# Find the absolute extremes of the function f(x,y)=x^2-3y^2-2x+6y on the square R with vertices (0,0),(0,2),(2,2),(2,0).

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### 1 Answer

You need to determine partial derivatives to find the local extrema of function such that:

`f_x (x,y) = 2x - 2` (differentiate the function f with respect to x only and consider y as a constant term)

`f_y (x,y) = - 6y + 6`

You need to remember that you may find the critical points of the function solving the system of simultaneous equations such that:

`f_x (x,y) = 0`

`f_y (x,y) = 0`

Substituting `2x - 2` for `f_x (x,y)` and `- 6y + 6 ` for `f_y (x,y)` yields:

`2x - 2 = 0 =gt x - 1 = 0`

`=gt x = 1`

`- 6y + 6 = 0 =gt y - 1 = 0 `

`=gt y = 1`

You should verify if the point (1,1) is a critical point or a saddle point using discriminant D such that:

`D = f_(x x) (x,y)*f_(y y) (x,y) - [f_(x y) (x,y)]^2`

`D = 2*(-6) - 0 =gt D = -12 lt 0` , hence (1,1) is a saddle point

**Hence, since the point `(1,1)` is not an extreme but a saddle point, you may not go any further from here to find the extrema on the bounded region.**