Maximal Rectangle

Suppose that the diagonals of the quadrilateral make an acute angle $\alpha$ with each other, so that $A = \tfrac12pq\sin\alpha$ .

Consider a rectangle that circumscribes the quadrilateral. If the diagonal of length $p$ makes an angle of $\theta$ with the side of the rectangle that it touches, then the rectangle must have height $p\sin\theta$ and width $q\cos(\theta-\alpha)$ . so that the rectangle has area $\Delta \; = \; pq\sin\theta \cos(\theta-\alpha) \; = \; \tfrac12pq\big[\sin(2\theta-\alpha) + \sin\alpha\big] \; = \; \tfrac12pq\sin(2\theta - \alpha) + A$ so that $\Delta \le \boxed{\tfrac12pq + A}$ , with the maximum achieved when $\theta = \tfrac12\alpha + \tfrac14\pi$ .