Maximum Possible Area of Given Rectangle

A rectangle has its base on the x-axis and its upper two vertices
on the curve of y = 1 - x^4. What is the maximum possible area for such a rectangle?

Top right vertex of rectangle has coordinates (x, 1-x^4), where 0 ≤ x ≤ 1
Top left vertex of rectangle has coordinates (-x, 1-x^4)

Rectangle has width 2x, and height 1-x^4

Now we can express area as function of x:
A = 2x (1-x^4)
A = 2x - 2x^5

To find maximum area, find x when A'(x) = 0
A' = 2 - 10x^4 = 0
10x^4 = 2
x^4 = 1/5
x = 1/∜5 ------> we take positive root, since x > 0

A'' = -40x^3 < 0 when x > 0
A has maximum value at x = 1/∜5 (on interval [0, 1])

A = 2/∜5 (1 - (1/∜5)^4) = 2/∜5 (4/5) = 8/(5∜5) = 1.069984488

y = 1-x^4 = (1+x^2)(1-x^2) = (1+x^2)(1+x)(1-x)

So the zeros are at 1 and -1. Also, this graph is symmetric about the y axis.

Then this rectangle will have a side along the x axis with length 2x (from -x to x)
and a height of 1-x^4.

Then the area of this rectangle is A = 2x(1-x^4) = 2x - 2x^5

A = 2x - 2x^5, so A' = 2 - 10x^4

A' = 2(1-5x^4)

So A' = 0 when 5x^4 = 1

x = (1/5)^(1/4) will give the maximum area.

Then Amax = A((1/5)^(1/4)) = 2(1/5)^(1/4) - 2(1/5)^(1/4)^5 = 2/[5^(1/4)] - 2/[5^(5/4)]