Results 1 to 3 of 3

Math Help - find extreeemes

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    sthlm
    Posts
    53

    find extreeemes

    f(x,y)=x+xy^2
    only def for x^2+y^2 \leq 4
    find the highest and lowest value for f.

    Attempt: https://dl.dropbox.com/u/586465/Phot...2 09 04 42.jpg
    I find (0,0) and (±√4 ; ±2/3*√6)

    The book says the answer is: Skärmavbild 2012-09-12 kl. 09.08.23.png
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Bangalore, India
    Posts
    142
    Thanks
    14

    Re: find extreeemes

    In your post g'(x) calculation is incorrect. In case you want to verify see the solution below.
    Spoiler:

    Quote Originally Posted by jacob93 View Post
    f(x,y)=x+xy^2
    only def for x^2+y^2 \leq 4
    find the highest and lowest value for f.
    Consider
    f(x,y)=x+xy^2 = x(1+y^2) f(x,y) \ge 0 x \ge 0 and f(x,y) < 0 x < 0 one conclusion that can be drawn is f(x,y) reaches its extremes under the restriction x^2+y^2 \leq 4 at precisely on the circle x^2+y^2 = 4. So substitute y^2 = 4 -x^2 in f(x,y). We have g(x) = x(5-x^2) Equating the derivative to zero we have g'(x) = 5- 3x^2 = 0 \implies x = \pm \sqrt{\frac{5}{3}}
    Last edited by kalyanram; September 12th 2012 at 12:27 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2012
    From
    sthlm
    Posts
    53

    Re: find extreeemes

    thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 04:57 PM
  2. Replies: 2
    Last Post: July 5th 2010, 08:48 PM
  3. Replies: 1
    Last Post: February 17th 2010, 03:58 PM
  4. Replies: 0
    Last Post: June 16th 2009, 12:43 PM
  5. Replies: 2
    Last Post: April 6th 2009, 08:57 PM

Search Tags


/mathhelpforum @mathhelpforum