Optimize Peripheral Vision

$θ=α+β$

$tgθ=tg(α+β)=\frac{tgα+tgβ}{1-tgα{\cdot}tgβ}{\implies}tg(α+β)=\frac{\overline{AP}+\overline{PB}}{1-\overline{AP}{\cdot}\overline{PB}}=\frac{\frac{2}{\sqrt{3}}}{1-\overline{AP}{\cdot}\overline{PB}}$

By the AM GM inequality:

$\frac{a+b}{2}{\geq}\sqrt{ab}$

we know that, to maximize the product $\overline{AP}{\cdot}\overline{PB}$ , we must have $\overline{AP}=\overline{PB}$ .

Now it's easy to calculate $θ$ :

$tgθ=\frac{\frac{2}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}})^2}=\sqrt{3}{\implies}θ=60º$

Set $x = AP$ and $Q \in [CD]$ such as $(PQ)//(AC)$ .

Set $\theta _1 (x)= \angle CPQ = \arctan (x)$ and $\theta _2 (x)= \angle QPD = \arctan (\frac{2}{\sqrt3} - x)$

Then $\theta (x) = \theta _1 (x) + \theta _2 (x) = \arctan (x) + \arctan (\frac{2}{\sqrt3} - x)$

$\theta'(x)= \frac{2(\frac{1}{\sqrt3}-x)(\frac{2}{\sqrt3})}{(1+x^2)(1+(\frac{2}{\sqrt3}-x)^2)}$

$\theta$ is then maximum when $x = \frac{1}{\sqrt3}$

Finally, $\boxed{\theta_{max} = \theta (\frac{1}{\sqrt3}) = 2 \arctan(\frac{1}{\sqrt3}) = 60 ^\circ }$

PS : According to the figure, it is rectangle ABDC, not ABCD.