Prove that the maximum area of a rectangle inscribed in an ellipse is $2ab$ Prove that the maximum area of a rectangle inscribed in an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is $2ab$.

My attempt:
Equation of ellipse: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.
Assume that $a&gt;b$ &amp; let $l$ &amp; $w$ are length and width of rectangle then area will be $$A=l\times w \tag{1}$$
Now, I substituted the point $\left(\dfrac {l}{2}, \dfrac {w}{2}\right)$ in the equation of ellipse
$$\frac{(l/2)^2}{a^2}+\frac{(w/2)^2}{b^2}=1,$$
$$b^2l^2+a^2w^2=4a^2b^2.\tag{2}$$
I am not sure how to proceed from here.
Consider the coordinate transformation $$(u,v) = (x/a, y/b),$$ which maps the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ to the unit circle $$u^2 + v^2 = 1.$$  Such a transformation preserves the ratios of areas; in particular, if a quadrilateral inscribed in the ellipse has maximal area among all such quadrilaterals, then the transformed quadrilateral under this mapping will also have maximal area; and vice versa.  This is not difficult to show.
Therefore, it suffices to consider all quadrilaterals inscribed in the unit circle in the transformed coordinate system.  Since we know that the quadrilateral of maximal area in a circle is a square (proof below), it follows that the family of quadrilaterals of maximal area inscribed in the ellipse are those parallelograms that arise from the inverse transformation.  These squares all have diagonal with length $2$, thus area of $2$, and upon the inverse transformation, the corresponding parallelograms all have area $2ab$.  Of these, there is only one that is a rectangle, corresponding to the configuration where the sides are parallel to the coordinate axes; although merely observing existence is sufficient (uniqueness is irrelevant).

For the proof that a quadrilateral of maximal area inscribed in a circle is a square, let $ABCD$ be any quadrilateral inscribed in the circle.  Consider a diagonal, say $AC$.  Then $\triangle ABC$ is an inscribed triangle whose area is maximized when the height $h_B$ perpendicular to $AC$ is maximized, since $AC$ is fixed.  This corresponds to choosing the midpoint of $AC$ as the foot of the perpendicular.  Similarly, the area of $\triangle ADC$ is maximized in the same manner.  But this means $BD$ must be a diameter, since the feet of the perpendiculars coincide, both being the midpoint of $AC$, and the perpendicular to any chord through its midpoint is a diameter.  Thus among all quadrilaterals with fixed diagonal $AC$, the one that maximizes the area is one where the other diagonal $BD$ is a diameter.  Then among all such quadrilaterals with $BD$ as diameter, the one that maximizes the area is the one where $AC$ is also a diameter.  Since $AC \perp BD$, it follows $ABCD$ is a square.
Let one vertex of the rectangle be $(a\cos t, b\sin t)$. Then, the other three are known as well and the area is
$$A= 4ab|\sin t \cos t |\le 2ab (\cos^2t+\sin^2t)=2ab$$
where the inequality $2uv\le u^2+v^2 $ is used.