How to reduce MaxUNSAT to MaxSAT? Is it possible to reduce MaxUNSAT to MaxSAT in a polynomial way ?
When considering the MaxSAT problem, one often considers also the MinUNSAT problem, which is almost the same. And for a propositional formula f in CNF it holds:
|f| = MaxSAT(f) + MinUNSAT(f)

where |f| is the number of clauses of f.
When considering MaxUNSAT and the corresponding MinSAT problem, the same relationship holds:
|f| = MaxUNSAT(f) + MinSAT(f)

Now, I was wondering if there is also a relationship between those two pairs, e.g. to reduce MaxSAT to MaxUNSAT or MinSAT (or the other way round) ?
Unfortunately, I could not figure out one by myself. And maybe there is none ?
Update 1: Inspired by Yuval Filmus's answer, I will give a reduction for my question.
Reduction from MaxUNSAT to its corresponding decision problem:
Let $\phi = {C_1, ..., C_m}$ a set of clauses over the variables $x_1, ..., x_n$, then it holds:
$$MaxUNSAT(\phi) = BinarySearch(0, |\phi|, MaxUNSAT(\phi, k) )$$
with

BinarySearch(start, end, CompareProcedure( )):
Searches for the element e between start and end so that CompareProcedure(e)=true and CompareProcedures(e+1)=false

and
$$MaxUNSAT(\phi, k) := \exists v\in\{0, 1\}^n:\sum_{i=1}^m 1 - I_v(C_i) \geq k$$
where $I_v$ is the interpretation of a propositional formula under assignment $v$.
Reduction from decision problem $MaxUNSAT(\phi, k)$ to SAT:
One can reduce the devision problem $MaxUNSAT(\phi, k)$ to the SAT problem by adding blocking variables to each clause and adding a cardinality constraint as propositional formula to limit the number of used clauses with help of the blocking variables.
I can describe this in more detail, if needed.
Conculsion:
One can reduce the MaxUNSAT problem to the SAT problem and then solve the SAT problem with the MaxSAT problem. This is a reduction that works in polynomial time.
Both MaxSAT and MaxUNSAT are NP-complete, and so can be reduced to one another. MaxSAT is NP-hard by (trivial) reduction from SAT. To see that MaxUNSAT is NP-hard, we follow Creignou [1] and reduce from MaxCUT. Given an instance $G=(V,E)$ of MaxCUT, consider the instance
$$ \bigwedge_{(x,y) \in E} (x \lor \lnot y) \land (\lnot x \lor y). $$
Each cut corresponds to a $0/1$ assignment to the vertices. If an edge $(x,y)$ is separated by a cut, then one of the two corresponding clauses will be unsatisfied, and otherwise none will be unsatisfied.

[1] Creignou, A dichotomy theorem for maximum generalized satisfiability problems, JCSS 51, 511-522 (1995).