Calculus BC: Rectangle inside an ellipse

In summary, The Mean Value Theorem states that for a continuous function f(x) on the interval [a,b], there exists a value c in [a,b] such that the average value of f(x) on the interval [a,b] is equal to f(c). So in this case, the average value of f(x) on [a,b] is given by the integral of f(x) over [a,b] divided by the length of the interval (b-a).Now, looking at the options given, we can see that II and III are simply restatements of the Mean Value Theorem for integrals. So we just need to focus on I. Can you think of a counterexample where the integral of f(x
  • #1
yeahyeah<3
27
0

Homework Statement


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


Homework Equations


Must be done using optimization and first derivitive


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 ..
 
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  • #2
yeahyeah<3 said:
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?
 
  • #3
Tom Mattson said:
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 :)
 
  • #4
yeahyeah<3 said:
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.
 
  • #5
Tom Mattson said:
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

=/
 
  • #6
yeahyeah<3 said:
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 [itex]\sqrt{16-x^2}[/itex] come from?
 
  • #7
Tom Mattson said:
OK



Where does [itex]\sqrt{16-x^2}[/itex] come from?

the derivative of A(x)?
it might be rad 4-x^2 instead... but regardless my answer still isn't there.
 
  • #8
yeahyeah<3 said:
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.
 
  • #9
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?
 
  • #10
12 is what I got.
 
  • #11
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.
 
  • #12
This problem is meant to direct your attention to the Mean Value Theorem for integrals.
 

Related to Calculus BC: Rectangle inside an ellipse

1. What is Calculus BC and how does it relate to rectangles inside an ellipse?

Calculus BC is an advanced level mathematics course that builds upon the concepts learned in Calculus AB. It covers topics such as derivatives, integrals, and applications of these concepts. The concept of rectangles inside an ellipse is related to the application of calculus in finding the area of a shape that is bounded by an ellipse.

2. How do you find the area of a rectangle inside an ellipse using calculus?

To find the area of a rectangle inside an ellipse using calculus, we need to use the concept of double integrals. We first set up a double integral that represents the area of the rectangle and then use the properties of ellipses to convert the integral into a form that can be solved using calculus.

3. What is the significance of finding the area of a rectangle inside an ellipse?

The area of a rectangle inside an ellipse can have various applications in real-world scenarios. For example, it can be used to calculate the area of a field that is bounded by an elliptical fence. It can also be used in engineering and architecture to find the area of a curved surface.

4. Can the concept of rectangles inside an ellipse be extended to other shapes?

Yes, the concept of using calculus to find the area of a shape bounded by an ellipse can be extended to other shapes as well. For example, it can be used to find the area of a rectangle inside a parabola or a hyperbola. However, the equations and methods used may differ depending on the shape.

5. Are there any real-world applications of the concept of rectangles inside an ellipse?

Yes, there are many real-world applications of this concept, especially in fields such as engineering, architecture, and physics. For example, it can be used to calculate the area of a curved surface in a building or to find the volume of a tank that has an elliptical base. It can also be used in physics to find the area enclosed by an elliptical orbit of a planet around the sun.

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