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?


  • $\begingroup$ Using Parametric Coordinate of Ellipse, Put $x=\cos \phi$ and $y=\frac{1}{2}\sin \phi.$ $\endgroup$ – juantheron Jan 12 '16 at 15:27
  • $\begingroup$ 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 $4xy$. $\endgroup$ – πr8 Jan 12 '16 at 15:28
  • $\begingroup$ @πr8 how ? can you be more explicit $\endgroup$ – Taylor Ted Jan 12 '16 at 15:30
  • $\begingroup$ 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. $\endgroup$ – πr8 Jan 12 '16 at 15:32
  • $\begingroup$ @πr8 And why will it be xy? $\endgroup$ – Taylor Ted Jan 12 '16 at 15:34

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:




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


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.





For maxima,




Corresponding to this,








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