This question is related to the earlier question  https://mathoverflow.net/questions/78103/weighted-area-of-a-voronoi-cell .  As in that question, let $X = \{ x_1,\dots,x_n\} $ denote a set of $n$ points in the unit square $S = [0,1]\times[0,1]$, and let $w = \{w_1,\dots,w_n\}$ denote a set of weights corresponding to the $n$ points in $X$.  Define the "power diagram" of $X$ in $S$ to be a partition of $S$ into at most $n$ pieces $V_i(\mathbf{w})$, where

$V_i(\mathbf{w}) = \{x\in S: \|x - x_i\|^2 + w_i \leq \|x - x_j\|^2 + w_j \forall j \neq i \}$

i.e. a "weighted Voronoi diagram".  Thus, larger values of $w_i$ correspond to smaller cells $V_i$.  The cells are always convex, and the partition does not change if we add a constant to all terms of a weight vector $\mathbf{w}$.

My question is:  let's consider some quantity $Q(V_i(\mathbf{w}))$ associated with the Voronoi cells with the following monotonicity property:  for any two weight vectors $\mathbf{w}$ and $\mathbf{w}'$, if it turns out that $V_i(\mathbf{w})\subsetneq V_i(\mathbf{w}')$, then $Q(V_i(\mathbf{w})) < Q(V_i(\mathbf{w}'))$.  Examples of $Q$ would be the area, perimeter, diameter, or width of the cells.  I want to know:  is it always the case that $Q(V_i(\mathbf{w}))$ is somehow "equivalent" to the gradient of some other function?

My reasoning for this question is as follows:  let's say we want to select weights $\mathbf{w}$ such that $Q(V_i(\mathbf{w}))$ is equal for all $i$.  This means that we want to *increase*(resp. decrease) values of $w_i$ for those regions where $Q(V_i(\mathbf{w}))$ is large (resp. small).  A sensible way to do this would be to iteratively set $w_i \mapsto w_i + \epsilon Q(V_i(\mathbf{w})) $, where $\epsilon$ is some tiny stepsize.  I have tried numerical experiments with this scheme for the case where $Q$ measures area or perimeter and this seems to work.