Area of $\triangle ABC$ whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is The maximum area of a triangle whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is
$\bf{My\; Try::}$ Area of $\displaystyle \triangle ABC = \frac{1}{2}ab\sin C = \frac{1}{2}ab\cdot \sqrt{1-\cos^2 C}\;,$ bcz Largest side has largest angle.
Now Using $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab}.$ So we get area of $\displaystyle \triangle ABC = \frac{1}{2}ab\cdot \sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$
Now How can i find Maximum area.

You would be better off using Heron's formula. But there is an even better way: what is the maximum value that $\frac12ab\sin \theta$ can attain, if $a \le 1$, $b \le 2$, and $\theta$ can be any angle? Is this value attainable given the additional constraint $2 \le c \le 3$?
