, 2020

The parabola $y=\frac14x^2$ and the parabolic rectangle are each symmetrical about the $y$-axis, and thus a second vertex of the rectangle lies on the parabola and has coordinates $(-6,9)$. A third vertex of the parabolic rectangle lies on the $x$-axis vertically below $(6,9)$, and thus has coordinates $(6,0)$. Similarly, the fourth vertex also lies on the $x$-axis vertically below $(-6,9)$, and thus has coordinates $(-6,0)$. If one vertex of a parabolic rectangle is $(-3,0)$, then a second vertex has coordinates $(3,0)$, and so the rectangle has length 6. The vertex that lies vertically above $(3,0)$ has $x$-coordinate 3. This vertex lies on the parabola $y=\frac14x^2$ and thus has $y$-coordinate equal to $\frac14(3)^2=\frac94$. The width of the rectangle is equal to this $y$-coordinate $\frac94$, and so the area of the parabolic rectangle having one vertex at $(-3,0)$ is $6\times\frac94=\frac{54}{4}=\frac{27}{2}$. Let a vertex of the parabolic rectangle be the point $(p,0)$, with $p>0$. A second vertex (also on the $x$-axis) is thus $(-p,0)$, and so the rectangle has length $2p$. The width of this rectangle is given by the $y$-coordinate of the point that lies on the parabola vertically above $(p,0)$, and so the width is $\frac14p^2$. The area of a parabolic rectangle having length $2p$ and width $\frac14p^2$ is $2p\times\frac14p^2=\frac12p^3$. If such a parabolic rectangle has length 36, then $2p=36$, and so $p=18$. The area of this rectangle is $\frac12(18)^3=2916$. If such a parabolic rectangle has width 36, then $\frac14p^2=36$ or $p^2=144$, and so $p=12$ (since $p>0$). The area of this rectangle is $\frac12(12)^3=864$. The areas of the two parabolic rectangles that have side length 36 are 2916 and 864. Let a vertex of the parabolic rectangle be the point $(m,0)$, with $m>0$. A second vertex (also on the $x$-axis) is thus $(-m,0)$, and so the rectangle has length $2m$. The width of this rectangle is given by the $y$-coordinate of the point that lies on the parabola vertically above $(m,0)$, and so the width is $\frac14m^2$. The area of a parabolic rectangle having length $2m$ and width $\frac14m^2$ is $2m\times\frac14m^2=\frac12m^3$. If the length and width of such a parabolic rectangle are equal, then $$\begin{align*} \frac14m^2 & = 2m\\ m^2 & = 8m\\ m^2-8m & = 0\\ m(m-8) & = 0 \end{align*}$$ Thus $m=8$ (since $m>0$), and so the area of the parabolic rectangle whose length and width are equal is $\frac12(8)^3=256$.