What is the maximum possible area of a triangle if the sides $a, b, c$ satisfy the following inequalities:

$$
0<a \leqq 1 \leqq b \leqq 2 \leqq c \leqq 3
$$
The height $m$ perpendicular to side $a$ cannot be greater than the smaller of sides $b$ and $c$:

$$
m \leq b
$$

Thus, we get an upper bound for the area $t$ of the triangle:

$$
t=\frac{a m}{2} \leq \frac{a b}{2} \leq \frac{1 \cdot 2}{2}=1
$$

The area can also reach the upper limit of 1 if, on the one hand, $a=1$ and $b=2$, and on the other hand, $m=b$, meaning the triangle is a right triangle with legs $a$ and $b$. This triangle also satisfies the condition for $c$, because in this case $c=\sqrt{5}$, and this is a number between 2 and 3.