Maximum area of a triangle

Question

In the given drawing an isosceles triangle is inscribed in parabola whose equation is y=6-x^2. what is the maximum area that the triangle can have?

Answer 1

y=6-x^2 is a parabola that opens downward with vertex at (0,6)

That suggests your drawing probably has the top of the triangle at the parabola's vertex and the other two points where the parabola intersects the x axis, at (sqr6,0) and (-sqr6, 0)

You could just find the area of the right triangle bounded by the y-axis, x-axis and the line from the parabola's vertex to the point (sqr6,0). Then double it to get the area of the isosceles triangle inscribed inside the parabola.

The area of that right triangle is half of base times height. height is 6. base is sqr6

Their product is 6sqr6 = twice the right triangle's area = area of the isoscelese triangle.

square root of 6 = about 2.45.

6 x 2.45 = 14.7

exact most compact answer might be 6^3/2