Circle Inscribed in Bell Curves 2

Let $r$ be the distance between a point on the bell curve and the origin. Then by Pythagorean's Theorem, $r^2 = x^2 + y^2$ . Using implicit differentiation, $2r \frac{dr}{dt} = 2x + 2y \frac{dy}{dx}$ which means $\frac{dr}{dx} = \frac{x + y \frac{dy}{dx}}{r}$ .

A circle inscribed by the bell curve would have a minimum radius, so $\frac{dr}{dx} = 0$ , and as the bell curves are symmetrical we only need to examine the first quadrant when $y = ae^{-ax^2}$ (and therefore when $\frac{dy}{dx} = -2a^2xe^{-ax^2}$ ), so $0 = \frac{x + ae^{-ax^2} \cdot -2a^2xe^{-ax^2}}{r}$ , which solves to $x^2 = \frac{\ln{2a^3}}{2a}$ for $a \geq \frac{1}{\sqrt[3]{2}}$ . When $0 < a \leq \frac{1}{\sqrt[3]{2}}$ , $\frac{dr}{dx}$ is always greater than $0$ in the first quadrant, and so its minimum radius is when $x = 0$ , or when $y = ae^{-ax^2} = ae^{-a0^2} = a$ , or when $r = \sqrt{x^2 + y^2} = \sqrt{0^2 + a^2} = a$ .

When $0 < a \leq \frac{1}{\sqrt[3]{2}}$ , $r = a$ , and the area of the circle is $A = \pi r^2 = \pi a^2$ , and since $a$ is always increasing in the first quadrant, the maximum area would be when $a$ is at its biggest value at $a = \frac{1}{\sqrt[3]{2}}$ , and so the maximum area is $A = \pi a^2 = \frac{\pi}{\sqrt[3]{4}}$ .

When $a \geq \frac{1}{\sqrt[3]{2}}$ , since $A = \pi r^2$ , $r^2 = x^2 + y^2$ , $y = ae^{-ax^2}$ , and $x^2 = \frac{\ln{2a^3}}{2a}$ , $A = \pi(x^2 + (ae^{-ax^2})^2) = \pi(\frac{\ln{2a^3}}{2a} + (ae^{-a(\frac{\ln{2a^3}}{2a})})^2)$ which simplifies to $A = \frac{\pi}{2a}(\ln{2a^3} + 1)$ . When the area is maximized, $\frac{dA}{da} = 0$ , and taking the derivative of $A$ gives $\frac{dA}{da} = \frac{3\pi}{2a^2} - \frac{\pi}{2a^2}(\ln{2a^3} + 1)$ . Solving $0 = \frac{3\pi}{2a^2} - \frac{\pi}{2a^2}(\ln{2a^3} + 1)$ gives $a = \frac{e^{2/3}}{\sqrt[3]{2}}$ . This value of $a$ gives an area of $A = \frac{\pi}{2a}(\ln{2a^3} + 1) = \frac{\pi}{2(\frac{e^{2/3}}{\sqrt[3]{2}})}(\ln{2(\frac{e^{2/3}}{\sqrt[3]{2}})^3} + 1) = \frac{3 \pi \sqrt[3]{2}}{2e^{2/3}}$ .

Since $\frac{3 \pi \sqrt[3]{2}}{2e^{2/3}} > \frac{\pi}{\sqrt[3]{4}}$ , the value of $a$ that gives the maximum area of a circle inscribed in the given bell curve is $a = \boxed{\frac{e^{2/3}}{\sqrt[3]{2}}}$ .