Mr Davis 97 said:
Homework Statement
For positive x, y, z where ##x \le y \le z## such that x + y + z = 3, what is the maximum value of ##xy##?
Homework Equations
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.
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.