Find the maximum area of a triangle

Question

The maximum area of a triangle whose sides $a,b,c$ satisfy $0\le a\le1,1\le b\le2,2\le c\le3$

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We can clearly see that $1,2,3$ as sides does not make a triangle so we can't just choose the maximum values.

Then I tried using triangle inequalities but in vain.

Any help is greatly appreciated. Also there is an exact same question asked on this site, about $9$ years ago, but no one there gave a correct, satisfactory answer and also the OP did not accept the two so-called answers that were posted there.

Area of $\triangle ABC$ whose sides $a,b,c$ satisfy $0\leq a \leq1;1\leq b \leq2;2\leq c \leq3$ is

Answer 1

A similar argument: The area will be $(1/2)ab\sin\angle C$ where $a\le1,b\le2\sin\angle C\le1$. We can simultaneously maximize all these factors, and then from the Pythagorean Theorem $c=\sqrt5<3$. So we choose the right triangle with legs $a=1$ and $b=2$.

Answer 2

Suppose we found the optimal value of $b$. Then the area is $\frac 12hb$ where $h$ is the height and clearly $h\le a$. It follows that $a$ should be perpendicular to $b$ and as long as possible. With this in mind, $b$ should also be maximal, and we end up with the right triangle with sides $a=1$, $b=2$, and $c=\sqrt5$.