The maximum area of the triangle whose sides a,b and c satisfy $0\le \mathrm{a}\le 1,1\le \mathrm{b}\le 2,$ and $2\le \mathrm{c}\le 3$ is 

Let the vertices be $\mathrm{O}\left(0,0\right),\mathrm{A}\left(\mathrm{\alpha },0\right),$ and $\mathrm{B}\left({\mathrm{\alpha }}_1,{\mathrm{\beta }}_1\right).$ Where $0\le \mathrm{\alpha }\le 1$ and $1\le {\mathrm{\alpha }}_1^2+{\mathrm{\beta }}_1^2\le 4.$

So, the area of $\mathrm{\Delta OAB}$ is maximum where $\mathrm{\alpha }=1$ and $\left({\mathrm{\alpha }}_1,{\mathrm{\beta }}_1\right)$ is $\left(2,0\right)$.

In this case, $\mathrm{a}=1,\mathrm{b}=2,$ and $\mathrm{c}=\sqrt{5},$ which satisfies $2\le \mathrm{c}\le 3.$

Therefore, the maximum area is 1.