A rectangle is inscribed into the region bounded by the graph of f(x)=(x^2-1)^2 and the x-axis, in such a way that one side of the rectangle lies on the x-axis and the two vertices lie on the graph of f(x). what is the maximum possible area of such a rectangle?

Question

Please help I have no idea how to solve this problem.

Answer 1

Your graph looks like a W and I assume you want the rectangle placed inside the centre part.

Let the two other vertices be (x,y) and (-x,y)
So the length is 2x and the height is y

Area = 2xy = 2x(x^4 - 2x^2 + 1)
= 2x^5 - 4x^3 + 2x

d(area)/dx = 10x^4 - 12x^2 + 2
= 0
5x^4 - 6x^2 + 1 = 0
treating this as a quadratic in x^2
x^2 - 1)(5x^2 - 1) = 0
x^2 = 1 or x^2 = 1/5
x = ±1 or x = ±1/√5

clearly, if x = 1, area = 0 , must be the minimum
and
when x = 1/√5
area = 2x(x^2 - 1)^2
= 2/√5(-4/5)^2
= 2/√5 (16/25)
or appr .572 units^2

check my arithmetic