# Optimization problem

1. Aug 11, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
For positive x, y, z where $x \le y \le z$ such that x + y + z = 3, what is the maximum value of $xy$?

2. Relevant equations

3. The attempt at a solution
First, before I attempt a solution, isn't it the case that $x=y=z=1$, since the only partition of 3 into three terms is 1+1+1? I feel like if this would the case the problem would be way too simple, which is why I'm asking.

2. Aug 11, 2017

### andrewkirk

Only if x,y,z have to be integers. In optimisation problems, one usually assumes they can be any real numbers, unless advised otherwise. So here are three other possibilities:

x=0.1, y=0.2, z=2.7
x=0.5, y=1, z=1.5
x=0.7, y=1.1, z=1.2

The set of allowable points will be a solid polygon.

3. Aug 11, 2017

### Ray Vickson

You have found the globally optimal solution over the region $\{x >0,y \geq x,z \geq y, x+y+z=3 \}$ but proving that this is the case may be tricky (because you have a non-convex optimzation problem, so in principle could have numerous local maxima). However, a bit of further re-writing shows this to not be the case: there is only one solution, and you have found it.

First, re-write your final constraint as $x+y+z \leq 3$; it will be satisfied as an equality at the maximum, (because for any point where the inequality is strict, we can move it out a bit to the boundary and keep the other constraints satisfied, while also increasing $x y$).

So now we want to maximize a non-convex function $xy$ but over a convex region. As is done in so-called Geometric Programming, we can change variables to $x=e^u, y = e^v, z = e^w$. We want to minimize $x^{-1} y^{-1}$, so our problem becomes to minimize $f(u,v) = -u-v$, subject to constraints $u-v \leq 0,$ $v-w \leq 0$ and $e^u + u^v + e^w \leq 3$. The function $g(u,v,w) = e^u + u^v + e^w$ is strictly convex, so the region $g(u,v,w) \leq 3$ is convex. Altogether, we have a convex programming in $(u,v,w)$, and so any local constrained minimum (of $f(u,v)$) is automatically a global constrained minimum. In other words, the solution is unique and can be found from the Karush-Kuhn-Tucker conditions.

Last edited: Aug 11, 2017