [/tex]
    This is a problem, because if z is negative x according to the question, they must both be 0 or positive. This isn't a problem if x = 0, and that's what I get as 1 of two solutions by differentiating P when written in terms of x, and setting it to 0.
    [tex]
    \begin{array}{l}
    P = x(30 - x)( - x) = \left( { - x} \right)^2 (30 - x) = - 30x^2 + x^3 \\
    P = x^3 - 30x^2 \\
    \\
    P' = 3x^2 - 60x \\
    P' = x(3x - 60) \\
    \end{array}
    [/tex]
    [tex]
    x = 0
    [/tex]
    and
    [tex]
    3x - 60 = 0\,\, \Rightarrow \,\,3x = 60\,\, \Rightarrow \,\,x = \frac{{60}}{3} = 20
    [/tex]
    The z=-x isn't a problem in the x=0 solution, but it is a problem in the x=20 solution. This gives me x=20, y=10, z=-20 P=xyz=-4000.

    The back of the book gives x=20, y=10, z=20 P=4000.

    If z >= 0 and I get -20 for z, I can't just switch it to positive, can I? Don't I have to discard that solution?
     
  2. jcsd
  3. May 20, 2006 #2

    nrqed

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    Watch out!!! 30-y is NOT 30-30-x!!
    (you will go :cry: when you will see what I mean)

    ;-)

    Pat
     
  4. May 20, 2006 #3

    tony873004

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    :cry:

    30-(30-x)
    30-30+x

    :approve:
     
  5. May 20, 2006 #4

    nrqed

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    :biggrin: Yes...so z=x (which was clear from x+y = 30 and z+y = 30, wasn't it? :shy: )
     
  6. May 20, 2006 #5

    tony873004

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    yes it was, thanks!!
     
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