In the Wiki article on adjunction https://en.wikipedia.org/wiki/Adjoint_functors, there is a motivation section that talks about how adjunctions can be viewed as "Solutions to optimization problems", or "a way of giving the most efficient solution to some problem via a method which is formulaic". The subsequent discussion, and later definition spawned from this motivation do make sense in terms of the free-forgetful adjunctions, like yes, I see how the free group construction is the "most efficient way to create a group out of a set", and hence is the left adjoint of the forgetful functor, which is the "most efficient way to create a set out of a group".
Another example of this "optimization motivation" in a situation that is not exactly a free-forgetful adjunction is the diagonal functor being adjoint to the product — there's still some intuition there about $X \mapsto (X,X)$ being the "most natural/efficient way" of going from one object to a pair. For situations involving both left and right adjoints, the Wiki page has a very nice interpretation of this as well:
The notion that F is the most efficient solution to the problem posed by G is, in a certain rigorous sense, equivalent to the notion that G poses the most difficult problem that F solves. This gives the intuition behind the fact that adjoint functors occur in pairs: if F is left adjoint to G, then G is right adjoint to F.
A very nice example from A bestiary about adjunctions is the idea that quantifiers can be thought of as adjoints to the inverse image function. The inverse image functor applied to $f: X\to Y$ can be thought of as "the shadow of $Y$ on $X$ when projected through $f$", so it is reasonable that the "most efficient solution" to this problem is just to find all points of $Y$ that could have (existential quantifier) cast that shadow. Not sure yet how to phrase the right adjointness of the universal quantifier. Anyhow, many examples in the above "bestiary about adjunctions" link have some intuitive sense of being an "efficient and formulaic conversion" …but I have no sense of this intuition at all regarding the tensor-hom adjunction.
My question is this: is there some interpretation of the tensor-hom adjunction https://en.wikipedia.org/wiki/Tensor-hom_adjunction that is in a similar line of thinking to this "optimization motivation"? I suppose implicitly my question is asking for a more formal/rigorous understanding of what the phrase "the problem $G$ poses" means in the Wiki page's sentence "$F$ is the most efficient solution to the problem posed by $G \iff F$ is the left adjoint of $G$".
EDIT 12/19/22: Indeed, trouble understanding "in which sense a functor poses a question" has already been asked on MO to the writer of the Wikipedia article, with no answer.
EDIT 12/18/22: I came across https://mathoverflow.net/questions/6551/what-is-an-intuitive-view-of-adjoints-version-1-category-theory, with I think this answer being the one I like more than the others; in part it reads:
In essence, if two notions are related to eachother closely enough that giving a definition of one notion (in the presence of sufficient ambient structure) defines the other notion completely, they can be expressed as a (possibly $\infty$-) adjoint pair of functors. Conversely, when we see that two notions can be expressed as an adjoint pair of functors it means we can think about one by thinking about the other plus some canonical additional structure.
(this makes some sense for e.g. the Group-Set forgetful-free functor; going from Group to Set we have no canonical additional structure to add, but going back the canonical additional structure is the group structure formed by the words generated by the letters — although I still don't really see the hom functor being "the tensor functor plus some canonical additional structure" or vice versa). I also think this comment (about the inclusion $\iota: \mathbb Z \to \mathbb R$ having right/left adjoint floor/ceil function resp.) is cute:
The right adjoint is conservative, safe, it does not want to over estimate, hence the floor [rounding down]. The left adjoint is liberal, risky, it wants to make sure it gets credit for partial work, hence the ceil [rounding up].
See also this MO thread about the prevalence of forgetful functors having left adjoints.
Maybe one key difficulty in connecting the 2 concepts is that forgetful-free adjunctions are most intuitive through the unit-counit definition of adjunction; while the tensor-hom adjunction is most intuitive through the natural isomorphism of hom functors perspective.
EDIT 12/20/22: the previous statement is too subjective (as pointed out in the comments), and I'm starting to change my mind about it. However, another key difficulty may be that (the most common examples of) forgetful-free adjunctions are between two "tangibly" different categories (Group vs. Set for instance), so the free group generated by a set $X$ can be thought of as a "groupified version of the set $X$" in the precise sense that the Group-maps outward from the groupification $F(X)$ "are essentially" Set-maps outward from $S$ to a restricted class of sets, namely those that are "set versions" of groups. So $F(X)$ and $X$ are really the "same object" in different categories, at least in the eyes of all other groups $Y$ (and their alter egos $G(Y)$ in Set). This "outsider POV" on $F(X)$ and $X$ somehow encodes the "ambient structure" of Group and Set resp. discussed in the abovementioned MO answer.
