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 $2 \le c \le 3$?
As the above user said a better approach would be to look at the area of a triangle as $\frac12 ab \sin{x}$, where $a$ and $b$ are the lengths of two adjacent sides and $x$ is the measure of the angle between these two consecutive sides. The maximum value of $\sin{x}$ occurs when $x=\pi/2$ which gives a value of 1. We now try to maximize the length of the legs and get that the maximum area is $2\cdot 1/2=1$. We make sure that we can have 2 and 1 as leg lengths by checking if $c$ fits in the inequality that was given in the problem, and indeed $\sqrt{5}$ is between 2 and 3.