# Find the local minimums and maximums and saddle points of the function:f(x,y)=3(x^2)*y+(y^3)-3(x^2)-3(y^2)+2What exactly is a saddle point?

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### 1 Answer

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

`f_x (x,y) = 6xy - 6x`

`f_y (x,y) = 3x^2 + 3y^2 - 6y`

You need to remember that critical points check simultaneously the equations `f_x (x,y) = 0` and `f_y (x,y) = 0` such that:

`6xy - 6x = 0`

`3x^2 + 3y^2 - 6y = 0`

You need to divide by 6 in first equation such that:

`xy - x = 0 =gt xy = x =gt x = 0; y=1`

You need to divide by 3 the second equation such that:

`x^2 + y^2 - 2y = 0`

You need to substitute 1 for y in the second equation such that:

`x^2 + 1 - 2 = 0 =gt x^2 = 1 =gt x_(1,2) = +-1`

The solution to the system of equations is `(1;1),` since x=0 and x=-1 do not check the simultaneous equations.

You need to find the second partial derivatives `f_(x x) (x,y), f_(yy) (x,y)` and `f_(xy) (x,y)` such that:

`f_(x x) (x,y) = 6y-6`

`f_(yy) (x,y) = 2y-` `2`

`f_(xy) (x,y) = 6x`

You need to evaluate `f_(x x) (x,y)*f_(yy) (x,y) - [f_(xy) (x,y)]^2 ` at (1;1) such that:

`f_(x x) (1,1)*f_(yy) (1,1) - [f_(xy) (1,1)]^2 = 0*0 - 36 = -36 lt 0`

Since `f_(xx) (1,1)*f_(yy) (1,1) - [f_(xy) (1,1)]^2 = -36 lt 0` , then the function has a saddle point at (1,1).

**Since a saddle point is not an extreme point, the function has a saddle point at (1,1).**