Find a matrix which maximizes expression Assume I have column vectors $x,y\in\mathbb{R}^n$, and the following expression
$$
A\in M_n(\mathbb{R}),\ |\det A|\leq 1,\ K(A) = x^tAy
$$
How can I find the matrix $A$ that maximizes expression $K(A)$  (which is a scalar)? I searched the net on trying to do this with calculus but I yielded no relevant information.
 A: For your expression we have
$$
K(A) = x^t A y = \sum_{i,j} x_i a_{ij} y_j = \sum_{i,j} a_{ij} x_i y_j 
$$
which is a quadratic form. You can make $K(A)$ arbitrary large by choosing large matrix elements,
there is no bound, no maximum.
Update: 
A constraint restricting $A$ to $\lvert \mbox{det }(A) \rvert \le 1$ was added to the problem.
This will not change the situation as:
$$
d = \mbox{det }(A) = \frac{10^n}{10^n} \mbox{det }(A) = \mbox{det}(10^n a_1, 10^{-n} a_2, \ldots, a_n) = \mbox{det }(B)
$$
so we can amplify $a_1$ (or any other column vector) and thus increase $K(A)$ to any large value $K(B)$ while still keeping the determinant value $d$ for that new matrix $B$.
A: Let $x = (x_i)_{i=1}^n$ and $y = (y_i)_{i = 1}^n$ be vectors in $\Bbb R^n$. You can think of the situation as $K: \Bbb R^{n \times n}\to \Bbb R$, given by: $$K(a_{11},\cdots,a_{1n},a_{21},\cdots, a_{nn}) = \sum_{i,j=1}^n a_{ij}x_iy_j,$$and now solve the system: $$\frac{\partial K}{\partial a_{ij}} = 0, \quad 1\leq i,j \leq n$$ to find the critical points, and such.
A: Hint.
we have $x^TAy=\langle x,Ay\rangle$ and this scalar product has is maxim value wan $Ay$ is parallel to $x$. 
A: This expression attains no maximum or minimum for $n \geq 2$ and non-zero $x,y$. We can prove that this is the case as follows:
For any $x,y$ and $C \neq 0$, we can construct an $A$ such that $Ay = Cx$ and $\det(A) = 1$.  For this $A$, we have
$$
K(A) = x^TAy = x^T(Cx) = C\|x\|^2
$$
Thus, $K(A)$ can be made arbitrarily large or arbitrarily small.
A: Many have already said it, the problem is unbounded. I want to just give an explicit example:
For arbitrary $x,y\ne 0$, without loss of generality norm 1. Extend $x$ and $y$ to orthonormal bases $X$ and $Y$. 
For $n\in\mathbb N$ define 
$$ D_n = \operatorname{diag}(n, 1/n, 1, \dotsc, 1) $$
and
$$A_n = X D_n Y^T. $$
Then, we have 
$$|\det(A_n)| = |\det X||\det Y||\det D_n| = |\det D_n| = 1.$$
Further, we have
$$ K(A_n) = x^T A_n y = x^T X (X^T A_n Y) Y^T y = e_1^T D_n e_1 = n \to \infty. $$
