Why is the variation of position in the boundary of configuration space not reversible? 
[...] we have investigated the question of an extremum under the condition that we  are inside the boundaries of configuration space. A function which does not assume any extremum inside a certain region may well assume it on the boundary of that region. On the boundary of the configuration space the displacements are no longer reversible and hence our argument that the first variation must vanish$-$because otherwise  it can be made positive as well as negative$-$no longer holds. For non-reversible displacements a function may well assume an extremum without having a stationary value at that point. In that case an extremum exists without the vanishing of the first variation. 

The above is excerpted from Stationary value versus extreme value from Cornelius Lanczos' The Variational Principles of Mechanics; here the author explains 

The extremum of a function requires a stationary value only for reversible displacements. On the boundary of the configuration space, where the variation of position is not reversible, an extremum is possible without a stationary value.

I couldn't comprehend what the author meant by non-reversible variation of position; for the very time in the book he did use the term reversible displacement and its antonym. 
But what are actually they?
What does make a "variation of position" reversible or non-reversible?
Why is the displacement not reversible in the boundary of a configuration space?
Could anyone shed light on these terms?
 A: This is strange wording, but it's trying to explain the fact that at the boundary of a function's domain of definition, it can be extremal without being stationary - because the very idea of "stationary" doesn't make sense at the boundary.
By definition, a "stationary" point is one where the first derivative (or "variation", which is just a disguised functional derivative) vanishes. But standard differentiability is only defined on open sets (see this math.SE question) - although one might try to extend the derivative to the boundary by continuity, it doesn't make sense in general to speak of the derivative at the boundary. 
That the variation is "not reversible" means that you can vary in one direction (inwards) at the boundary, but not outwards, since that variation would carry the function out of its domain of definition. Supposing that Lanzcos' argument for why the first variation has to vanish at a extremum is a variant of the standard argument, it's now clear why "non-reversibility" of the variation destroys the argument: The variation would acquire one sign under variations inwards, and the opposite sign under variations outwards, but the latter don't exist else we wouldn't be at the boundary, so we simply can't derive the vanishing of the variation/derivative.
