A geometry problem by Nikkil V

Let $\alpha=\angle ABE$ and $\beta = \angle CBF$ , we want to maximize $AB\times BC = 31\cos \alpha \times 40\cos \beta = 1240 \cos \alpha \cos \beta$ .

We are given that $\sin \theta = \frac{1}{5} \implies \cos (\alpha + \beta) = \frac{1}{5}$ . We will perform a lagrange multiplier since we have one constraint and need to maximize one expression.

$f(x,y) = \cos \alpha \sin \beta - \lambda (\cos (\alpha + \beta) - \frac{1}{5})$
$f(x,y)_x = -\sin \alpha \cos \beta + \lambda \cos (\alpha + \beta)$
$f(x,y)_y = -\cos \alpha \sin \beta + \lambda \cos (\alpha + \beta)$

We need $f(x,y)_x = f(x,y)_y = 0 \implies \tan \alpha = \tan \beta$ . Since $\alpha, \beta < 90^\circ$ , we must have $\alpha = \beta$ . Substitute the result,

$\cos \alpha \cos \beta = \cos ^2 \alpha = \frac{\cos 2\alpha + 1} {2} = \frac{3}{5}$ .

Finally, the answer is $1240 \times \frac{3}{5} = 744$ .