I recently encountered the following problem:
Let $M$ be the finite set of 4 by 4 matrices $A$ with entries $a_{ij} \in \{1,-1\}$. Restrict the determinant viewed as a function on this set $M$ as follows: $$ det: M \rightarrow \mathbb{Z}. $$
Find $A \in M$ so that $det(A)$ is maximal.
My idea is to view the determinant of $A$ as the volume of the parallelotope in four dimensions. Say we have the four vectors $a, b, c $ and $d$ (for example $a = \sum_{i=1}^4a_ie_i$), each of which specifies a different side of the parallelotope, then the determinant would be $$ det A= \sum_{\sigma\in S_4}sgn(\sigma)\prod_{i=1}^4a_{\sigma(i),i}= \begin{vmatrix} a_1 & a_2 & a_3 & a_4 \\ b_1 & b_2 & b_3 & b_4 \\ c_1 & c_2 & c_3 & c_4 \\ d_1 & d_2 & d_3 & d_4 \end{vmatrix} $$ with every entry taking on the value 1 or -1. Viewing the determinant as a function, I would like to take its derivative and compute the entry values for which it gets a maximal value. I'm not sure how to approach this though.
Does anybody have any tips? :)
It is known as Hadamard's maximal determinant problem, for $n=4$ the maximum value is equal to $16$.