Question

What is the area of the largest rectangle that can be inscribed in the ellipse 4x^2+9y^2=36?

Answer 1

have you made any progress yet?

Answer 2

All I know is A=4xy and x=sqrt(36-9y^2/4)
I don't know what to do now.

Answer 3

where do you get A=4xy from?

Answer 4

oh I see

Answer 5

right, so substitute for x into A

Answer 6

if you want the hard way, that is...

Answer 7

Well, if looks like what you can do is let \(A = f(y)\). Then maximize \(f(y)\).

Answer 8

By finding \(dA/dy = 0\).

Answer 9

So I don't use A=4xy?

Answer 10

You use it, you substitute x with the other equation for x. This makes it entirely a function of y.

Answer 11

Ok so A=4(sqrt(36-9y^2/4))(y)

Answer 12

I'm not sure what to do now.

Answer 13

So you've written your area function in terms of one variable so far?
Ok good.

So you've successfully applied your `constraint` (the ellipse) to your area function.
Now we want to maximize the area function with respect to y.

Remember how to find max/min?

Answer 14

Nope. That's my problem. I don't remember how to do these kinds of problems. :/

Answer 15

To find max/min of our area function we'll need to first find `critical points` of the function. Then we'll be able to do some quick tests to classify those points.

Mmm before we do that though, I think maybe your x is wrong. Lemme check.

Answer 16

\[\Large\bf\sf 4x^2+9y^2=36\qquad\to\qquad x=\sqrt{\color{red}{9}-\frac{9}{4}y^2}\]I think you forgot to divide the 36 by 4 as well, yes? :o

Oh oh maybe you didn't. Maybe it's the way you wrote it.
Use brackets to be clear XD

A=4(sqrt([36-9y^2]/4))y

Mmm so many brackets -_-

Answer 17

So we want to take the derivative of our function, then set that derivative equal to zero.
Solving for those y's will give us critical points of our y variable.

Answer 18

So ummm.... this one is a little messy.
We'll need to apply the product rule.

Answer 19

\[\Large\bf\sf A=4y\sqrt{9-\frac{9}{4}y^2}\]
\[\Large\bf\sf A'\quad=\quad \color{royalblue}{(4y)'}\sqrt{9-\frac{9}{4}y^2}+4y\color{royalblue}{\left(\sqrt{9-\frac{9}{4}y^2}\right)'}\]

Answer 20

We need to take the derivatives of these blue parts.
This process looks familiar I hope? :o

Answer 21

Yes

Answer 22

would that be \[(4y')(\sqrt{9-(9/4)y^2})+(4y)(((18/4)y')/2\sqrt{9-(9/4)y^2}) \]

Answer 23

\[\Large\bf\sf A'\quad=\quad \color{royalblue}{(4y)'}\sqrt{9-\frac{9}{4}y^2}+4y\color{royalblue}{\left(\sqrt{9-\frac{9}{4}y^2}\right)'}\]

\[\Large\bf\sf A'\quad=\quad \color{orangered}{(4)}\sqrt{9-\frac{9}{4}y^2}+4y\color{orangered}{\left(\frac{-\dfrac{18}{4}y}{2\sqrt{9-\frac{9}{4}y^2}}\right)}\]Umm close, yes.

Answer 24

We're taking our derivative `with respect to y`, so we shouldn't end up with any `y'` terms.

Answer 25

That y in the first part will turn into a 1.
The y^2 in the second part will turn into a y, right?

Answer 26

Yes. I understand.

Answer 27

Then we set our first derivative equal to zero, and solve for y.

Answer 28

Am I on the right track?
\[8\sqrt{9-(9/4)y^2}+18y^2=0\]

Answer 29

I think your second term is -18y^2, yes?

Answer 30

We picked up a negative from the derivative of the square root part.

Answer 31

Oh ok

Answer 32

should I do \[8\sqrt{9-(9/4)y^2}/18=y^2\]

Answer 33

Oh and also umm.. If you multiplied through by that denominator, shouldn't your first term become, \(\Large\bf\sf 8(9-(9/4)y^2)\) ? Without the root?

Answer 34

Ohh right..

Answer 35

After you distribute the 8, it should work out a little nicer.

Answer 36

Does y= sqrt(2) ?

Answer 37

Hmm that looks right.
I got something different when I did it.

Lemme check again real quick D:

Answer 38

Ya sqrt2 looks good!
We should get two solutions since y is squared.

We actually get \(\Large\bf\sf y=\pm sqrt2\)
But -sqrt2 doesn't make sense for a rectangles dimensions so we can ignore that value.

Answer 39

oh ok. now do I plug in y to x=sqrt(9-(9/4)y^2)?

Answer 40

Ya that's a good idea. We've found the y coordinate which maximizes area.
That will give us the corresponding x value.

Answer 41

so x= sqrt(9/2)?

Answer 42

Ok good. so now how do you find that maximized area given x and y?

Answer 43

plug it into 4xy?

Answer 44

yes

Answer 45

so the answer is 12?

Answer 46

Mmm yes that seems right!
Good job! \c:/

Answer 47

Thank you so much!!