The basic fact is that one can express the required extreme values of the areas in the generic case (our use of this term is explained below). The simple formulae involved can be found below.
The decisive fact is that for any value $F$ which satisfies conditions to be described below there are precisely two quadrilaterals $ABCD$ which have area $F$ and side lengths $a,b,c,d$. To be precise, I am using directed areas (i.e. $A\wedge B+B\wedge C+C\wedge D+D\wedge A$) and so the area can be negative or zero in non-trivial ways (example —the quadrilateral with vertices $(0,0),(1,0),(0,1),(1,1)$). The side lengths are $|AB|=a$ and so on cyclically. The required max and min are the largest and smallest values for $|F|$ which satisfy the given condition.
The condition which $F$ must fulfill is the positivity of an explicit sextic polynomial withcoefficients functions of the side lengths. Hence the required optimal values are roots of this polynomials. The polynomial is, in fact, a cubic in $F^2$ and so can be solved by radicals.
The computations are rather intricate (I used Mathematica) but the result is quite simple:the squares of the two areas are $$\frac 1 {16}(-a^4+2 a^2b^2-b^4+2a^2c^2 +2b^2c^2+2 c^2d^2-c^4\pm 8abcd+2a^2d^2+2b^2 d^2-d^4).$$
Of course, it can happen that one of these expressions is negative. This means is that there is no quadrilateral with the assigned side lengths.
The plus sign gives the maximum.
Caveat. Of the three related topics—triangles, tetrahedra and quadrilaterals—from elementary geometry, the latter displays some subtleties not present in the other two. This is due to non-rigidity. For example, one can lose uniqueness. This is often in a relatively weak sense—instead of one solution, there are two (typically one convex, the other non-convex—example, Brahmagupta). This has been taken care of here. The second potential problem, which is relevant here, is that there are singular cases which have to be dealt with separately This can arise when terms in the denominator vanish. One important class—kites (in particular, rhomba) and parallelograms— can be subsumed under the pythagorean quadrilaterals (the sums of the squares of the lengths of two opposite pairs of adjacent sides coincide). I have also computed this case but will spare the reader the details.
Explanatory remarks about the relation to Brahmagupta’s formula for the area of a cyclic quadrilateral: There are, in fact, two such formulae, one for the convex case, one for the non-convex one (this distinction is sometimes not explicitly formulated in the literature). They correspond exactly to the above two expressions for the sup and inf. Since the validity of either of these expressions is equivalent to the quadrilateral being cyclic, this shows that (despite comments here) both the sup and inf are attained and, indeed, cyclic quadrilaterals.
The following is off topic but I would like to add it since I feel that it makes what is going on here more transparent. One can prove the Brahmagupta formulae with a few calculations (which can be done by hand—I checked them with Mathematica to be sure). The idea is very simple— wlog one can assume that $A_1=(\cos (\theta_1),\sin(\theta_1))$, etc., and then compute the corresponding side lengths and the (square of) the area. Showing that the corresponding differences vanish then reduces to the kind of manipulation of trigonometric identities that we all learned in high school. The difference between the convex and non-convex cases arises through the occurrence of square roots of expressions of the form $1-\cos(\theta_2-\theta_1)$ and so on. This is a sine expression, with choice of sign depending on the orientation of $A_1$ and $A_2$ as one traverses the circle. In the convex case one can assume that this is always $+$, otherwise that there is one $-$. This explains the dichotomy between the convex and the non-convex case.