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 $V_i$, where
$V_i = {x\in S: \|x - x_i\|^2 + w_i \leq \|x - x_j\|^2 + w_j \forall j \neq i }$
i.e. a "weighted Voronoi diagram". Now let's consider varying the weight $w_1$ while fixing the other weights; specifically, consider the function
$f(w_1) = w_1\cdot \text{Area}(V_1)$
Clearly as $w_1 \rightarrow 0$ we have $f(w_1) \rightarrow 0$ and as $w_1 \rightarrow \infty$ we have $f(w_1) \rightarrow 0$ as well. My question: is $f(w_1)$ unimodal? Convex? Is the answer different if I only have $n=2$ points? What if I define my cells slightly differently, such as
$V_i = {x\in S: \|x - x_i\| + w_i \leq \|x - x_j\| + w_j \forall j \neq i }$ ?
