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 $(0,0),(1,0),(-1,1),(1,1)$). 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.