# find the critical points, and use second partial test to classify as relative max. relative min or saddle point f(x,y)=x^(2)-4xy+3y^2+2x-4y

*print*Print*list*Cite

### 1 Answer

`f(x,y) = x^2 - 4xy + 3y^2 + 2x-4y`

First we look for the critical points by finding the first partials and setting them equal to 0:

`f_x = 2x - 4y + 2`

`f_y = -4x + 6y - 4`

Critical points occur when both `f_x` and `f_y` are 0. Thus:

`0 = 2x - 4y + 2`

`0 = -4x + 6y - 4`

We can solve this by multiplying the first equation by 2, and adding the equations together:

`0 = 4x - 8y + 4`

`0 = -4x + 6y - 4`

`0 = 0 -2y + 0`

We get that `y=0`. Substituting this into either equation, we get `x=-1`

So `(-1,0)` is our critical point.

To see if it is a minimum, maximum, or saddlepoint, we find the second partials:

`f_(xx) = 2`

`f_(yy) = 6`

`f_(xy)=f_(yx) = -4`

Let `D = f_(xx) f_(yy) - (f_(xy))^2`

If `D >0 ` and `f_(xx) > 0` then the point is a local minimum.

If `D >0 ` and `f_(xx) < 0` then the point is a local maximum.

If `D <0 ` then the point is a saddle point.

If `D =0 ` then we have no information.

For us:

`D = 2*6-(-4)^2 = 12-16=-4` Thus we have a saddle point.