stationary points of a function of 2 variables

**a1007** · May 10th 2008, 08:38 AM

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

**Soroban** · May 10th 2008, 09:52 AM

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

**a1007** · May 10th 2008, 10:38 AM

Thanx! made it look so easy

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