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Math Help - Constrained Optimization again

  1. #1
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    Constrained Optimization again

    Hi. I still need help with a constrained optimization problem. A couple days ago I posted a similar problem. MHF Member "Anonymous" helped me to get farther along, but now I'm stuck again. Here's the problem and how far I've gotten:

    Maximize f(x,y,z)=\frac{x}{y+z} subject to the constraint x+z=c


    I.e. \underset{\left \{ \left. x,y,z \right \} \right.}{\max}\left [ \frac{x}{y+z} \right ]


    s.t.\: \: \: \: x+z=c



    The Lagrangian is \Lambda(x,y,z,\lambda )=\frac{x}{y+z}+\lambda(x+z-c)

    Here are the first order conditions:

    \frac{\partial \Lambda}{\partial x}=\frac{1}{y+z}+\lambda=0

    \frac{\partial \Lambda}{\partial y}=\frac{-x}{(y+z)^2}=0

    \frac{\partial \Lambda}{\partial z}=\frac{-x}{(y+z)^2}+\lambda=0

    \frac{\partial \Lambda}{\partial \lambda }=x+z-c=0


    My confusion is that I see two solutions to this system of equations. First:

    Setting all of the equations equal to each other...

    \frac{-x}{(y+z)^2}= \frac{-x}{(y+z)^2}+\lambda= \frac{1}{y+z}+\lambda=0

    and then simplifying gives us...

    \frac{1}{y+z}=0

    The second solution comes when you solve for \lambda in one of the equations and then substitute this in for lambda in one of the other equations...

    Solving for lambda in equation three:

    \lambda=\frac{x}{(y+z)^2}

    Substituting this in for lambda in equation one:

    \frac{1}{y+z}+\frac{x}{(y+z)^2}=0

    Finally, setting this equal to equation two and simplifying:

    \frac{1}{y+z}+\frac{x}{(y+z)^2}=-\frac{x}{(y+z)^2}

    y+z+2x=0


    So, on the one hand I get \frac{1}{y+z}=0 while on the other hand the same system of equations simplifies to y+z+2x=0. What gives? Do I have to now set these two equations equal to each other or what?
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  2. #2
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    Quote Originally Posted by rainer View Post
    Maximize f(x,y,z)=\frac{x}{y+z} subject to the constraint x+z=c


    I.e. \underset{\left \{ \left. x,y,z \right \} \right.}{\max}\left [ \frac{x}{y+z} \right ]


    s.t.\: \: \: \: x+z=c



    The Lagrangian is \Lambda(x,y,z,\lambda )=\frac{x}{y+z}+\lambda(x+z-c)
    Never mind Lagrange multipliers. Think about this problem from an elementary viewpoint. If you want to maximise f(x,y,z)=\frac{x}{y+z} subject to the constraint x+z=c then the obvious thing to do is to put x=c, z=0 and y equal to some very small quantity, say y = 1/N for a large number N. Then x+z=c and f(x,y,z)=Nc. That tells you that f(x,y,z) can be made arbitrarily large while subject to the given constraint. So it has no maximum value.
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  3. #3
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    Quote Originally Posted by rainer View Post
    Hi. I still need help with a constrained optimization problem. A couple days ago I posted a similar problem. MHF Member "Anonymous" helped me to get farther along, but now I'm stuck again. Here's the problem and how far I've gotten:

    Maximize f(x,y,z)=\frac{x}{y+z} subject to the constraint x+z=c


    I.e. \underset{\left \{ \left. x,y,z \right \} \right.}{\max}\left [ \frac{x}{y+z} \right ]


    s.t.\: \: \: \: x+z=c



    The Lagrangian is \Lambda(x,y,z,\lambda )=\frac{x}{y+z}+\lambda(x+z-c)

    Here are the first order conditions:

    \frac{\partial \Lambda}{\partial x}=\frac{1}{y+z}+\lambda=0

    \frac{\partial \Lambda}{\partial y}=\frac{-x}{(y+z)^2}=0

    \frac{\partial \Lambda}{\partial z}=\frac{-x}{(y+z)^2}+\lambda=0

    \frac{\partial \Lambda}{\partial \lambda }=x+z-c=0


    My confusion is that I see two solutions to this system of equations. First:

    Setting all of the equations equal to each other...

    \frac{-x}{(y+z)^2}= \frac{-x}{(y+z)^2}+\lambda= \frac{1}{y+z}+\lambda=0

    and then simplifying gives us...

    \frac{1}{y+z}=0

    The second solution comes when you solve for \lambda in one of the equations and then substitute this in for lambda in one of the other equations...

    Solving for lambda in equation three:

    \lambda=\frac{x}{(y+z)^2}

    Substituting this in for lambda in equation one:

    \frac{1}{y+z}+\frac{x}{(y+z)^2}=0

    Finally, setting this equal to equation two and simplifying:

    \frac{1}{y+z}+\frac{x}{(y+z)^2}=-\frac{x}{(y+z)^2}

    y+z+2x=0


    So, on the one hand I get \frac{1}{y+z}=0 while on the other hand the same system of equations simplifies to y+z+2x=0. What gives? Do I have to now set these two equations equal to each other or what?
    A fraction is equal to 0 if and only if its numerator is equal to 0. \frac{1}{y+ z} is never 0 so this problem has no solution just as Opalg says.
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