The area, call it $K$, is indeed maximized when the quadrilateral is cyclic, in which case $K$ is given by Brahmagupta's remarkable generalization of Heron's formula: $K = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where as in Heron $s = (a+b+c+d)/2$ is the semiperimeter. This is a known consequence of Bretschneider's formula $$ K^2 = (s-a)(s-b)(s-c)(s-d) - abcd \cos^2 \theta $$ where $2\theta$ the sum of two opposite angles, so $\cos^2 \theta = 0$ iff opposite angles add up to $\pi$, which is a classical condition for a quadrilateral to be cyclic. (There are two pairs of opposite angles, but the two choices yield values of $\theta$ that sum to $\pi$ and thus have the same $\cos^2 \theta$.)

For proofs and references see for example the Wikipedia pages for the Brahmagupta and Bretschneider formulas.

The area, call it $K$, is indeed maximized when the quadrilateral is cyclic, in which case $K$ is given by Brahmagupta's remarkable generalization of Heron's formula: $K = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where as in Heron $s = (a+b+c+d)/2$ is the semiperimeter. This is a known consequence of Bretschneider's formula $$ K^2 = (s-a)(s-b)(s-c)(s-d) - abcd \cos^2 \theta $$ where $2\theta$ is the sum of two opposite angles, so $\cos^2 \theta = 0$ iff opposite angles add up to $\pi$, which is a classical condition for a quadrilateral to be cyclic. (There are two pairs of opposite angles, but the two choices yield values of $\theta$ that sum to $\pi$ and thus have the same $\cos^2 \theta$.)

For proofs and references see for example the Wikipedia pages for the Brahmagupta and Bretschneider formulas.