# Finding Maximum Area of a Rectangle in an Ellipse [duplicate]

Question: A rectangle and an ellipse are both centred at $(0,0)$. The vertices of the rectangle are concurrent with the ellipse as shown

Prove that the maximum possible area of the rectangle occurs when the x coordinate of point $P$ is $x = \frac{a}{\sqrt{2}}$

What I have done

Let equation of ellipse be

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

Solving for y

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

Let area of a rectangle be $4xy$

$$A = 4xy$$

$$A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}})$$

$$A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right)$$

$$A'(x) = 4\sqrt{ b^2 - \frac{b^2x^2}{a^2}} + \frac{-8x^2b^2}{\sqrt{ b^2 - \frac{b^2x^2}{a^2}}a^2} = 0$$

$$4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 , \sqrt{ b^2 - \frac{b^2x^2}{a^2}a^2} \neq 0$$

$$4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0$$

$$4a^2b^2 - 4b^2x^2 - 8x^2b^2 = 0$$

$$4a^2b^2 - 12x^2b^2 = 0$$

$$12x^2b^2 = 4a^2b^2$$

$$x^2 = \frac{a^2}{3}$$

$$x = \frac{a}{\sqrt{3}} , x>0$$

Where did I go wrong?

edit:The duplicate question is the same but both posts have different approaches on how to solve it so I don't think it should be marked as a duplicate..

Your mistake is here $$A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) +\color{red}{\frac12} \times4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right).$$

• Ah yes! Thank you very much Olivier! – bigfocalchord Mar 6 '16 at 2:26
• @dydxx You are welcome. – Olivier Oloa Mar 6 '16 at 2:26
• Well, I answered just a minute earlier... but that doesn't matter anyway. :) – Win Vineeth Mar 6 '16 at 2:28

Its easier to solve this question using parametric points.

Let one vertex of the rectangle be $(a\cos\theta,b\sin\theta)$.

The other vertices are $(a\cos\theta,-b\sin\theta)$, $(-a\cos\theta,b\sin\theta)$, $(-a\cos\theta,-b\sin\theta)$

The area of rectangle formed is $$A(\theta)=4ab\cos\theta\sin\theta=2ab\sin2\theta$$

Maximum area is $2ab$ and it occurs when $\theta=\frac{\pi}{4}$ (or when $\sin2\theta$ is maximum).

When $\theta=\frac{\pi}{4}$, $x$-coordinate $=a\cos\frac{\pi}{4}=\frac{a}{\sqrt{2}}$

Let equation of ellipse be

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

Solving for y

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

Let area of a rectangle be $4xy$

$$A = 4xy$$

$$A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}})$$

$$A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-b^2x}{a^2} \right)$$

$$A'(x) = 4\sqrt{ b^2 - \frac{b^2x^2}{a^2}} + \frac{-4x^2b^2}{\sqrt{ b^2 - \frac{b^2x^2}{a^2}}a^2} = 0$$

$$4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 4x^2b^2 = 0 , \sqrt{ b^2 - \frac{b^2x^2}{a^2}a^2} \neq 0$$

$$4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 4x^2b^2 = 0$$

$$4a^2b^2 - 4b^2x^2 - 4x^2b^2 = 0$$

$$4a^2b^2 - 8x^2b^2 = 0$$

$$8x^2b^2 = 4a^2b^2$$

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

$$x = \frac{a}{\sqrt{2}} , x>0$$

The mistake is in third step while differentiating.

differentiating $\sqrt x$ will give you $\frac{1}{2\sqrt x}$

The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is a circle of radius $a$ in $(\hat x,y)$ coordinates, where $\hat x=\dfrac{a}{b}x$. This transformation multiplies areas by the constant $\dfrac{a}{b}$, so the problem is equivalent to finding the rectangle of maximum area in a circle, which is well-known to be a square.

Or, looked at another way (pun intended), this ellipse is what you see if you look at the circle of radius $a$ in the $x-y$ plane from just the right angle instead of from directly above. When you see what appears to be an inscribed rectangle in the ellipse of maximum area, what youâ€™re looking at is an inscribed rectangle in the circle of maximum area.

• I like this, because it means you can "see" the answer without calculation. Worth having a duplicate question for. – Calum Gilhooley Mar 6 '16 at 10:17