Box bounded by parabolas

Let the bottom edge of the box be $2x$ , then the height of the box is the difference between both parabolas: $-x^2+1-x^2 = -2x^2 + 1$

Now the area of the box is:

$A(x) = \big( 2x \big) \big( -2x^2 + 1 \big) =-4x^3 + 2x$

Differentiating and setting to zero:

$A'(x) = -12x^2+2 = 0$ $x = \frac{1}{\sqrt{6}}$

So the box with the largest area is:

$A = -4 \bigg( \frac{1}{\sqrt{6}} \bigg)^3 + 2 \bigg( \frac{1}{\sqrt{6}} \bigg) = \frac{2 \sqrt{6}}{9} \approx 0.544$

Let's shift the whole diagram down by $\frac{1}{2}$ a unit so that the center of the rectangle is at the origin. Then the equations of the two parabolas are $y = -x^2 + \frac{1}{2}$ and $y = x^2 - \frac{1}{2}$ . Now by symmetry we only need to examine the maximum area of the fourth of the rectangle that is in the first quadrant that has one vertex on the parabola $y = -x^2 + \frac{1}{2}$ and one vertex at the origin.

The area of the rectangle in the first quadrant is $A_q = xy = x(-x^2 + \frac{1}{2}) = -x^3 + \frac{1}{2}x$ . Setting its derivative to zero gives $A_q' = -3x^2 + \frac{1}{2} = 0$ , which solves to $x = \frac{\sqrt{6}}{6}$ , which makes $A_q = -(\frac{\sqrt{6}}{6})^3 + \frac{1}{2} \cdot \frac{\sqrt{6}}{6} = \frac{\sqrt{6}}{18}$ .

The area of the whole rectangle is $4$ times that much or $A = 4 \cdot \frac{\sqrt{6}}{18} = \frac{2 \sqrt{6}}{9} \approx \boxed{0.544}$ .