I was able to do most of the question, up until it asked for a maximum and minimum for the total enclosed area.
I got two equations:
$4 = 4x + 2r\cdot\pi$
for the perimeter and,
$A = x_2 + \pi(r_2)$
for the area.
I isolated $r$ in the first equation plugged it into the second equation, isolating $x$. I rearranged it a bit to make it a little easier for myself and then took the derivative. Next, I set $\frac {dA}{dx}$ to $0$ to find my critical points. I happen to find that I had only one critical point which was $\frac 1{\pi +4}$.
I have no idea how to find the maximum and minimum from here on out. I know I could draw a sign diagram, but that would not help since I only have one critical point and the graph I got was linear, meaning it doesn't really have a max or min anyways. Did I just do my algebra wrong or did I do something else wrong? Thanks in advance!
Let the length of the first piece be $x$, then the length of the second piece is $4-x$. The side of a square will be $x/4$ and the radius of circle will be $\frac{4-x}{2\pi}$. Thus, the area function will be the sum of areas of square and circle: $$A=\frac{x^2}{16}+\pi\cdot \frac{(4-x)^2}{4\pi^2}$$ Can you finish and find the maximum/minimum of $A(x)$, on the interval $0\le x \le 4$?