Let $S(t/2,0)$, $R(1,t(2-t))$. The region $A$ is the triangle $SQR$, and $P\in SR$, thus the triangles of largest possible area are $\triangle SPQ$ and $\triangle RPQ$. But $\operatorname{area} \triangle SPQ = (1/2)t^2(1-t/2) \leq (1/4)t < 1/4$ (equality for $(1/4)t$ is for $t=1$, disallowed). $\operatorname{area} \triangle RPQ = (1/2)(1-t)t(2-t) \leq 1/3\sqrt{3} < 1/4$ (equality for $1/3\sqrt{3}$ is for $t=1-1/\sqrt{3}$). EDIT. Referring to the next post - it is interesting to figure out the envelope of the lines $y=t(x-t)$ is the parabola $y=x^2$, but in some way it is irrelevant, since the problem functions for a fixed $t$, with $P(t,t^2)$ fixed on the parabola for some $0<t<1$, and the triangular area $A$ being delimited by the tangent to that parabola at $P$.

$l_t : y=t(2x-t)=-tx^2+2tx=-(t-x)^2+x^2\leq x^2$, thus the parabola $C : y=x^2$ is envelope of the lines $l_t$ Which touches $C$ at $x=t\ (0<t<1)$. The domain $D$ of the family of the lines swept by $l_t$ is shown by the shaded region, excluded two points $(0, 0),(1, 1)$. Edit: I was misreading the context of the problem, thank you for pointing put it, mavropnevma. I have just attached another figure. Needless to say, mavropnevma's solution is perfect. P.S. i remember that the similar problem has been posed in 1970's in Tokyo University entrance exam/Science. According to my memory, let $f(t)$ be the maximum area of any triangle which is involved in the region $A$ in original problem, then draw the graph of $y=f(t)$ to find the extrema of $f(t)$.

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Here is the similar problem posed in Tokyo University entrance exam/Science, second round, 1978 In the $xy$-plane, let $L$ be the part which is correspond to $0<x<1$ of the parabola $y=x^2$, that is to say, $L=\{(x,\ x^2)|0<x<1\}$. Let the tangent Line of $L$ at $P(x,\ x^2)$ intersects with the line $y=0$ at $A$ and intersects with the line $x=1$ at $B$. Let $O(0,\ 0), \ C(1,\ 0),\ D(1,\ 1)$. We are to consider the questions as below in the range of $0<x<1$. (1) Let denote the area of triangles $PAC,\ PCB$ by $g(x),\ h(x)$, respectively. Find the range of $x$ such that $g(x)\leq h(x)$. (2) Let $M$ be the domain enclosed by line segments $OC,\ CD$ and $L$. Note : $M$ contains line segments $OC,\ CD$ and $L$. Let $f(x)$ be the maximum area of the triangle with a vertex $P(x,\ x^2)$ which is contained in $M$. Find the function $f(x)$ and draw the graph, then find the exterme value. Note : A function $f$ has local minimum (or local maximum) at a point $a\in{I=\{x|0<x<1\}}$, which means for all points $x$ which is closed to $a$, $f(a)\leq f(x)\ (or\ f(a)\geq f(x))$ holds. We call local maximum, local minimum as extreme value.

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