Question

Let \( f(x,y) = x^2 + 3y^2 - \frac{2}{3} xy \) at \( (x,y) \in \mathbb{R}^2 \). Then

• A. \( f \) has a point of local minimum
• B. \( f \) has a point of local maximum
• C. \( f \) has a saddle point
• D. \( f \) has no point of local minimum, no point of local maximum, and no saddle point

Answer 1

The Correct Option is C

1. Critical Points: - To find critical points, compute partial derivatives: \[ \frac{\partial f}{\partial x} = 8x^3, \quad \frac{\partial f}{\partial y} = -6y. \] - Setting these derivatives to zero: \[ 8x^3 = 0 \quad \Rightarrow \quad x = 0, \quad -6y = 0 \quad \Rightarrow \quad y = 0. \] - Thus, the only critical point is \((0, 0)\). 
2. Second Partial Derivatives: - Compute second partial derivatives: \[ f_{xx} = 24x^2, \quad f_{yy} = -6, \quad f_{xy} = f_{yx} = 0. \] 3. Hessian Determinant: - The Hessian determinant is: \[ H = f_{xx} f_{yy} - (f_{xy})^2 = (24x^2)(-6) - 0 = -144x^2. \] 4. Analyze the Critical Point: - At \((0, 0)\), \( H = 0 \), and \( f_{xx} = 0 \). - Since \( f_{xx} \) changes sign for different values of \( x \), \((0, 0)\) is a saddle point.