Weird optimization problem (Calculus) I am given the following question:

given a piece of wire with length $l$, one of which will be transformed in a circle and the other in a square, how should we cut the wire to have a minimum total area?

The weird thing about that question is that all the alternatives DO NOT depend on $l$, as it would be expected. It seems like there must have been a picture along the exercise (which does not):

What I found as an answer is that the lenght used for the circle should be a third that the one used for the square.
Am I missing something?
 A: I think that the $1$ at the numerator should be the length $l$.
We have that $l=2\pi r+4a$ and the total area as a function of $r$ is
$$A(r)=\pi r^2+a^2=\pi r^2+\frac{(l-2\pi r)^2}{16}$$
with $r\in[0,l/(2\pi)]$.
By taking the derivative we get
$$A'(r)=2\pi r-\frac{\pi(l-2\pi r)}{4}$$
and it is easy to see that the function $A$ attains the minimum value at $r=\frac{l}{8+2\pi}$.
Hence $a=\frac{l-2\pi r}{4}=\frac{l}{4+\pi}$ and
$\frac{2\pi r}{l}\approx 0.44$ is the fraction of the wire to be transformed in a circle. 
A: Using Maxima the instructions are:
E1 : -L + 2*pi*r + 4*s;
E2 : -A + pi*r^2 + s^2;

R : solve([E1,E2],[A,s]);

dAdr : diff(R[1][1],r);

rvalue : solve([dAdr],[r]);

the results are:
(%o1)                        (- L) + 4 s + 2 pi r
                                       2       2
(%o2)                         (- A) + s  + pi r
                 2                   2           2
                L  - 4 pi r L + (4 pi  + 16 pi) r       L - 2 pi r
(%o3)     [[A = ----------------------------------, s = ----------]]
                                16                          4
(%i4) 
                                  2
                           2 (4 pi  + 16 pi) r - 4 pi L
(%o4)                  0 = ----------------------------
                                        16
(%i5) 
                                        L
(%o5)                           [r = --------]
                                     2 pi + 8

