<s>[INST] A triangle has the maximum possible area on the parabola $y = x^2$ Given points $P(−1, 1)$ and $Q(2, 4)$ on the parabola $y = x^2$, where should the point $R$ be on the parabola (between P and Q) so that the triangle $PQR$ has the maximum possible area?

How should I do this question? 
 [/INST] Let $R(x,x^2),$ where $-1\leq x\leq2.$
We have
$$m_{PQ}=1,$$ which gives the equation of $PQ:$
$$y-1=1(x+1)$$ or $$x-y+2=0.$$
Thus, the altitude of the triangle from $R$ is
$$\frac{|x-x^2+2|}{\sqrt2}$$ and since $$PQ=\sqrt{3^2+3^2}=\sqrt{18},$$ by AM-GM we obtain:
$$S_{\Delta PQR}=\frac{\sqrt{18}\cdot\frac{|x-x^2+2|}{\sqrt2}}{2}=\frac{3}{2}\cdot(2-x)(x+1)\leq\frac{3}{2}\left(\frac{2-x+x+1}{2}\right)^2=\frac{27}{8}.$$
The equality occurs for $x+1=2-x$, which says that we got a maximal value.
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