It is easy to observe that the domain is a closed triangle. Its three vertices are $B\left(\frac{t}{2}, 0\right)$, $Q(1, 0)$ and $C(1, t(2-t))$. Pick a point $X$ inside the $\triangle BQC$, then the area of $\triangle PQX$ is equal to half of the product of $PQ$ with the distance from $X$ to $PQ$. So the area of $PQX$
takes its maximum value when the distance from $X$ to $PQ$ is maximized, i.e., when $X$ coincides with $B$ or $C$.
The area of $\triangle PQB$ is

$$\begin{aligned} \frac{1}{2}\left(1-\frac{t}{2}\right)t^2 &= \frac{1}{4}(2-t)t^2 \le \frac{1}{4}(2-t)t \\ &\le \frac{1}{4}\left(\frac{2-t+t}{2}\right)^2 = \frac{1}{4}; \end{aligned}$$
the area of $\triangle PQC$ is
$$\begin{aligned} \frac{1}{2}(1-t)(2t-t^2) &= \frac{1}{4}2t(1-t)(2-t) \\ &\le \frac{1}{4}\left(\frac{2t+1-t+2-t}{3}\right)^3 = \frac{1}{4}. \end{aligned}$$
Hence, in the domain $A$, any triangle with $P, Q$ as two of its vertices cannot have an area that exceeds $\frac{1}{4}$. $\square$