Maximum rectangle within a parallelogram There is a quadrilateral with equal-length for opposite sides but the diagonals are different (and I hope the word parallelogram is correct here), what would be the biggest rectangle I can inscribe, and why?
To put this in practical terms:
Someone cut a board and checked that the opposite sides have equal length. However when checking the diagonals, there are deviations. How to get back to a rectangular shape with minimum area to be cut off?
Note: reassembly is not intended here. If it were, there is a trivial solution that I am not interested in: https://en.wikipedia.org/wiki/Parallelogram#/media/File:Parallelogram_area_animated.gif
 A: I'll try to expand the comment written under the question:
Consider the figure below:

We call $P$ the parallelogram of base $B$ and height $H$ fixed, and $R$ will be the rectangle inscribed in $P$. Of course $\gamma$ is fixed.
Let's evaluate the area of the triangle $1$
Thanks to a well known property of triangles we have $\frac{l}{\sin\gamma} = \frac{x}{\sin( 180 - \gamma - \alpha)}$
then
$$h = l \sin \alpha = \frac{x \sin\alpha\sin\gamma}{\sin(180 - \gamma - \alpha)}$$
and the area of that triangle will be
$$\text{Area}_1 = \frac{x^2 \sin\alpha\sin\gamma}{2\sin(180 - \gamma - \alpha)}$$
For the trinagle $2$ we have that his height is $H - h$ and the base is $B - x$. In conclusion:
$$\text{Area}_2 = \frac{1}{2}(B-x)\Big(H - \frac{x \sin\alpha\sin\gamma}{\sin(180 - \gamma - \alpha)} \Big)$$
In conclusion, to maximise $R$ area we have that $\text{Area}_1 + \text{Area}_2$ must reach its minimum depending on $x$ and $\alpha$.
A: @Gabrielek actually delivered a really nice start, which I believe can be improved, and I will take a try on this. Let me know if there is a fault with it. I do not mind criticism, as this problem really is of interest for me. (And it has been a long time since I joined maths lessons.)
To inscribe a rectangle into a parallelogram there is more than one solution. Two probably simple cases are to take opposite sides as 'authorative' and cut the remainder off so we get a rectangle.

But how many more solutions are out there? If the rectangle would not have to be maximized, we all can easily imagine that there are an infinite number of solutions. But it could still be an infinite number when we go for maximized rectangle area. Just hitting one of the maxima would be satisfactory for me.
Because we inscribe the rectangle into the parallelogram, I assume that the vertices of the rectangle would be one on each of the four edges of the parallelogram. @Gabrielek said there is an angle $\alpha$ describing how much the rectangle is turned versus the parallelogram. While increasing the angle the rectangle's vertices should be seen wandering on the edges of the parallelogram.
I believe it is sufficient to look at a 90° turn, as after that the solutions found would be repetitive. Over the course of these 90°, the vertices would not make their way over one complete edge as there are positions where they cannot exist. In my painting that would be the lower left corner, for example.
So let's carefully select a possible value, maybe in the middle of the lower edge. At this time I care about the vertex location and not yet about the angle. Because of symmetry, the vertex on the upper edge should then also be in the middle of the upper edge. Knowing one vertex immediately sets the position for the other one. Just two points left.
Also for them we have some restrictions: They must be on the left and right edges. And for both our known vertices, looking at one of the unknowns and then turning to the see other we have to get a 90° angle. Also for the two unknown vertices the known ones must appear on a 90° angle.
I tried to draw this but somehow fail. So I start believing that the two unknowns do not have to be part of the edges, but the 90° constraint remains as after all I am searching for a rectangle.
Edit: I moved forward with my paint job and am now convinced not all rectangles can have all their vertices on the parallelogram's edges. But at the same time I can clearly see that the rectangle cannot be as big as the one I had drawn as the left solution.
And with a lengthy enough parallelogram this would mean the rectangle would only touch two sides and can be moved within the parallelogram - that would require an offset x to describe exactly which rectangle I am looking at. Which proves @Gabrielek was right all the time. I accepted his answer as the right one.
Let me show what I faced:
I have a parallelogram and choose one vertex. Due to symmetry, the opposite vertex has a known position as well. I draw a line through the both known points. It is one of the diagonals of my intended rectangle. Both halfs of the rectangle are right triangles. So by creating a circle around the middle of the diagonal I know all possibilities for the remaining two vertices - and get them by just choosing another line going through the center point that cuts the circle within the bounds of the parallelogram.

Whichever line I choose, it will have the same length as the first diagonal. To maximize my rectangle's area, I want these vertices to be as much apart as possible. And that seems done when looking at my left solution. It would mean the rectangle that is parallel to the longer of the edges (and thus matches a fair part of them) is the largest possible one already. Should the solution I labelled as 'trivial' be the one and only biggest one?
In numbers it would mean $\alpha$ = 90°, x = cos($\lambda$)*H.
Coming back to the wooden board it would mean there is no other way than to cut off the gray parts on the left and right. Luckily we spotted the deviation quite early so it is not too much of lost material...
Thanks for all those who joined the puzzle game!
