## Friday, April 17, 2015

### Maximize The Area Of The Lune

In the $xy$ plane, there are two circles, a larger one of radius 1 unit and a smaller one of radius $r$. The two circle intersect such that the two points of intersection are on a diameter of the smaller circle. Find the value of $r$ which maximizes the area that is inside the smaller circle, but is outside the larger circle.

Okay, now the area $A$ in question may be described by:

$A=2 \int_{0}^{r} \sqrt{r^2-x^2}-\sqrt{1-x^2}+\sqrt{1-r^2}\,dx$

The first term in the integrand represents the upper half of the smaller circle, which has radius $r$, and so rather than use the calculus, we may simply use the formula for the area of a circle, and so we have:

$A=\frac{\pi}{2}r^2+ 2\int_{0}^{r}-\sqrt{1-x^2}+\sqrt{1-r^2}\,dx$

The last term does not depend on the variable $x$, and so we may use the anti-derivative form of the FTOC (Fundamental Theorem of Calculus) and write:

$A=\frac{\pi}{2}r^2+ 2r\sqrt{1-r^2}-2\int_{0}^{r}\sqrt{1-x^2}\,dx$

So, differentiating term by term, and using the derivative form of the FTOC on the remaining integral, we obtain:

$A'(r)=\pi r+2\left(r\frac{1}{2\sqrt{1-r^2}}(-2r)+(1)\sqrt{1-r^2}\right)-2\sqrt{1-r^2}$

Simplifying, we obtain:

$A'(r)=\pi r+2\left(-\frac{r^2}{\sqrt{1-r^2}}+\sqrt{1-r^2}\right)-2\sqrt{1-r^2}$

$A'(r)=\pi r-\frac{2r^2}{\sqrt{1-r^2}}$

$A'(r)=\frac{r\left(\pi \sqrt{1-r^2}-2r\right)}{\sqrt{1-r^2}}$

Now, the critical values $r=0$ and $r=1$ are at the minimum ($A=0$), and so we are left with the critical value that comes from:

$\pi \sqrt{1-r^2}-2r=0$

So, we want to solve for $r$:

$\pi\sqrt{1-r^2}=2r$

Square:

$\pi^2\left(1-r^2\right)=4r^2$

$\pi^2-\pi^2r^2=4r^2$

$\pi^2=4r^2+\pi^2r^2$

$\pi^2=r^2\left(4+\pi^2\right)$

$r^2=\frac{\pi^2}{4+\pi^2}$

Taking the positive root, we find:

$r=\frac{\pi}{\sqrt{\pi^2+4}}$

Since $A(0)=A(1)=0$, we know this critical value is at a maximum.