Largest rectangle problem
Unfortunately this problem is NP-hard, so no computer scientist knows (or is admitting that they know! :-P) a way to solve it in time polynomial in the number of sticks.  Don't feel bad about using brute-force here -- it's essentially the best you can do.

First, let's reframe the rectangle problem as a decision problem, that is, a problem to which the answer is either YES or NO:

Rectangle: Given a multiset $Y = (y_1, \dots, y_m)$ of stick lengths and an area threshold $k$, is it possible to build a rectangle from the sticks in $Y$ having area at least $k$?

Squares are optimal solutions when they exist

Before getting into the reduction, we will need a property about what a maximal solution to the Rectangle problem looks like.  Under the constraint that $a+b=c$, $ab$ is maximised precisely when $a=b$, i.e., squares have uniquely maximum area among all rectangles of fixed perimeter.  You can verify this by substituting $b=c-a$ into $ab$ and then differentiating w.r.t. $a$: the derivative becomes zero at $a=c/2=b$, and this is clearly a maximum since the coefficient of the $a^2$ term is negative.  The practical significance of this fact is: If and only if it is possible to partition the stick lengths into 4 equal-size subsets, then the maximum-size area achievable is $s^2/16$, where $s$ is the sum of all stick lengths.

Reduction from Partition

I'll show this is NP-hard by reducing the NP-hard problem Partition to it.  In Partition, we are given a multiset of $n$ integers $X=(x_1, \dots, x_n)$, and our task is to partition this multiset into two parts having equal sum.

Given an instance $X=(x_1, \dots, x_n)$ of Partition, with total sum $t=\sum x_i$, we construct an instance $Y = (y_1, \dots, y_{n+2})$ of Rectangle by taking each $x_i$ as a stick length $y_i$, and adding two more "big" sticks, $y_{n+1}$ and $y_{n+2}$, both of length $t/2$.  (That is, $m=n+2$.)  Finally, we set $k$ to $t^2/4$.

If the answer to the constructed Rectangle problem instance is YES, then by the maximal-square property, it must be possible to partition $Y$ into four equal-sum parts, each having sum $t/2$.  Let $(A, B, C, D)$ be such a partition.  Two of these four multisets must each contain a "big" stick and nothing else, since the big sticks have to go somewhere, and combining one of them with any other sticks would mean that one side exceeds $t/2$, meaning that some other side must fall short of $t/2$, meaning that the rectangle is not square and thus of size strictly less than $t^2/4$ -- a contradiction.  Remove these two big sticks, leaving two remaining multisets of sticks, which must also be of equal sum (that is, also $t/2$): these two multisets represent a valid partition of the original $X$ into two equal-sum parts, showing that the answer to the original Partition instance is also YES in this case.

In the other direction, if the answer to the original Partition problem is YES, meaning that there is a way to partition $X$ into two equal-sum parts $U$ and $V$, then clearly the answer to the constructed Rectangle problem instance is also YES: We could simply use the four multisets $U$, $V$, $\{y_{n+1}\}$ and $\{y_{n+2}\}$.

Since a YES to either problem instance implies a YES to the other, it must also be that a NO to either instance implies a NO to the other: in other words, the answers to both problems are always the same.  And since the Rectangle problem instance can be constructed in polynomial time from the Partition instance, any algorithm that can solve Rectangle in polynomial time could also be used as a subroutine to solve Partition (any, by extension, every other NP-hard problem) in polynomial time.  Since Partition is an NP-hard problem, this implies that Rectangle is too.