The area of the cut-off parabolic segment contained between
the curves

(11)

(12)

can be found by eliminating ,

(13)

so the points of intersection are

(14)

with corresponding -coordinates
.
The area is therefore given by

(15)

(16)

(17)

The maximum area of a triangle inscribed in this segment will have two of its polygon
vertices at the intersections and , and the third at a point to be determined. From the general equation for a
triangle, the area of the inscribed triangle is given by
the determinant equation

(18)

Plugging in and using gives

(19)

To find the maximum area, differentiable with respect to
and set to 0 to obtain