Solution:

Let $\beta = \angle BAQ$ and $\gamma = \angle CAS,$ so $\beta + \gamma = 90^\circ - A.$ From the right triangles $AQB$ and $ASC,$ the sides of the rectangle are $AQ = 40\cos\beta$ and $AS = 31\cos\gamma,$ so its area is $$40 \cdot 31 \cos\beta\cos\gamma = 620\bigl(\cos(\beta - \gamma) + \cos(\beta + \gamma)\bigr) = 620\bigl(\cos(\beta - \gamma) + \sin A\bigr),$$ using the product-to-sum identity and $\cos(90^\circ - A) = \sin A.$

This is maximized when $\beta = \gamma,$ which the constraint allows, giving area $620\left(1 + \frac{1}{5}\right) = 744.$