△ABC中，B=60°，AC=√3,则AB+2BC的最大值是多少？

最大值是2√7。
过程：因为a/sin A=b/sin B=c/sin C=2，所以a=2sin A c=2sin C
又因为B=60°，所以 A+C=120°，
所以AB+2BC=2sin C+4sin A=4sin A+√3cos A+sin A=√28sin(A+φ)=2√7sin(A+φ)
所以最大值为2√7

By sine rule
AC/ sinB = AB/sinC = BC/sinA
2 = AB/sinC = BC/sinA
AB = 2sinC and BC= 2sinA

S = AB+2BC
= 2sinC + 4sinA
= 2sinC + 4sin(180°-60°-C)
=2sinC+4sin(120°-C)
S' = 2cosC -4cos(120°-C) =0
2cosC - 4(cos120°cosC +sin120°sinC) =0
2cosC - 2(-cosC + √3sinC)=0
2cosC-√3sinC =0
tanC = 2/√3
S''( arctan(2/√3) ) <0 ( max )

S= 2sinC+4sin(120°-C)
= 2sinC +4[sin120°cosC - cos120°sinC]
= 4sinC +2√3cosC
when C= arctan(2/√3)
max C
= 4(2/7) + 2√3 (√3/7)
= 8/7 + 6/7
=2