We have been that length of both the diagonals are equal to $x$. What can be the maximum value of the product of length of sides?
It is obvious that an upper bound exists, but I can't get the maximum value. Maybe it involves some formula. Can you help?
I would be surprised if any solution was not equivalent to the vertices $(0,0)$, $(\frac12,1)$, $(1,0)$ and $(\frac12,0)$. Each diagonal is obviously $1$. The product of the sides is $\frac{\sqrt{5}}{2} \times \frac{\sqrt{5}}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac5{16}$.
If you do not regard this as strictly a convex quadrilateral then it provides an approachable supremum.
It is not difficult to show that moving any individual vertex within the constraints will reduce the product. It is harder to show this is a global optimum, since even fixing one diagonal there would be four possible equivalent solutions. It might be possible to show that the diagonals must be orthogonal and that the quadrilateral is symmetric, i.e. an equidiagonal kite, which would simplify the demonstration of global optimality.