Let us look at a one dimensional piecewise constant case, namely the domain is $(-R,R)$, and the coefficient $A(x)=a$ when $x<0$, and $A(x)=b$ when $x>0$.

Let us now consider the first eigenvalue/eigenvector of the problem
$$
-\frac{d}{dx}(A(x)\frac{d}{dx} u) = \lambda u \mbox{ in }H^1_0(-R,R),
$$ 
normalised by $u(0)=1$. A simple comparison using the Raleigh quotient shows that
$$
\lambda = \frac{\pi^2}{4R^2} \nu \mbox{ with }  \min(a,b)\leq \nu \leq \max(a,b) 
$$

An explicit computation shows that $u,\nu$ are given by
$$
u=\frac{\displaystyle\cos\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)\sin\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{a}R}\right)}{\displaystyle\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}+\cos\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{a}R}\right)\mbox{ when }x<0,
$$ 
and 
$$
u=-\frac{\displaystyle\cos\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)\sin\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{b}R}\right)}{\displaystyle\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)}+\cos\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{b}R}\right)\mbox{ when }x>0,
$$ 
and $\nu$ is the smallest positive solution of 
$$
\tan\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)\sqrt{\nu}\sqrt{a}=-\tan\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)\sqrt{\nu}\sqrt{b}.
$$
It is easy to see that for the solution to stay positive, it must stay fairly close to the $\min(a,b)$. For example, when $b/a \gg1$, $\nu\approx 4a$.
Near the boundaries(say, -R), we have
$$
u\approx \frac{\cos^2\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}{\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}(1+\frac{x}{R})
$$
So the scaling in $R$ (or $r$) of the dependence is not the one you hope for, I think.