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

My Work:

y=(x^2-1)^2
A=xy
A=2xy
A=2x(x^2-1)^2
Idk how to find the max

Answer 1

the max or min area occurs where dA/dx = 0

dA/dx = 2(5x^4-6x^2+1) = 2(5x^2-1)(x^2-1)
So, dA/dx = 0 at x = ±1/√5, ±1

Naturally, A=0 at x=1, so the obvious choice is x = 1/√5.

Thus, the maximum possible area is 2/√5(1/5-1)^2 = 32/(25√5)

Answer 2

Thanks so much i understand it now :)

Answer 3

yep. nice feeling when it clicks, no?