$A$, $B$, $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$?
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$
Let pile $A$ have $A$ rocks, and so on.
The total weight of $A$ and $C$ can be expressed as $44(A + C)$.
To get the total weight of $B$ and $C$, we add the weight of $B$ and subtract the weight of $A$: $44(A + C) + 50B - 40A = 4A + 44C + 50B$
Therefore, the mean of $B$ and $C$ is $\frac{4A + 44C + 50B}{B + C}$, which is simplified to $44 + \frac{4A + 6B}{B + C}$.
We now need to eliminate $A$ in the numerator.  
Since we know that $40A + 50B = 43(A + B)$, we have $A = \frac{7}{3}B$
Substituting,
$44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}=44 + \frac{46}{3}*\frac{B}{B + C}$
In order to maximize $\frac{B}{B + C}$, we can minimize the denominator by letting $C = 1$ (C must be a positive integer). Since $\frac{46}{3}$ must cancel to give an integer, and the only fraction that satisfies both conditions is $\frac{45}{46}$
Plugging in, we get
$44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59$, which is choice E