1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $1$.

$\endgroup$
-2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.