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? 
 A: $Area = Base \times Height /2$
To maximize the height, the tangent at $R$ is parallel to the base $PQ$.
A: Do you know how you calculate triangle area with coordinates of vertices? 
Clearly $R$ has coordinate $(x,x^2)$. You have to find a maximum of the following function $$f(x)=\left(%
\begin{array}{cc}
  2+1 & 4-1 \\
  2-x & 4-x^2 \\
\end{array}%
\right) = 3(4-x^2)-3(2-x) = -3x^2+3x+6 
$$
So $f$ achieve maximum at $x=-{b\over 2a} = {1\over 2}$, so $R(1/2,1/4)$.
A: i would use the Heron formula $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ with
$$PR=\sqrt{(x+1)^2+(x^2-1)^2}$$
$$QR=\sqrt{(2-x)^2+(4-x^2)^2}$$
$$PQ=\sqrt{3^2+3^2}$$
A: 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.
A: First you calculate $r$ the straight line passing in through $P$ and $Q$. You write the distance between your point $R$ and the line $r$, and finally you could derivate this respect to $x$ and find the point of your maximum area.
$$m=\frac{\Delta y}{\Delta x}=\frac{4-1}{2-(-1)}=1$$
$$y-y_P=m(x-x_P)\quad\rightarrow\quad y-1=x-(-1) \quad\rightarrow\quad y-x-2=0$$
$$d(R, r)=\frac{y-x-2}{\sqrt2}=\frac{x^2-x-2}{\sqrt2}$$
its derivate is equal to $2x-1$ so you have a zero in $x_R=\frac{1}{2}$, and now you can calculate $y_R=x_R^2=\frac{1}{4}$.
A: The area of triangle is $\frac{b*h}{2}$. The base, in this case, is given as $3*\sqrt{2}$. The height will be shortest distance from line y=x+2 to point(x,x^2). Using point-line distance formula, the area you are looking for is the maximum of $$\frac{(3*\sqrt{2})*\frac{(|x-x^2+2|)}{\sqrt{2}}}{2}$$ where x=1/2. The maximum area is thus $\frac{3*\sqrt{2}}{2}*\frac{2.25}{\sqrt{2}}=3.375$
