Are bounds known for the maximum determinant of a (0,1)-matrix of specified size and with a specifed number of 1s? The problems of determining the maximum determinant of an $n \times n$ $(0,1)$-matrix and the spectral problem of determining exactly which other determinants can possibly occur are both reasonably well studied.
What I'd like to know is whether there are bounds known if the number of 1s in the matrix is specified.
In other words, if we let $f(n,k)$ denote the maximum determinant of an $n \times n$ $(0,1)$-matrix with exactly $k$ ones (in total, not per row), then are there any known bounds on the values for $f(n,k)$?
I've searched Math Reviews and Googled it, but all the papers that I have found refer to the maximum over the entire set of $n \times n$ matrices.
Edit: Here is a plot of the maximum determinant of a binary $6 \times 6$ matrix with $k$ 1s, where $0 \leq k \leq 36$ (as long as I haven't messed up the computation).

Edit 2: Here is the analogous plot for $7 \times 7$ matrices. 

Edit 3: Here is a plot of the bound given by Peter Mueller's post, together with the actual values for n=7.

 A: The argument used in Hadamard's inequality works here too: If $A$ is the matrix, then ${\rm tr}(A^{t}A) = k,$ so if $\lambda_{1},\lambda_{2}, \ldots, \lambda_{n}$ are the (real, non-negative) eigenvalues of $A^{t}A$, then $\sum_{i=1}^{n}\lambda_{i} =k.$ Hence $|{\rm det}(A)| = \prod_{i=1}^{n} \lambda_{i}^{\frac{1}{2}}  \leq \left(\sqrt{\frac{k}{n}} \right)^{n}.$  The determinant is obviously $0$ if $k < n.$
( Later edit: I don't know whether this might be useful, but if we let $J$ be the all $1$'s matrix, note that if the $\alpha$-eigenspace of $A$ has dimension greater than $1,$ it non-trivially intersects the $0$ eigenspace of $J,$ which has dimension $n-1.$ Hence $-\alpha$ is an eigenvalue of $J-A,$ which is also a $\{0,1\}$-matrix, but with $n^{2}-k$ non-zero entries. Even later edit, somewhat related (indirectly, I think)  to Peter Mueller's comment: in particular, if ${\rm rank}(J-A) <n-1,$ then $J$ and $J-A$ have a common non-zero eigenvector with eigenvalue $0,$ so also ${\rm det}(A) = 0).$
A: In this paper Gasper, Pfoertner and Sigg prove an interesting upper bound for a real $n\times n$ matrix $M$: Let $n\alpha$ and $n\beta$ be the sum of the entries or the squares of the entries of $M$, respectively. Set $\delta=(\alpha^2-\beta)/(n-1)$. If $\delta\ge0$, then $\lvert\det M\rvert\le \alpha(\beta-\delta)^{(n-1)/2}$.
So $\alpha=\beta=k/n$ for a $(0,1)$-matrix. In contrast to the Hadamard bound, this bound is attained quite often for $(0,1)$-matrices, namely for the incidence matrices of symmetric block designs. Looking at their (somewhat involved) proof, this result is not surprising, for they show that upon fixing $\alpha$, $\beta$ and $n$, the maximum of the determinant is attained if $MM^t$ is a linear combination of the identity matrix and the all $1$ matrix.  
