Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Multivariable calculus problem (Global Max/Min)

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Multivariable calculus problem (Global Max/Min)

    Find the global maximum and minimum of f(x,y) = (x^2) -2xy + 2y on the domain x ≥ 0, y ≥ 0 and y ≤ 2-x

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Multivariable calculus problem (Global Max/Min)

    Quote Originally Posted by bagels0 View Post
    Find the global maximum and minimum of f(x,y) = (x^2) -2xy + 2y on the domain x ≥ 0, y ≥ 0 and y ≤ 2-x

    Thanks.
    Use the second partial derivatives test.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Re: Multivariable calculus problem (Global Max/Min)

    Thanks Alex,

    but could you please give a little more detail? I get very confused with the shading and all that...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Multivariable calculus problem (Global Max/Min)

    Quote Originally Posted by bagels0 View Post
    Thanks Alex,

    but could you please give a little more detail? I get very confused with the shading and all that...
    Where do f_x and f_y vanish?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Re: Multivariable calculus problem (Global Max/Min)

    What do you mean by 'vanish'?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Multivariable calculus problem (Global Max/Min)

    Quote Originally Posted by bagels0 View Post
    What do you mean by 'vanish'?
    equal zero
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Re: Multivariable calculus problem (Global Max/Min)

    2x - 2y = 0 [fx]
    -2x + 2 = 0 [fy]

    So, x = 1, and y = -1?

    What does this mean? Now what do I do?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Multivariable calculus problem (Global Max/Min)

    Quote Originally Posted by bagels0 View Post
    2x - 2y = 0 [fx]
    -2x + 2 = 0 [fy]

    So, x = 1, and y = -1?

    What does this mean? Now what do I do?

    Thanks.
    It means that there is a local extremum at (1, -1). Determine whether it is a maximum or minimum by using the second derivative test.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Re: Multivariable calculus problem (Global Max/Min)

    Cool. But what about the Domain? How do I take care of that?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Multivariable calculus problem (Global Max/Min)

    Quote Originally Posted by bagels0 View Post
    Cool. But what about the Domain? How do I take care of that?
    In the end, we will have to evaluate the function at the end-points of the domain.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Re: Multivariable calculus problem (Global Max/Min)

    Ok, so lets do the second derivative test.

    fxx = 2

    fyy = 0

    Also, my previous points were wrong. It was (1,1), not (1,-1).

    So, what do I do now? Now that I've taken the second derivatives? What do they show?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Multivariable calculus problem (Global Max/Min)

    Quote Originally Posted by bagels0 View Post
    Ok, so lets do the second derivative test.

    fxx = 2

    fyy = 0

    Also, my previous points were wrong. It was (1,1), not (1,-1).

    So, what do I do now? Now that I've taken the second derivatives? What do they show?
    Don't you know what the Second partial derivatives test states?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Re: Multivariable calculus problem (Global Max/Min)

    Oh, so find the D? D = fxx*fyy - fxy^2

    D = 0-4 = -4

    So, that means...there is a saddle point at (1,1)?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Multivariable calculus problem (Global Max/Min)

    Quote Originally Posted by bagels0 View Post
    Oh, so find the D? D = fxx*fyy - fxy^2

    D = 0-4 = -4

    So, that means...there is a saddle point at (1,1)?
    Yes. So there is no local extremum in the interior of the domain (the triangle). So you only need to evaluate the function at the endpoints of the domain (the sides of the triangle) to find the extrema.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Sep 2010
    Posts
    9

    Re: Multivariable calculus problem (Global Max/Min)

    Ok, so here is what I did.

    Found the critical point (1,1). f(1,1) = 1

    The Discriminant showed it is a saddle point.

    Found the edge points to be (2,0), (0,2) and (0,0) (Is this correct?)

    Evaluating the f(edge points), I got 4, 4 and 0 respectively.

    So.....global maximum is 4 at (2,0) and (0,2)? Global minimum is 0 at (0,0)?

    Saddle point isn't max or min right? Also, is it ok if the maximum happens at 2 points? Is all this correct?
    Last edited by bagels0; October 16th 2011 at 02:07 PM.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. An interesting problem for Multivariable Calculus Students
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: April 25th 2012, 04:52 AM
  2. Replies: 4
    Last Post: January 16th 2011, 06:45 AM
  3. Global min and max multivariable
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 4th 2010, 01:51 PM
  4. Hard multivariable calculus problem..
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 3rd 2010, 10:15 AM
  5. Replies: 4
    Last Post: February 19th 2009, 04:27 PM

Search Tags


/mathhelpforum @mathhelpforum