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
 A: 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 $1$.
A: 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. 
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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
