Find the extremum of $f(x, y)$ subject to the given constraint, and state whether it is a maximum or a minimum. \[ f(x, y)=3 x^{2}+3 y^{2}-x y ; x+y=2 \] Find the Lagrange function $F(x, y, \lambda)$. \[ F(x, y, \lambda)=\square-\lambda( \] Find the partial derivatives $F_{x}, F_{y}$, and $F_{\lambda}$. \[ \begin{array}{l} F_{x}=\square \\ F_{y}=\square \\ F_{\lambda}=\square \end{array} \] There is a value of $\square$ located at $(x, y)=$ (Type an integer or a fraction. Type an ordered pair, using integers or fractions.)

Step 1 ：Form the Lagrange function, which is the original function plus a new variable (lambda) times the constraint. The Lagrange function is \(F(x, y, \lambda) = 3x^{2} - xy + 3y^{2} - \lambda(x + y - 2)\).

Step 2 ：Find the partial derivatives of the Lagrange function with respect to each variable (x, y, and lambda). The partial derivatives are \(F_{x} = -\lambda + 6x - y\), \(F_{y} = -\lambda - x + 6y\), and \(F_{\lambda} = -x - y + 2\).

Step 3 ：Set these partial derivatives equal to zero to get a system of equations. Solve this system to find the values of x, y, and lambda that give the extremum of the original function. The solution to the system of equations is x = 1, y = 1, and lambda = 5.

Step 4 ：Check the second derivatives of the function to determine whether this is a maximum or a minimum. The second derivatives of the function at the point (1, 1) are \(F_{xx} = 6\), \(F_{yy} = 6\), and \(F_{xy} = -1\).

Step 5 ：Calculate the determinant of the Hessian matrix to determine whether the extremum is a maximum or a minimum. The determinant of the Hessian matrix is 35. Since the determinant is positive, this means that the extremum is a minimum.

Step 6 ：\(\boxed{\text{The extremum of the function } f(x, y)=3 x^{2}+3 y^{2}-x y \text{ subject to the constraint } x+y=2 \text{ is a minimum and it is located at the point } (1, 1).}\)