Let \(y\) amount of mass should be transferred. Then:
\(F^{\prime}=\frac{G(M-y)(m+y)}{r^2}\)
\(=\frac{G(3 x-y)(2 x+y)}{r^2}\)
\(\text { For this to be maximum, } \frac{d F^{\prime}}{d y}=0\)
\(\frac{d F^{\prime}}{d y}=\frac{d\left[6 x^2+3 x y-2 x y-y^2\right]}{d y}\)
\(=3 x-2 x-2 y=0\)
\(x-2 y=0\)
\(y=\frac{x}{2}\)