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

Algebra ->  Triangles -> 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       Log On


   

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) About Me  (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+=+%281%2F2%29%282x%29y+=+xy+=+x%283-x%5E2%2F12%29+=+3x+-+x%5E3%2F12
==> A' = 3+-+x%5E2%2F4
Setting this to 0, we get
3+-+x%5E2%2F4+=+0
==> x%5E2+=+12 ==> x=2sqrt%283%29, the critical value of the function.
Now A" = -x/2 ==> A" = -sqrt%283%29+%3C+0, and so by the 2nd derivative test there is a local max at x+=+2sqrt%283%29.
The two vertices are (-2sqrt%283%29,2) and (2sqrt%283%29,2).