Answer
$\frac{9b}{\sqrt 3\pi+9}$, $\frac{\sqrt 3\pi b}{\sqrt 3\pi+9}$
Work Step by Step
Step 1. Assume the length of the first part is $x$; the length for the second part is then $b-x$.
Step 2. The first part is used to form an equilateral triangle with side length of $\frac{x}{3}$ and height of $\frac{\sqrt 3x}{6}$; thus the area is $A_1=\frac{1}{2}\frac{x}{3}\frac{\sqrt 3x}{6}=\frac{\sqrt 3x^2}{36}$
Step 3. The second part is used to form a circle with radius $r$; thus we have $2\pi r=b-x$ and $r=\frac{b-x}{2\pi}$ and the area for this part is $A_2=\pi r^2=\pi (\frac{b-x}{2\pi})^2=\frac{b^2-2bx+x^2}{4\pi}$
Step 4. The sum of the two areas is given by $A=A_1+A_2=\frac{\sqrt 3x^2}{36}+\frac{b^2-2bx+x^2}{4\pi}$
Step 5. The above area will reach an extreme when $A'=0$; we have $A'=\frac{\sqrt 3x}{18}+\frac{x-b}{2\pi}=0$ which gives $x=\frac{9b}{\sqrt 3\pi+9}$ and the length of the second part is $b-x=\frac{\sqrt 3\pi b}{\sqrt 3\pi+9}$
Step 6. Check $A''=\frac{\sqrt 3}{18}+\frac{1}{2\pi}\gt0$ and we know the region is concave up with a local minimum at the above point.
