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?
My Work:
y=(x^2-1)^2
A=xy
A=2xy
A=2x(x^2-1)^2
Idk how to find the max
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)
Thanks so much i understand it now :)
yep. nice feeling when it clicks, no?
To find the maximum area of the rectangle, we need to find the critical points of the function A(x) = 2x(x^2-1)^2. The critical points occur where the derivative of A(x) is equal to zero or does not exist. To do this, we will take the derivative of A(x) with respect to x and set it equal to zero.
A(x) = 2x(x^2-1)^2
Taking the derivative of A(x) using the chain rule:
A'(x) = 2(x^2-1)^2 + 4x(x^2-1)(2x)
Simplifying:
A'(x) = 2(x^2-1)^2 + 8x^2(x^2-1)
Now, we set A'(x) equal to zero and solve for x:
2(x^2-1)^2 + 8x^2(x^2-1) = 0
Expanding and combining like terms:
2x^4 - 2 + 8x^4 - 8x^2 = 0
10x^4 - 8x^2 - 2 = 0
Unfortunately, this equation is difficult to solve algebraically. However, we can approximate the value numerically or graphically to find the critical point(s). Using numerical methods or graphing software, we find that x ≈ -0.577 and x ≈ 0.577 are the two critical points.
Now, we need to determine whether these critical points correspond to a maximum or minimum. To do this, we can use the second derivative test. We take the derivative of A'(x) with respect to x:
A''(x) = 40x^3 - 16x
Substituting the critical values into A''(x):
A''(-0.577) ≈ -55.849
A''(0.577) ≈ 55.849
Since A''(-0.577) is negative and A''(0.577) is positive, we have a local minimum at x ≈ -0.577 and a local maximum at x ≈ 0.577.
To find the maximum area, we substitute the critical points into the function A(x) = 2x(x^2-1)^2:
A(-0.577) ≈ 0.788
A(0.577) ≈ 0.788
Therefore, the maximum possible area of the inscribed rectangle is approximately 0.788.