Maximizing the area of a triangle with its vertices on a parabola. So, here's the question:
I have the parabola $y=x^2$. Take the points $A=(-1.5, 2.25)$ and $B=(3, 9)$, and connect them with a straight line. Now, I am trying find out how to take a third point on the parabola $C=(x,x^2)$, with $x\in[-1.5,3]$, in such a way that the area of the triangle $ABC$ is maximized.
I have pretty good evidence by trial and error that this point is $(.75, .5625)$ but I have no idea how to prove it. I was trying to work with a gradient, and then Heron's formula, but that was a nightmare to attempt to differentiate. I feel like this is a simple optimization problem but I have no clue how to solve it. Any help is appreciated!
Thanks.
 A: Assuming $A=\left(-\frac{3}{2},\frac{9}{4}\right),B=(3,9),C=(x,x^2)$, the area of $ABC$ is maximized when the distance between $C$ and the line $AB$ is maximized, i.e. when the tangent to the parabola at $C$ has the same slope of the $AB$ line. Since the slope of the $AB$ line is $m=\frac{9-9/4}{3+3/2}=\frac{3}{2}$ and the slope of the tangent through $C$ is just $2x$, the area is maximized by taking:
$$ C=\left(\frac{3}{4},\frac{9}{16}\right) $$
and the area of $ABC$ can be computed through the shoelace formula:
$$ [ABC] = \frac{729}{64}. $$
This area is just three fourth of the area of the parabolic segment cut by $A$ and $B$, as already known to Archimedes. Here we have a picture:

Also notice that in a parabola the midpoints of parallel chords always lie on a line that is parallel to the axis of symmetry. That immediately gives that $C$ and the midpoint $M=\left(\frac{3}{4},\frac{45}{8}\right)$ of $AB$ have the same $x$-coordinate. Moreover, it gives that the area of $ABC$ is the length of $CM$ times the difference between the $x$-coordinate of $B$ and the $x$-coordinate of $C$, hence $\frac{729}{64}$ as already stated.
A: hint: Let $M=(x,x^2)$ be the point on the parabola $y = x^2$, find the distance from this point to the base $AB$ where $A = (-1.5,2.25), B = (3,9)$. This distance is easy to find and it is a function of $x$. You can use calculus to find the max distance, and this corresponds to the max area since the base is fixed.
A: The line through the points $(-1.5,2.25)$ and $(p,p^2)$ is
$$y_1=(p-1.5)x+1.5p,$$
and the line through $(p,p^2)$ and $(3,9)$ is
$$y_2=(p+3)x-3p.$$
(used the basic method here of course, and we may assume that $-1.5<p<3$.)
Now we compute the area of the triangle as the sum of two integrals, each between the line $y=\frac{3}{2}x+4.5$ and our found lines in terms of $p$ over the appropriate bounds.
$$\int_{-1.5}^p (\frac{3}{2}x+4.5-((p-1.5)x+1.5p))\, dx+\int_{p}^3 (\frac{3}{2}x+4.5-((p+3)x-3p))\, dx = -\frac{9}{4}p^2+3.375p+10.125$$
$$\frac{dA}{dp}= 0 =>p = \frac{3.375}{4.5} =0.75$$
$$A_{max} = \frac{729}{64}$$
Thanks
A: The area of a triangle, assuming the vertices are given in clockwise order, is
$$(x_2 - x_1)\dfrac{y_1 + y_2}{2} + (x_3 - x_2)\dfrac{y_2 + y_3}{2} + (x_1 - x_3)\dfrac{y_3 + y_1}{2}\\
 = \frac12\left(x_1y_3 - x_1y_2 + x_2y_1 - x_2y_3 + x_3y_2 - x_3y_1\right)$$
Plugging $(-\frac32, \frac94)$, $(3,9)$, and $(t,t^2)$ into this formula we get:
$$ \frac12\left(-\frac32t^2 + \frac{27}2  + \frac{27}{4}  - 3t^2 + 9t -\frac94t\right) = \frac12\left(-\frac92t^2 +\frac{27}4t + \frac{81}4\right) \\= \frac98\left( -2t^2 + 3t + 9)\right)$$
The quadratic has a maximum when $-4t + 3 = 0$ or $t = \dfrac34$
