To find the critical points of the function f(x,y), we need to find where the partial derivatives of the function with respect to x and y are zero: fx = -8 + 4x - 4y + 2xy = 0 fy = 8 - 4x + 2xy + 4y = 0 To solve for the critical points, we can use either the substitution or elimination method. Let's use the substitution method: fx = -8 + 4x - 4y + 2xy = 0 => y = (8 - 4x)/(2x - 4) = 2 - 1/(x - 2) fy = 8 - 4x + 2xy + 4y = 0 => y = (-4 + 2x)/(2x - 1) = -2 + 3/(2x - 1) Equating the two expressions for y, we get: 2 - 1/(x - 2) = -2 + 3/(2x - 1) => 4x^2 - 13x + 10 = 0 => (4x - 5)(x - 2) = 0 So, the critical points are (5/4, 3/2) and (2, 2). To determine whether the critical points are maxima, minima, or saddle points, we need to use the second partial derivative test. The Hessian matrix of f(x,y) is: H = [4 + 2y, -4 + 2x - 2y] [-4 + 2x - 2y, 2 + 2x] At the critical point (5/4, 3/2), we have: H = [7, -1] [-1, 13/2] The determinant of H is positive and the trace is positive, so the critical point (5/4, 3/2) is a local minimum. At the critical point (2, 2), we have: H = [6, 0] [0, 6] The determinant of H is positive and the trace is positive, so the critical point (2, 2) is a local minimum. Therefore, the critical points of f(x,y) are both local minima.