Reformed Disk

Relevant wiki: Second Derivative Test

Here's a calculus approach:

Like what Christopher Boo has written up, we have

$r^2 + R^2 = 1 ,\quad (r,R) \in [0,1]\times[0,1] .$

And we want to maximize $f = r+R$ .

Substituting $R = \sqrt{1-r^2}$ gives $f = r + \sqrt{1-r^2}$ .

This is a 1-variable calculus problem with $f' = 1 - \dfrac{r}{\sqrt{1-r^2}}$ .

At the extrema point, we have $f' = 0 \Rightarrow r = R = \dfrac1{\sqrt2}$ .

To prove that it is a local maximum, we apply the second derivative test , to show that $f'' < 0$ when $r = \dfrac1{\sqrt2}$ .

$f'' = -\dfrac1{(1-r^2)^{3/2}} < 0 \text{ when } r = \dfrac1{\sqrt2}.$

Since we have established that it is a local maximum, we still need to prove that it's a global maximum. It's trivial to show that at the boundary conditions $f(r,R) = f(0,1)$ or $f(1,0)$ yields a lower value than $f\left( \dfrac1{\sqrt2},\dfrac1{\sqrt2} \right)$ .

Thus we have proved that the global maximum of the function $r+R$ occurs when $r=R = \dfrac1{\sqrt2}$ . Our answer is $r + R = \dfrac2{\sqrt2} = \boxed{\dfrac{\sqrt2}2 }$ .

Relevant wiki: Cauchy-Schwarz Inequality

Let the radius of the inner region be $r$ and the radius of the reformed green area be $R$ .

We are given that their area is equal to the area of the original unit disk, hence

$\pi r^2 + \pi R^2 = \pi \\ r^2 + R^2 = 1$

We want to maximise the combined circumference $2\pi r + 2\pi R \rightarrow 2\pi(r+R)$ . Since $2\pi$ is constant, we only have to maximise $r+R$ . By Cauchy-Schwarz Inequality,

$(r^2+R^2)(1+1)\geq (r+R)^2 \\ r+R\leq \sqrt{2}$

The condition to satisfy the maximum is when $r=R$ , hence $r = \frac{\sqrt{2}}{2}$ .