Find max + min $P$

With $x^{2} + y^{2} = 8$ we can let $x = 2\sqrt{2}\cos(\theta)$ and $y = 2\sqrt{2}\sin(\theta)$ , in which case

$P = 4xy - 3 = 4 \times 2\sqrt{2}\cos(\theta) \times 2\sqrt{2}\sin(\theta) - 3 = 32\cos(\theta)\sin(\theta) - 3 = 16\sin(2\theta) - 3$ ,

since $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ . The maximum of $P$ is then $16 - 3 = 13$ when $2\theta = \dfrac{\pi}{2} + 2n\pi$ and the minimum is $-16 - 3 = -19$ when $2\theta = \dfrac{3\pi}{2} + 2n\pi$ .

The desired sum of max + min is then $13 + (-19) = \boxed{-6}$ .

The constraint suggests circles and trigonometry. Put $x=\sqrt8 \cos{\theta}$ , $y=\sqrt8 \sin{\theta}$ . All such $x$ and $y$ satisfy the constraint, and all $(x,y)$ pairs satisfying the constraint can be written in this form.

We then have $P=16\sin{2\theta}-3$ , which clearly ranges from $-19$ to $13$ , with sum $\boxed{-6}$ .