Top right corner on the ellipse (centered on origin): $x = a \cos \theta$ and $y=b\sin\theta$
where $0\le \theta \le \frac{\pi}{2}, a=\frac{width}{2}$ and $b = \frac{height}{2}$
So area:$A(\theta) = 4xy=4ab\sin\theta\cos\theta = 2ab\sin2\theta$ Maximum of sin is at $2\theta=\frac{\pi}{2}$. So, we have $\theta = \frac{\pi}{4}$. Therefore the 4 corners are: $(a\cos \frac{\pi}{4},b\sin \frac{\pi}{4}), (a\cos \frac{3\pi}{4},b\sin \frac{3\pi}{4}), (a\cos \frac{5\pi}{4},b\sin \frac{5\pi}{4}), (a\cos \frac{7\pi}{4},b\sin \frac{7\pi}{4}),$
which are of course equal to:
$(\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}), (-\frac{a\sqrt{2}}{2},\frac{b\sqrt{2}}{2}), (-\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}), (\frac{a\sqrt{2}}{2},-\frac{b\sqrt{2}}{2}),$
And it's area is: $a\sqrt{2}\cdot b\sqrt{2}=2ab$