# Math Help - multivariable maximization

1. ## multivariable maximization

Hello all, I got another maximization problem I would like to go over.
Let 0<a<b. Find points (x, y, z) where a<x<y<z<b that maximize

$u=\frac{xyz}{(a+x)(x+y)(y+z)(z+b)}$

First, I said since a>0, and a is the smallest in the domain, let

$v=ln(u)$

and the maximum should be the same.

Hence,

$v(x, y, z, a, b)=lnx+lny+lnz-ln(a+x)-ln(x+y)-ln(y+z)-ln(z+b)$

where
$v_1(x, y, z, a, b)=\frac{1}{x}-\frac{1}{a+x}-\frac{1}{x+y}$

$v_2(x, y, z, a, b)=\frac{1}{y}-\frac{1}{x+y}-\frac{1}{y+z}$

$v_3(x, y, z, a, b)=\frac{1}{z}-\frac{1}{y+z}-\frac{1}{z+b}$

$v_4(x, y, z, a, b)=\frac{-1}{a+x}$

$v_5(x, y, z, a, b)=\frac{-1}{z+b}$

Does this seem reasonable thus far?

2. ## Re: multivariable maximization

Originally Posted by quantoembryo
Hello all, I got another maximization problem I would like to go over.
Let 0<a<b. Find points (x, y, z) where a<x<y<z<b that maximize

$u=\frac{xyz}{(a+x)(x+y)(y+z)(z+b)}$

First, I said since a>0, and a is the smallest in the domain, let

$v=ln(u)$

and the maximum should be the same.

Hence,

$v(x, y, z, a, b)=lnx+lny+lnz-ln(a+x)-ln(x+y)-ln(y+z)-ln(z+b)$

where
$v_1(x, y, z, a, b)=\frac{1}{x}-\frac{1}{a+x}-\frac{1}{x+y}$

$v_2(x, y, z, a, b)=\frac{1}{y}-\frac{1}{x+y}-\frac{1}{y+z}$

$v_3(x, y, z, a, b)=\frac{1}{z}-\frac{1}{y+z}-\frac{1}{z+b}$

$v_4(x, y, z, a, b)=\frac{-1}{a+x}$

$v_5(x, y, z, a, b)=\frac{-1}{z+b}$

Does this seem reasonable thus far?
a and b are constants, the last two partial derivatives should not appear.

CB

3. ## Re: multivariable maximization

Okay, given I have done that, is there an easier way to evaluate this that you can see? It doesn't look like a very nice system to solve.

4. ## Re: multivariable maximization

The equations were not too bad, however, just one question. When I plug the original equation into wolfram it states there is no maxima.

5. ## Re: multivariable maximization

Originally Posted by quantoembryo
The equations were not too bad, however, just one question. When I plug the original equation into wolfram it states there is no maxima.
The extrema are either calculus like or occur on the boundary of the feasible region, but the way your problem is set up the boundary is not part of the feasible region, so there may well be no solution. If the constraints were $a \le x \le y \le z \le b$ and there is no calculus like maximum in the region, there will be a maximum on the boundary.

CB

6. ## Re: multivariable maximization

I managed to get it, thanks for the help.