1. ## 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?

2. Originally Posted by rainer
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.

3. 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 $\displaystyle f(x,y,z)=\frac{x}{y+z}$ 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.