AMC 12 Mock Exam 5 Problem 23

By the law of sines, $\frac{BC}{\sin A}=\frac{6}{\sin60^\circ}=4\sqrt3$, so $AB=4\sqrt3\sin C$ and $AC=4\sqrt3\sin B$, where $B+C=120^\circ$ and $0^\circ<B<120^\circ$. Then $$AB+2AC=4\sqrt3\left(\sin(120^\circ-B)+2\sin B\right)=4\sqrt3\left(\tfrac{\sqrt3}{2}\cos B+\tfrac52\sin B\right).$$ The expression $\frac{\sqrt3}{2}\cos B+\frac52\sin B$ has maximum $\sqrt{\frac34+\frac{25}4}=\sqrt7$, attained when $\tan B=\frac{5}{\sqrt3}$, i.e., $B\approx70.9^\circ$, which lies in $(0^\circ,120^\circ)$. The maximum is $4\sqrt3\cdot\sqrt7=\boxed{4\sqrt{21}}$.