Finding the radii that maximizes and minimizes the area of four inscribed circles in an equilateral triangle. 

An equilateral triangle with side length $1$ unit contains three identical circles $C_1$, $C_2$ and $C_3$ of radius $r_1$, each touching two sides of the triangle. A fourth circle $C4$ of radius $r_2$ touches each of $C_1$, $C_2$ and $C_3$ as shown. Except for the contact points with $C_4$, none of the circles have any points in common with any of the other circles. Determine the values of $r_1$ and $r_2$ which minimize and maximize the sum $S$ of the areas of the four circles.

My attempt:
To maximize the area, I figured that I had to maximize the equation $S = 3\pi r_{1}^2 + \pi r_{2}^2$ by setting its derivative to zero and solving for $r_{1}$, then $r_{2}$. In order to change this formula so that I only had to work with $r_{1}$, I concluded after some experimentation that I could use $\frac{1}{\sqrt{3}} = 3r_{1}+r_{2}$ to substitute terms, where $\frac{1}{\sqrt{3}}$ is the distance between the vertex and the circumcenter. I got that $r_{2} = 3r_{1} - \frac{1}{\sqrt{3}}$. Following this process, I had $S = \pi(12r_{1}^2-2\sqrt{3}r_{1} + \frac{1}{3}), \frac{dS}{dr_{1}} = 24r_{1}-2\sqrt{3}$, and finally got a final answer of $r_{1} = \frac{\sqrt{3}}{12}$. 
However, this result made my $r_{2}$ value negative, and then I got really stuck with what I was doing. I tried thinking it out, such as trying to figure out the domains and finding other equations to use, but I seem to be blanking out with this question. I'm now even more confused how I am supposed to find the values that result in the minimum $S$. Any help would be very appreciated!
 A: It is true that $\frac{1}{\sqrt3} = 3r_1+r_2$, although you seem not to have a proof of this fact.
But it does not follow that $r_2 \stackrel?= 3r_1 - \frac{1}{\sqrt3}.$
You have made a mistake in your algebra there, although once you square the expression for $r_2$ the mistake is canceled out.
A more serious mistake is when you thought that $r_1 = \frac{\sqrt{3}}{12}$ would maximize $\pi(12r_1^2-  2\sqrt3 r_1 + \frac13)$.
Instead, it gives a minimum.
But the real trick here is to figure out how large $r_1$ can be before the figure can no longer satisfy the description.
Here I would argue that the problem is ill-stated, because the configuration where the $r_1$ circles just touch each other is forbidden by the problem statement, yet it has a smaller area than any configuration in which the circles do not touch, and any of those configurations can be given less area by making the $r_1$ circles a little larger (just not quite large enough to touch). Hence no allowed configuration minimizes the area; it's like asking for the smallest value of $x$ such that $x > 1.$
On the other end, the problem does not say if the $r_2$ circle is required to be entirely contained in the triangle. If it is not, the area is maximized when 
$r_1 = 0,$ provided that you consider a circle of radius $0$ to be a circle.
If the $r_2$ circle is limited to be inside the triangle then the minimum value of $r_1$ is larger than zero.

If I had to guess what the author meant, I would guess that they did not really mean to prevent the $r_1$ circles from touching, merely that they could not overlap, and I would guess that they meant the $r_2$ circle to be contained within (but possibly tangent to) the triangle.
That gives you minimum and maximum values of $r_1.$
You will note that the value of $r_1$ that sets $dS/dr_1$ to zero is not in that range.
You should know something about finding minima and maxima of a function when the domain is bounded, which you can then apply to this problem.
A: I think you should formulate an optimization problem and solve it using KKT conditions.
The objective function $f(r_1,r_2)=\pi(3r_1^2+r_2^2)$
The equality constraint $h(r_1,r_2)=3r_1+r_2-1/\sqrt 3=0$
The inequality constraint $g(r_1,r_2)=r_2-\sqrt{5/4}+1/\sqrt 3 \leq 0$.
The last constraint aims to limit the circle to be inside the triangle.
A: $r_1$ varies from $\min(r_1)=\tfrac{\sqrt3}{18}$,
when $r_2$ is equal the inradius $r=\tfrac{\sqrt3}6$,
to $\max(r_1)=\tfrac14\,(\sqrt3-1)$, when all the four circles 
are mutually tangent to each other.
The value of $r_2$ can be expressed in terms of 
$r_1$ and $r$ as
\begin{align}
r_2(r_1)&=2r-3r_1
,
\end{align} 
hence the objective function 
is
\begin{align} 
f(r_1)&=
\pi\,(3r_1^2+r_2(r_1)^2)
=4\pi\,(3r_1^2-3\,r\,r_1+r^2)
,\\
f'(r_1)&=12\,\pi\,(2r_1-r)
,
\end{align} 
so the objective function is a parabola 
and on the interval of interest 
it has the minimum at $r_1=\tfrac12\,r=\tfrac{\sqrt3}{12}$
(all the four circles has the same radius)
and the maximum at $r_1=\min(r_1)=\tfrac{\sqrt3}{18}$.
