Calculus BC: Rectangle inside an ellipse

[yeahyeah<3]

1. The problem statement, all variables and given/known data
What is thea are of the largest rectangle that can be inscribed in the ellipse 4x^2 +9y^2 = 36
A) 6 rad 2 B) 12 C)24 D) 24 rad 2 E) 36


2. Relevant equations
Must be done using optimization and first derivitive


3. The attempt at a solution
I know I have to use A= 2x2y
But i don't know how to remove one of the variables ..

[Tom Mattson]

I know I have to use A= 2x2y


How about writing it as 4xy?


But i don't know how to remove one of the variables ..

Well, what equation do x and y satisfy if the corners of the rectangle are all points on the ellipse?

[yeahyeah<3]

How about writing it as 4xy?



Well, what equation do x and y satisfy if the corners of the rectangle are all points on the ellipse?

x = radical ((36-9y^2)/4)
but i'm not sure how that helps me.
even when i plug that back in and take the derivitve, my answer isn't one of the multiple choice answers.

thank you for the fast reply though :)

[Tom Mattson]

x = radical ((36-9y^2)/4)
but i'm not sure how that helps me.


It helps you because it eliminates one of the variables.


even when i plug that back in and take the derivitve, my answer isn't one of the multiple choice answers.


Taking the derivative is only one step.

Please post what you've done so far. That's the best way to see if you're making a mistake.

[yeahyeah<3]

It helps you because it eliminates one of the variables.



Taking the derivative is only one step.

Please post what you've done so far. That's the best way to see if you're making a mistake.
A= 4xy
A= 4 y (radical ((36-9y^2)/4))
A'(y) = 6 radical 16-x^2 - (6x^2/radical 16-x^2) = 0
when y = 2Rad2

plug that back in : x = (radical ((36-9y^2)/4)) = 3rad2

4(2rad2)(3rad2) = 48

=/

[Tom Mattson]

A= 4xy
A= 4 y (radical ((36-9y^2)/4))


OK


A'(y) = 6 radical 16-x^2 - (6x^2/radical 16-x^2) = 0
when y = 2Rad2


Where does \sqrt{16-x^2} come from?

[yeahyeah<3]

OK



Where does \sqrt{16-x^2} come from?

the derivative of A(x)?
it might be rad 4-x^2 instead... but regardless my answer still isn't there.

[Tom Mattson]

the derivative of A(x)?
it might be rad 4-x^2 instead...


Don't you think it makes a difference?? :eek:


but regardless my answer still isn't there.

Yes it is. You need to correct that error and proceed from there. Please try it and if you get stuck, post your work. That's the only way I can spot a mistake.

[yeahyeah<3]

Tom, can you just please tell me if the answer is 12.

I've done this problem at least 7 times, possibly 8.
I have quite a few pages of work (with crossouts of course)

I got y to be rad 2 and x to be rad 9/2

so A = 4 xy
A= 12?

[Tom Mattson]

12 is what I got.

[yeahyeah<3]

yay, thank you so much =]

would you like to give me a hint on another problem?

If f is the continuous strictly increasing function on the interval a is less than or equal to x and x is less than or equal to b, which of the following must be true.

I. integral from a to b f(x) dx < f(b) (b-a)
II. integral from a to b f(x) dx > f(a) (b-a)
III. integral from a to b f(x) dx = f(c) (b-a) for some number of c such that a<c<b

I have no idea where to start and any hint would be wonderful.

[Tom Mattson]

This problem is meant to direct your attention to the Mean Value Theorem for integrals.