The difficulty of the tensor-hom adjunction is perhaps then that because the categories the functors map between are similar, there is no intuition for what the "ambient structure" is. Qiaochu's answer on reframing adjunction in terms of representability seems to say that the "outsider POV" perspective is the "right" one to take, and so the fact that in the forgetful-free setting the outsider POV somehow encodes the ambient structure of each category is a "coincidence", and not really the heart of the matter.
Best Answer
Let $R, S$ be two rings and $f : R \to S$ a ring homomorphism. Pullback along $f$ induces a forgetful functor $\text{Mod}(S) \to \text{Mod}(R)$ from left $S$-modules to left $R$-modules, and this functor has a left adjoint $\text{Mod}(R) \to \text{Mod}(S)$ given by the tensor product $S \otimes_R (-)$, giving an adjunction which is simultaneously a free-forgetful and a tensor-hom adjunction:
$$\text{Hom}_S(S \otimes_R M, N) \cong \text{Hom}_R(M, \text{Hom}_S(S, N)).$$
Here $\text{Hom}_S(S, N)$ is a slightly confusing name for the forgetful functor; we are thinking of $S$ as an $(S, R)$-bimodule here. The left adjoint here is variously called change of rings, induction (in the sense of induced representations), or extension of scalars. It specializes, for example, to induced representations for groups if we take $R, S$ to be group algebras and $f$ to be a morphism induced by a group homomorphism, as well as to induced representations for Lie algebras if we take $R, S$ to be universal enveloping algebras.
I'm not aware of a (useful) free-forgetful interpretation of the tensor-hom adjunction in general.
Re: the idea of thinking of adjunctions in terms of "optimization," actually I do not recommend thinking in these terms. That part of the Wikipedia article was written by a single person over a decade ago (who happens to be an old friend of mine, incidentally) and does not exactly reflect the practice of a community of category theorists or anything like that. There are enough different ways of thinking about adjunctions that you can take your pick from the available options; I think the one closest to the "optimization" idea is to think in terms of representability.
Namely, every functor $F : C \to D$ poses a "representability problem": are the presheaves $\text{Hom}_D(F(-), d) : C^{op} \to \text{Set}$ representable? $F$ has a right adjoint $G$ precisely iff the answer is "yes" for every $d$, in which case the representing object is $G(d)$. (In particular, the existence of these representing objects alone uniquely determines $G$ as a functor; this is a nice exercise if you haven't done it already.) These representability problems sometimes resemble optimization problems (e.g. in the case of posets, as in your floor / ceiling example) but not always and thinking in terms of representability is fully general; then you can try to build intuition over time for what it really means to ask that a functor be representable. Dually for left adjoints, of course.
I already don't think it quite makes sense to think of, say, the free group functor as the "most efficient way to turn a set into a group." Let's consider a variation where we replace groups with magmas, which are sets $M$ equipped with a binary operation $M \times M \to M$ satisfying no axioms. Here is a very efficient way to turn any set $X$ into a magma: adjoin a new element $0$, and define every product to have value $0$. Easy! However, this is not the free magma on $X$, which is much more interesting; the free magma is the set of (rooted, planar) binary trees whose leaves are labeled by elements of $X$.
I don't think "efficiency" is the right word to describe what makes these constructions different. Here I think it's worth taking "freeness" seriously as a word: the free group (or free magma, etc.) is the freest way to turn a set into a group (or magma, etc.) in the sense that it imposes the fewest constraints on which elements of the group are equal (only the constraints imposed by the axioms). Representability, as it turns out, is exactly how to make this idea precise.
That is, let $U : \text{Grp} \to \text{Set}$ be the forgetful functor from groups to sets. What does it mean to ask that the left adjoint $F : \text{Set} \to \text{Grp}$ (the free group functor) exist? It means that we want all the functors $\text{Hom}_{\text{Set}}(X, U(G)) \cong G^X$ to be representable, as functors of the group variable $G$, for every set $X$. The representing objects $F(X)$ must satisfy
$$\text{Hom}_{\text{Grp}}(F(X), G) \cong \text{Hom}_{\text{Set}}(X, U(G)) \cong G^X$$
and what this means is exactly that $F(X)$ is a group containing elements labeled by $x \in X$ which are as "free" or unconstrained as possible, in the sense that they can map onto arbitrary elements of any other group $G$. (The map $X \to F(X)$, or more precisely $X \to U(F(X))$, is of course the unit of the adjunction.)