An isosceles triangle, whose base is the interval from (0,0) to (c,0), has its vertex on the graph of f(x)=12-x^2. For what value of c does the triangle have maximum area?

Answer 1

Let A(a,12-a�^2) be the point of the triangle which lies on the parabola.
Since it must be an isosceles triangle, c = 2a

The area of the triangel = 1/2(c)(12-a^2)
=1/2(2a)(12-a^2)
=12a - a^3

Then d(Area)/da = 12 - 3a^2
= 0 for a max area

3a^2 = 12
a = +- 2, c = 2a
so c = 4