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Thread: 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 $\displaystyle f(x,y,z)=\frac{x}{y+z}$ subject to the constraint $\displaystyle x+z=c$


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


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



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

    Here are the first order conditions:

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

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

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

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


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

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

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

    and then simplifying gives us...

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

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

    Solving for lambda in equation three:

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

    Substituting this in for lambda in equation one:

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

    Finally, setting this equal to equation two and simplifying:

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

    $\displaystyle y+z+2x=0$


    So, on the one hand I get $\displaystyle \frac{1}{y+z}=0$ while on the other hand the same system of equations simplifies to $\displaystyle y+z+2x=0$. 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 $\displaystyle f(x,y,z)=\frac{x}{y+z}$ subject to the constraint $\displaystyle x+z=c$


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


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



    The Lagrangian is $\displaystyle \Lambda(x,y,z,\lambda )=\frac{x}{y+z}+\lambda(x+z-c)$
    Never mind Lagrange multipliers. Think about this problem from an elementary viewpoint. If you want to maximise $\displaystyle f(x,y,z)=\frac{x}{y+z}$ subject to the constraint $\displaystyle x+z=c$ then the obvious thing to do is to put $\displaystyle x=c$, $\displaystyle z=0$ and y equal to some very small quantity, say $\displaystyle y = 1/N$ for a large number N. Then $\displaystyle x+z=c$ and $\displaystyle f(x,y,z)=Nc$. That tells you that $\displaystyle f(x,y,z)$ 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 $\displaystyle f(x,y,z)=\frac{x}{y+z}$ subject to the constraint $\displaystyle x+z=c$


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


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



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

    Here are the first order conditions:

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

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

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

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


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

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

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

    and then simplifying gives us...

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

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

    Solving for lambda in equation three:

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

    Substituting this in for lambda in equation one:

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

    Finally, setting this equal to equation two and simplifying:

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

    $\displaystyle y+z+2x=0$


    So, on the one hand I get $\displaystyle \frac{1}{y+z}=0$ while on the other hand the same system of equations simplifies to $\displaystyle y+z+2x=0$. 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. $\displaystyle \frac{1}{y+ z}$ is never 0 so this problem has no solution just as Opalg says.
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