Answer: 744

Small Hint:

With $\beta = \angle BAQ$ and $\gamma = \angle CAS,$ the area is $40\cos\beta \cdot 31\cos\gamma,$ and $\beta + \gamma = 90^\circ - A$

Big Hint:

Product-to-sum: $2\cos\beta\cos\gamma = \cos(\beta - \gamma)$ $+ \sin A,$ maximized when $\beta = \gamma$

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 $$\begin{aligned} 40 \cdot 31 \cos\beta\cos\gamma \\ &\tiny = 620\bigl(\cos(\beta - \gamma) + \cos(\beta + \gamma)\bigr) \\ &\tiny = 620\bigl(\cos(\beta - \gamma) + \sin A\bigr), \end{aligned}$$ 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 $620\left(1 + \frac{1}{5}\right) = 744.$