TITLE: Maximize area of a quadrilateral given three sides
QUESTION [8 upvotes]: What is the maximum possible area that a quadrilateral can have, if the lengths of three of its sides are given as 3, 4 and 5, while the fourth side can have arbitrary length? (Thinking of it as three fence segments arranged around a straight wall such that the area becomes maximal)
My approach: Since for given lengths a cyclic quadrilateral will have maximum area, I started out from here. According to Heron’s theorem the squared area of a cyclic quadrilateral equals $A^2 = (s-5)(s-4)(s-3)(s-2x)$ with $s = (5+4+3+2x)/2$ equaling half of the circumference and $x$ half of the unknown side. Rearrangement leads to $A^2=(3+x)(2+x)(1+x)(6-x)$. Setting the x-derivative to zero yields the maximum area $A = 20.495$ with $x = 4.0279$ (and the missing side $2x = 8.0558$). 
Is this correct? Is the maximal quadrilateral at all cyclic as I assumed?
And If yes, isn’t there a more elegant and intuitive geometric solution?

REPLY [3 votes]: Your assumption of being maximal happens when it is a cyclic quadrilateral seems right. Consider $ABCD$ your maximal quadrilateral. Let's say $AB=3$, $BC=4$, $CD=5$. If $\angle ACD\neq 90$, then you can rotate $CD$ around $C$ so that this angle $\angle ACD$ becomes $90$, and your area would increase, contradicting the maximality of your quadrilateral. So, $\angle ACD=90$. 
Similar analysis shows that $\angle ABD=90$. So, your maximal quadrilateral has to be cyclic. Even more, your other side $AD=2x$ has to be the diameter of the circumcircle. I think this should be enough info to actually find the value of $x$. And probably you'll get the same cubic formula to solve.