What are "best" polynoms f(x) g(x) of degree n ? I.e. ideal generated by them is as far from zero as possible ? (Best convolutional codes.) Consider polynoms f(x) g(x) of degree at most n. (I am mostly interested about F_2[x]).
Let us multiply them by arbitrary polynoms p(x)  i.e. consider ideal (p f , p g) in $F_2[x]\oplus F_2[x]$.
Let us delete zero from this ideal and calculate N(f,g) -  Hamming distance of I$\backslash${0} to {0}.
(Here Hamming distance is just the number of non-zero monoms).
Questions What is Max_{f,g of degree n} N(f,g) ? (Or at least some bounds on it ?)
What polynoms give this max ? 

What  seems not trivial:
It is clear that N(f,g) <= |f| + |g| (take p(x) =1 ) .
So it seems that we should take |f| and |g| of maximal possible Hamming norm i.e. 
g = f = x^n+x^{n-1} + ... x^2 +x +1
But it clear that N(f,g) =4 for this choice of f,g - i.e. very small. (Proof - just multiply them by p(x) = (x+1)).
So  these two effects are fighting each other - we want to take |f| , |g| big,
but multiplication by x+1 will spoil N(f,g) if these norms are too big....

Related question 
Find polynoms f,g such that for any polynom p(x):  |fp|+|gp|>=  |f|+|g| ? Where |*| is number of non-zero monoms.

From coding theory viewpoint 
map: p(x) -> (p f , p g) is encoder (non-recursive convolutional code of rate 1/2).
I am asking what is "best possible" code for deg f,g


Examples 
if n = 2, then f=x^2 +1 , g= x^2+x+1 gives max  which equal to 5 
(see Multiplication by polynomials x^2+1 ; x^2+x+1. Does minimal Hamming norm of image equal to 5 ?  )
If n=3 then max N(f,g) = 6 and can be realized by many choices
e.g. f=x^3+x^2+1, g=x^3+x+1 or f=x^3+x^2+x+1, g=x^3+x+1
(if I am not mistaking).
If n=4, then possibly(?) max N(f,g) = 8 
for e.g. f=(x+1)(x^3+1) ; g=(x+1)(x^3+x+1) ------ NOT TRUE : take p(x)=x^2+x+1,
we get N(f,g)<=6 for these pols. It might that max N(f,g) = 7.
achieved on e.g. f=x^4+x^3+x^2+1 g=x^4+x+1
 A: The following simple (suboptimal) bound will get you started. It is easier to give a lower bound for the degree $n$ of the polynomials in terms of the minimum distance $d$, so I will do that. Feel free to turn this into an upper bound on $d$ :-)
From the theory of biinary linear block codes we recall the so called Griesmer bound. It states that if a binary linear code has dimension $k$, minimum distance $d$ and length $N$, then we have the inequality
$$
N\ge\sum_{j=0}^{k-1}\lceil\frac{d}{2^j}\rceil.
$$
We apply this to convolutional codes of the type that you describe as follows. Consider the subspace gotten by restricting the multiplier $p(x)$ to be of degree $\le m$. The dimension of this subspace is $k=m+1$. The degrees of $pf$ and $pg$ are both at most $n+m$, so the pairs
$(pf,pg)$ have $N=2(n+m+1)$ coefficients. Thus Griesmer bound gives us the inequality
$$
2(n+m+1)\ge\sum_{j=0}^m\lceil\frac{d}{2^j}\rceil.
$$
How to select $m$? When we increase $m\to m+1$, the l.h.s. increases by two, and the r.h.s.
increases by $\lceil d2^{-m-1}\rceil$. Therefore we get the tightest bound on $n$, when
$m$ is the largest natural number with the property $d>2^{m+1}$, because this is the largest $m$ such that all the terms on the r.h.s. are larger than two.
As examples consider $d=5$. Then we select $m=1$, and get the bound
$$
2(n+2)\ge 5+3\implies n\ge 2.
$$
This bound is attained with the well known pair $f(x)=1+x^2$, $g(x)=1+x+x^2$. 
If we used $d=6$ instead, we would get $n\ge3$, so we can conclude that $d=5$ is the
best we can do with quadratic $(f,g)$. 
If we want
$d=9$, then we can select $m=2$, and
$$
2(n+3)\ge 9+5+3=17\implies n\ge 6
$$
(bearing in mind that $n$ is an integer). We see that $d=10$ also gives $n\ge 6$, but $d=11$ would give $n\ge 7$. This implies that with sextic $(f,g)$ the best we can hope for is $d=10$. This bound is actually achieved by the pair
$$
f(x)=1+x+x^3+x^4+x^6,\qquad g(x)=1+x^3+x^4+x^5+x^6.
$$
IIRC this code was used during the Voyager mission to transmit data (e.g. the pretty images) back to Earth.
The above bound is not always accurate. There are several reasons for this. When $k$ increases, the Griesmer bound is no longer tight. Also, there is no reason to think that
the type of binary linear codes by limiting the degree of $p$ would be optimal binary linear codes. Bounds like this will give you an idea, what to expect, and then you can start developing a search heuristic. If there is a precise answer to your question, then it will need to use some new machinery.
