Problem:
Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$. This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$. Find the maximum possible area of $AQRS$.
Solution:
Diagram - https://www.geogebra.org/calculator/ws78ucvv.
Let $\angle BAD = \theta$, and $\angle BAC = \alpha$.
Note that $AD = 40 \cos \theta$ and $AF = 31 \sin (\alpha + \theta)$. The area of the rectangle is then$$(40 \cdot 31)(\cos \theta \cdot \sin (\alpha + \theta))$$$$=(40 \cdot 31)\left(\frac{1}{2}\right)(\sin(\alpha + 2\theta) + \sin(\alpha))$$$$=(20 \cdot 31)\left(\sin(\alpha + 2\theta) + \frac{1}{5}\right)$$The maximum value of $\sin(\alpha + 2\theta)$ is $1$, so the maximum value of the area is $(20 \cdot 31)(1 + 1/5) = \boxed{744}$.
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