由a+b=1,∴a=1-b,
由S△ABC=absinC/2
=(1-b)bsin60°/2
=(-√3/4)b²+(√3/4)b
=(-√3/4)(b²-b+1/4)+√3/16
=(-√3/4)(b-1/2)²+√3/16.
当a=b=1/2时,S有最大值Smax=√3/16,
∴0<S≤√3/16.
由a+b=1,∴a=1-b,
由S△ABC=absinC/2
=(1-b)bsin60°/2
=(-√3/4)b²+(√3/4)b
=(-√3/4)(b²-b+1/4)+√3/16
=(-√3/4)(b-1/2)²+√3/16.
当a=b=1/2时,S有最大值Smax=√3/16,
∴0<S≤√3/16.