The largest rectangle inscribed in an ellipse

Symmetry. Let the vertices be $(x_1,y_1),(x_2,y_1),(x_1,y_2),(x_2,y_2)$ on the ellipse, with $x_1\neq x_2$ and $y_1\neq y_2$. From the first two vertices, $\frac{x_1^2}{a^2}=1-\frac{y_1^2}{b^2}=\frac{x_2^2}{a^2}$, so $x_2=-x_1$; likewise $y_2=-y_1$. So the rectangle is centred at the origin, and it is determined by its vertex in the first quadrant, $(a\cos\theta,b\sin\theta)$.

Area. The rectangle has sides $2a\cos\theta$ and $2b\sin\theta$:
$$A(\theta)=4ab\sin\theta\cos\theta=2ab\sin2\theta.$$

Maximum. $A'(\theta)=4ab\cos2\theta=0$ on $\left(0,\frac\pi2\right)$ only at $\theta=\frac\pi4$, where $A$ goes from increasing to decreasing. So
$$A_{\max}=2ab,\qquad\text{sides }a\sqrt2\text{ and }b\sqrt2.$$

Remark. The ellipse has area $\pi ab$: the best rectangle fills $\frac2\pi\approx64\%$ of it, whatever the ellipse. This is no coincidence: an ellipse is a stretched circle, and stretching multiplies all areas by the same factor $\frac ba$. For a circle ($a=b=r$), the best rectangle is a square of area $2r^2$.