The basic fact is that one 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. $A\wedge B+B\wedge C+C\wedge D+D\wedge A$) so the area can be negative or zero in non-trivial ways (for 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 on $F$ is the positivity of  an explicit  sextic polynomial whose coefficients are 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 can be solved by radicals.  

Then 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, of course, the maximum.

Caveat.  Of the three related topics—triangles, tetrahedra and quadrilaterals—the latter displays some subtleties not present in the other two due to non-rigidity.  Thus one loses uniqueness but in 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 fact, which is relevant here) is that there are singular cases which have to be dealt with separately sine expressions which appear in the denominator vanish there.  One important class (kites, in particular, rhomba, parallelograms) can be subsumed under the pythagorean quadrilaterals (sum 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.