Question 1026316
Let the x be the x-coordinate of the right-hand vertex of the triangle located on the parabola.
==> x > 0.

==> The base of the triangle has length 2x, while the height of the triangle would be y.

==> area is {{{A = (1/2)(2x)y = xy = x(3-x^2/12) = 3x - x^3/12}}}

==> A' = {{{3 - x^2/4}}}
Setting this to 0, we get
 {{{3 - x^2/4 = 0}}}
==> {{{x^2 = 12}}} ==> {{{x=2sqrt(3)}}}, the critical value of the function.

Now A" = -x/2 ==> A" = {{{-sqrt(3) < 0}}}, and so by the 2nd derivative test there is a local max at {{{x = 2sqrt(3)}}}.

The two vertices are (-{{{2sqrt(3)}}},2) and ({{{2sqrt(3)}}},2).