# Maximum area of a rectangle whose vertices lie on ellipse $x^2+4y^2=1$

Maximum area of a rectangle whose vertices lie on ellipse $x^2+4y^2=1$.

I try to do it by lagrange multiplier as

$F(x,y,t)= xy + t(x^2+4y^2-1=0)=0$. Differentiating w.r.t to x,y and solving i get $x=\frac{1}{\sqrt2}$ and $y=\frac{1}{2\sqrt2}$. So area=0.25. But textbook states answer =1. I like to know where i am wrong?

Thanks

• Using Parametric Coordinate of Ellipse, Put $x=\cos \phi$ and $y=\frac{1}{2}\sin \phi.$ – juantheron Jan 12 '16 at 15:27
• Your area will be $xy$ if you put your corner at the origin - if you extend your rectangle to also have a corner at the opposite side of the ellipse, your area will be $4xy$. – πr8 Jan 12 '16 at 15:28
• @πr8 how ? can you be more explicit – Taylor Ted Jan 12 '16 at 15:30
• Well, if your upper-right corner of the rectangle is at $(x,y)$, then you should have your other 3 corners at $(x,-y), (-x,y), (-x,-y)$. So your rectangle will be $2x$ wide, $2y$ tall, and have area $4xy$, all while lying inside the given ellipse. – πr8 Jan 12 '16 at 15:32
• @πr8 And why will it be xy? – Taylor Ted Jan 12 '16 at 15:34

## 2 Answers

The area of the rectangle is not $xy$. If the rectangle has a vertex at some point $(x,y)$, then the area will be $4xy$. Hopefully the crude drawing below will help you understand why. Note that, if you consider the rectangle whose bottom-left vertex is at the origin, then the sides have length $x$ and $y$, so the area would indeed be $xy$. This is the problem you solved, which gives $A=1/4$. Since this is a quarter of the total rectangle in question, we can multiply by $4$ to see that the answer is in fact $1$.

By symmetry, the rectangle with the largest area will be one with its sides parallel to the ellipse's axes. Consider any point B(x1,y1)B(x1,y1) on the ellipse located in the first quadrant.

​ You can easily see that A≡(−x1,y1)A≡(−x1,y1), D≡(x1,−y1)D≡(x1,−y1), and C≡(−x1,−y1)C≡(−x1,−y1).

So, Area=4x1y1Area=4x1y1

We also have the relation:

x21+4y21=1x12+4y12=1

⇒x21=1−4y21⇒x12=1−4y12

⇒x1=1−4y21−−−−−−√⇒x1=1−4y12

We've taken the positive value since we chose this point to be in the first quadrant.

Area=4y11−4y21−−−−−−√Area=4y11−4y12

The possible values of y1y1 for which there lies a point on the ellipse are [0,12][0,12] in the first quadrant. Let's differentiate the area to find its point of maxima.

dAdy=yddy1−4y2−−−−−−√+1−4y2−−−−−−√ddyydAdy=yddy1−4y2+1−4y2ddyy

dAdy=y121−4y2√(−8y)+1−4y2−−−−−−√dAdy=y121−4y2(−8y)+1−4y2

dAdy=−8y2+2−8y221−4y2√dAdy=−8y2+2−8y221−4y2

dAdy=2−16y21−4y2√dAdy=2−16y21−4y2

For maxima,

dAdy=0dAdy=0

⇒2−16y21−4y2√=0⇒2−16y21−4y2=0

⇒y=18√⇒y=18

Corresponding to this,

x=1−4y2−−−−−−√x=1−4y2

⇒x=1−12−−−−−√⇒x=1−12

⇒x=12√⇒x=12

Thus,

Areamax=412√18√Areamax=41218

Areamax=1sq.units