1. **Problem:** Find the maximum and minimum values of the function $$f(x,y) = x^2 - xy + y^2 - 2x + y.$$ 2. **Step 1: Find the critical points** by setting the partial derivatives equal to zero. \[\frac{\partial f}{\partial x} = 2x - y - 2 = 0,\quad \frac{\partial f}{\partial y} = -x + 2y + 1 = 0.\] 3. **Step 2: Solve the system of equations**: From the first equation: $$2x - y = 2 \implies y = 2x - 2.$$ Substitute into the second equation: $$-x + 2(2x - 2) + 1 = 0 \implies -x + 4x - 4 + 1 = 0 \implies 3x - 3 = 0 \implies x = 1.$$ Then, $$y = 2(1) - 2 = 0.$$ So the critical point is at $(1,0)$. 4. **Step 3: Classify the critical point using the second derivative test.** Calculate second derivatives: $$f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = -1.$$ Calculate the determinant of the Hessian matrix: $$D = f_{xx} f_{yy} - (f_{xy})^2 = (2)(2) - (-1)^2 = 4 - 1 = 3 > 0,$$ and since $f_{xx} = 2 > 0$, the critical point is a local minimum. 5. **Step 4: Find the minimum value** by evaluating the function at $(1,0)$: $$f(1,0) = 1^2 - (1)(0) + 0^2 - 2(1) + 0 = 1 - 0 + 0 - 2 + 0 = -1.$$ 6. **Step 5: Check for maximum values.** Since the quadratic form is positive definite around the minimum and there are no boundaries given, the function has no maximum (it tends to infinity as $x,y \to \infty$). **Final answer:** The function has a local minimum value of $-1$ at the point $(1,0)$ and no maximum.