The maximum area of the triangle whose sides a,b and c satisfy 0≤a≤1, 1≤b≤2, and 2≤c≤3 is

Let the vertices be O0,0, Aα,0, and Bα1,β1. Where 0≤α≤1 and 1≤α12+β12≤4.So, the area of ΔOAB is maximum where α=1 and α1,β1 is 2,0.In this case, a=1, b=2, and c=5, which satisfies 2≤c≤3.Therefore, the maximum area is 1.