Results 1 to 3 of 3

Math Help - stationary points of a function of 2 variables

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    7

    stationary points of a function of 2 variables

    hey i have this problem where

    z = xy(6-x-y) when multiplying out and differentiating with respect to x i get:

    dz/dx = 6y - 2xy -y^2 and y dz/dy = 6x - x^2 - 2xy

    I understand with the -2xy you can make the equations equal and find stationary points, which i got as (0,0) (6,0) and (0,6) but once checking the answers there was a further point (2,2) and I don't understand on how to obtain this stationary point, help would be much appreciated than you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,685
    Thanks
    616
    Hello, a1007!

    z \:= \:xy(6-x-y)
    We have: . \begin{array}{cccccc}z_x &=& 6y - 2xy -y^2 &=& 0 \\ z_y &=&6x - x^2 - 2xy &=& 0 \end{array}

    Subtract the equations: . x^2-y^2 - 6x + 6y \:=\:0

    Factor: . (x-y)(x+y) - 6(x-y) \:=\:0 \quad\Rightarrow\quad (x-y)(x+y-6)\:=\:0

    We have two cases to consider: . \begin{array}{ccccccc}x-y \:=\: 0 & \Rightarrow & y \:=\:x & {\color{blue}[1]}\\ x+y-6 \:=\: 0 & \Rightarrow & y \:=\: 6-x & {\color{blue}[2]}\end{array}


    Substitute [1] into z_x\!:\;\;6x - 2x^2-x^2\:=\:0\quad\Rightarrow\quad 6x-3x^2\:=\:0 \quad\Rightarrow\quad 3x(2-x) \:=\:0

    . . Hence: . x \:=\:0,\,2\quad\Rightarrow\quad y \:=\:0,\,2 \quad\Rightarrow\quad\boxed{ (0,0),\:(2,2)}


    Substitute [2] into z_x\!:\;\;6(6-x)-2x(6-x)-(6-x)^2\:=\:0\quad\Rightarrow\quad 3x(x-6) \:=\:0

    . . Hence: . x \:=\:0,\,6\quad\Rightarrow\quad y \:=\:6,\,0\quad\Rightarrow\quad\boxed{ (6,0),\:(0,6)}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    7
    Thanx! made it look so easy
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stationary points with 2 variables
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 3rd 2010, 12:15 PM
  2. Replies: 2
    Last Post: July 21st 2009, 10:49 AM
  3. Replies: 3
    Last Post: May 20th 2009, 08:35 AM
  4. Stationary points / two variables (help)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 3rd 2008, 06:08 PM
  5. Replies: 2
    Last Post: February 28th 2008, 05:36 AM

Search Tags


/mathhelpforum @mathhelpforum