Question

Consider an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.What is the area of the greatest rectangle that can be inscribed in the ellipse ?

Answer 1

Suppose that the upper righthand corner of the rectangle is at the point ⟨x,y⟩⟨x,y⟩.

Then you know that the area of the rectangle is, as you say, 4xy4xy,

and you know that

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ....... (1)

Thinking of the area as a function of xx, we have

$\frac{dA}{dx}=4x\frac{dy}{dx}+4y$

Differentiating (1)(1) with respect to xx, we have

$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{b^2}{a^2}\frac{x}{y}\therefore \frac{dA}{dx}=4y-\frac{4b^2{x}^2}{a^{2}y}$

Setting this to 00 and simplifying, we have

$y^2=\frac{b^2{x}^2}{a^2}$

From (1)(1), we know that

$y^2=b^2-\frac{b^2{x}^2}{a^2}$

thus,

$y^2=b^2-y^2{2y}^2=b^2\frac{y^2}{b^2}=\frac{1}{2}$

Then $\frac{x^2}{a^2}=\frac{1}{2}$

and area is maximised when

$x=\frac{a}{\sqrt{2}}$

$y=\frac{b}{\sqrt{2}}$

So, area of greatest rectangle $=4xy$

Area $=4\times \frac{a}{\sqrt{2}}\times \frac{b}{\sqrt{2}}$

$=2ab$