Maximum-minimum area from a fixed length rope

[Karol]

And from my answer to the book's, without the L:
$$a=\frac{9}{4\sqrt{3}+9}=\frac{9}{\sqrt{3}(4+3\sqrt{3})}=\frac{3\sqrt{3}\sqrt{3}}{\sqrt{3}(4+3\sqrt{3})}=\frac{9}{4+3\sqrt{3}}$$
Thank you kuruman, Mark, Ray and LCKurtz