Question

Let \( f: \mathbb{R}^2 \to \mathbb{R} \) be given by \[ f(x, y) = 4xy - 2x^2 - y^4. \] Then \( f \) has

• A. a point of local maximum and a saddle point
• B. a point of local minimum and a saddle point
• C. a point of local maximum and a point of local minimum
• D. two saddle points

Hint:

To classify critical points, use the second derivative test. The sign of the discriminant helps determine the nature of the critical points.

Answer 1

The Correct Option is A

To determine the critical points and classify them, we first compute the partial derivatives of \( f(x, y) \): \[ f_x = 4y - 4x, f_y = 4x - 4y^3. \] Setting both partial derivatives equal to zero: \[ 4y - 4x = 0 \text{and} 4x - 4y^3 = 0. \] Solving these equations, we find the critical points at \( (0, 0) \) and \( (1, 1) \). Next, we compute the second-order partial derivatives: \[ f_{xx} = -4, f_{yy} = -12y^2, f_{xy} = 4. \] At \( (0, 0) \), the discriminant is: \[ D = f_{xx} f_{yy} - (f_{xy})^2 = (-4)(0) - (4)^2 = -16. \] Since \( D < 0 \), \( (0, 0) \) is a saddle point. At \( (1, 1) \), the discriminant is: \[ D = (-4)(-12) - (4)^2 = 48 - 16 = 32. \] Since \( D > 0 \) and \( f_{xx} = -4 < 0 \), \( (1, 1) \) is a point of local maximum. Thus, the function has a point of local maximum and a saddle point, corresponding to option (A).