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

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