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-   -   Limits? (https://www.askmehelpdesk.com/showthread.php?t=531164)

  • Dec 2, 2010, 10:53 PM
    honeybear101
    Limits?
    I don't know if I got this correct or not. Find lim e^2x/x. x --> +infinity. The answer I got for this was positive infinity. Was that correct?
    I also have 2 more question that I can't seem to solve.
    (1) What is the area of the largest rectangle under the graph y = 3-x^2 and above the x-axis?
    (2) The total area of the region between the curve y=x^3 and the x-axis between x=-1 and x= 2 is?
  • Dec 3, 2010, 12:39 AM
    Unknown008

    Yes, you're right for the first one :)

    1.

    Let the height of the rectangle be y and the width be x.

    The area of the rectangle is then A = xy

    But you know that y = 3 - x^2, so you get:

    A = x(3 - x^2)

    This is the area of the rectangle in terms of x. Expand and simplify. You can find the derivative of this expression to get the value of x at the maximum area, or complete the square and find the optimum point.

    2. This is simply:



    I put minus because you know the graph is negative for x < 0, and finding the area from -1 to 0 will give a negative area.

    Post what you get! :)
  • Dec 3, 2010, 01:11 AM
    honeybear101
    Comment on Unknown008's post
    For number 2, I got 9 as my answer. As for number 1, I'm not sure if this is correct, but I got 4 as my answer. >_<
  • Dec 3, 2010, 01:26 AM
    Unknown008

    1. Wait, I realised I made a mistake in my explanation. The width is 2x, not x (though that doesn't really change the answer). This becomes:





    At optimum, A' = 0.









    A = 2(1)(2) = 4

    Right!

    2.



  • Dec 3, 2010, 08:09 AM
    honeybear101
    Comment on Unknown008's post
    Oh! That's how you do it. Thank you so much for helping me! :)
  • Dec 3, 2010, 08:16 AM
    Unknown008

    No problem! :)

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