# Find All local maxima and minima and all saddle points of the function

1. Mar 29, 2012

### knowLittle

1. The problem statement, all variables and given/known data
$f\left( x,y\right) =x^{2}-4xy+6x-8y+2y^{2}+10$

$f_{x}=2x-4y+6=0$
$f_{y}=-4x-8+4y=0$

$f_{y}=-4\left( x-y+2\right)$
-2=x-y, then solving fx and using this equality

$f_{x}=0=2\left( x-2y+3\right) =0$

2(-2 -y+3)=0
2(1-y)=0
y=1, then pluggin it to values

-2=x-y
-2=x-1
-1=x, So critical points at (-1,1)

fxx(-1,1)fyy(-1,1)-0^2=8, which is greater than zero and fxx is too. Therefore, there is a global minima at (-1,1)

There is no saddle point or global maxima?

2. Relevant equations

3. The attempt at a solution

Is this correct?

2. Mar 29, 2012

### LCKurtz

Ok until right there. How did you get $f_{xy}=0\$?

3. Mar 31, 2012

### knowLittle

You are right fxy= -4.

So,
fxx(-1,1)fyy(-1,1)-(-4^2)=
fxx=2;
fyy=4;
8-16=-8