Constrained Optimization again

**rainer** · April 4th 2010, 04:46 PM

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?

**Opalg** · April 5th 2010, 12:39 AM

Originally Posted by rainer

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.

**HallsofIvy** · April 5th 2010, 04:45 AM

Originally Posted by rainer

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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