Painters Painting Circles

The circumference of a circle scales as $r$ and the area scales as $r^2$ . So if our goal is to maximize the circumference, it's obvious that large circles are inefficient. We want as many small circles as possible, implying that the optimal condition is one in which $x = y$ (each individual paints his own circle). For one such circle:

$\large{\pi r^2 = 4 \implies r = \frac{2}{\sqrt{\pi}} \implies C = 4 \sqrt{\pi}}$

The total circumference is then:

$\large{C_{total} = 4 \sqrt{\pi} x = 4 \sqrt{\pi} \sqrt{xy} = k \sqrt{xy} \\ k^2 = 16 \pi }$

Let $G_i$ be the size of the $i$ th group. Note that $1 \leq i \leq y$ and $\sum_{i = 1}^y G_i = x.$

Now, group $i$ has $4G_i$ square meters of paint to work with. Let $r_i$ be the radius of the $i$ th group's circle. We have

$\begin{aligned} \pi r_i^2 &= 4G_i \\ r_i^2 &= \dfrac{4G_i}{\pi} \\ r_i &= 2 \sqrt{\dfrac{G_i}{\pi}}. \end{aligned}$

Thus, the circumference $C_i$ of the $i$ th group's circle is $C_i = 2\pi r_i = 4 \sqrt{\pi G_i}.$

We want to find the maximum value of the sum of all the circumferences of all the circles. In other words, we want to maximize $\sum_{i = 1}^y C_i.$ By the Cauchy-Schwarz Inequality,

$\begin{aligned} \sum_{i = 1}^y C_i &= \sum_{i = 1}^y 4 \sqrt{\pi G_i} \\ &= 4\sqrt{\pi} \sum_{i = 1}^y \sqrt{G_i} \\ &\leq 4 \sqrt{\pi} \sqrt{ \left (\sum_{i = 1}^y G_i \right) \left (\sum_{i = 1}^y 1 \right)} \\ &= 4 \sqrt{\pi} \sqrt{xy}. \end{aligned}$

Thus, $k = 4 \sqrt{\pi},$ and $\lfloor k^2 \rfloor = \lfloor 16 \pi \rfloor = \boxed{50}.$

Addendum: Equality occurs when $G_1 = G_2 = ... = G_{y - 1} = G_y$ i.e. the groups are all the same size. This can only occur if $y|x$ . If that is not the case, then equality is never attainable.