= Solution
An FCC crystal slips on $\{111\}\langle110\rangle$. For each <slip system>, <Schmid's law> gives $\tau=\sigma m$, with $m=|\hat{\mathbf l}\cdot\hat{\mathbf n}|\,|\hat{\mathbf l}\cdot\hat{\mathbf s}|$. Enumerating the twelve systems for $\mathbf l=(5,1,4)$ gives the maximum at
$$
\boxed{(1\bar11)[110].}
$$
Here $\mathbf n=(1,-1,1)$ and $\mathbf s=(1,1,0)$ are perpendicular, while $\mathbf l\cdot\mathbf n=8$, $\mathbf l\cdot\mathbf s=6$ and $|\mathbf l|^2=42$. Hence $m=48/(42\sqrt6)=0.46657$. Reversing both directed slip conventions describes the same physical system.
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