So here is the problem. I did it but just wanted to make sure it is correct... does the solution look ok to you guys. problem: an Isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the x-axis on the curve y = 27-x^2. find the largest area the triangle can have. solution:So I assumed the base of the triangle will be the points (x, y) and (-x, y). Therefore height of the triangle will be y length of the base will be x+x = 2x Area = 1/2 * base * height = 1/2 * 2x * y =xy now we know that y = 27 - x^2 so this area eqn becomes A = x(27 - x^2) A = 27x - x^3 now A' = 27 - 3x^2 = 0 so the critical points are x = 3 and x = -3 using the sign chart i figured out that i have to use x = 3 because that is the local maximum. and when I plug x = 3 back into A = x(27 - x^2) we get A = 54 unit^2 Can some one verify that my solution is correct? I would really appreciate it.