SOLUTION: A triangle has one side parallel to the x-axis, two vertices on the part of the parabola y=3-(x^2/12) above the x-axis and the third vertex at the origin. find the two vertices so

Question 1026316: A triangle has one side parallel to the x-axis, two vertices on the part of the parabola y=3-(x^2/12) above the x-axis and the third vertex at the origin. find the two vertices so that the triangle has the largest area.
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
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' =
Setting this to 0, we get

==> ==> , the critical value of the function.
Now A" = -x/2 ==> A" = , and so by the 2nd derivative test there is a local max at .
The two vertices are (-,2) and (,2).
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