- #1

knowLittle

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## Homework Statement

## 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?

## Homework Equations

## The Attempt at a Solution

Is this correct?