Answer
$\frac{4b}{4+\pi}$, $\frac{b\pi}{4+\pi}$
Work Step by Step
Step 1. Assuming 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 a square with side length of $\frac{x}{4}$; thus the area for this part is $A_1=\frac{x^2}{16}$
Step 3. The second part is used to form a circle with radius $r$; thus, we have $2πr=b−x$ and $r=\frac{b-x}{2\pi}$ and the area for this part is $A_2=πr^2=\frac{b^2-2bx+x^2}{4\pi}$
Step 4. The sum of the two areas is given by $A=A_1+A_2=\frac{x^2}{16}+\frac{b^2-2bx+x^2}{4\pi}$
Step 5. The above area will reach an extreme when $A′=0$; we have $A′=\frac{x}{8}+\frac{x-b}{2\pi}=0$, which gives $x=\frac{4b}{4+\pi}$ and the length of the second part is $b−x=\frac{b\pi}{4+\pi}$
Step 6. Check $A′′=\frac{1}{8}+\frac{1}{2\pi}\gt0$ and we know the region is concave up with a local minimum at the above point.
