The sides of the rectangle of greatest area that can be inscribed in the ellipse x² + 4y² = 64 are

Ellipse is $\frac{{\mathrm{x}}^2}{64}+\frac{{\mathrm{y}}^2}{16}=1$

$$\begin{gathered} where a^2=64,b^2=16 \\ \Rightarrow a=8,b=4 \end{gathered}$$
sides of the greatest rectangle are $(\mathrm{a}\sqrt{2},\mathrm{b}\sqrt{2})=(8\sqrt{2},4\sqrt{2})$