Results 1 to 2 of 2

Math Help - Max/Min Problem...Having trouble with it?

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    18

    Max/Min Problem...Having trouble with it?

    I have absolutely no idea how to approach this question...

    Draw a diagram showing the region enclosed between the parabola y^2=4ax and its latus rectum x=a

    The, find the dimensions of the rectangle of max area that can be inscribed in this region.

    I drew the diagram...but so far all i have for the area equation is A=2yx (y is doubled because the parabola extends below the axis as well XD)

    Im having trouble uploading a picture, so ill describe the diagram...its a sideways parabola extending on the right side of the y axis, and the latus rectum cuts the x axis at "a" and the parabola when y= 2a and -2a

    Basically, i have no idea what to do...could anyone help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    177
    The length along the x-axis is going to be a-x for some x. The length along the y-axis is 2\sqrt{4ax}=4\sqrt{ax}.

    That gives us the area A(x)=4\sqrt{ax}(a-x) to maximize.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Very simple problem I'm having trouble with
    Posted in the Statistics Forum
    Replies: 2
    Last Post: November 13th 2010, 04:26 PM
  2. Really having trouble with this chemistry problem
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: October 12th 2009, 12:24 AM
  3. Having A little Trouble with this problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2008, 06:23 PM
  4. trouble with problem
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: June 27th 2008, 07:51 AM
  5. sry for the trouble! another trigo problem!!
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 18th 2008, 10:20 AM

Search Tags


/mathhelpforum @mathhelpforum