Question

Let \( f : \mathbb{R}^2 \to \mathbb{R} \) be given by \[ f(x, y) = 4xy - 2x^2 - y^4 + 1. \] The number of critical points where \( f \) has local maximum is equal to ________.

Hint:

To determine the nature of critical points, compute the second-order partial derivatives and the discriminant. A positive discriminant and \( f_{xx}<0 \) indicate a local maximum.

Answer 1

Correct Answer: 2

To find the critical points of the function \( f(x, y) \), we first compute the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), and set them equal to zero.
The partial derivative of \( f \) with respect to \( x \) is: \[ f_x = \frac{\partial}{\partial x}(4xy - 2x^2 - y^4 + 1) = 4y - 4x \] Setting \( f_x = 0 \): \[ 4y - 4x = 0 \quad \Rightarrow \quad y = x \] The partial derivative of \( f \) with respect to \( y \) is: \[ f_y = \frac{\partial}{\partial y}(4xy - 2x^2 - y^4 + 1) = 4x - 4y^3 \] Setting \( f_y = 0 \): \[ 4x - 4y^3 = 0 \quad \Rightarrow \quad x = y^3 \] Now, substituting \( y = x \) into \( x = y^3 \), we get: \[ x = x^3 \] This simplifies to: \[ x(x^2 - 1) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = \pm 1 \] For \( x = 0 \), \( y = 0 \). For \( x = 1 \), \( y = 1 \), and for \( x = -1 \), \( y = -1 \). Thus, the critical points are: \( (0, 0), (1, 1), (-1, -1) \). To determine which of these critical points correspond to a local maximum, we check the second-order partial derivatives. The discriminant is given by: \[ D = f_{xx} f_{yy} - (f_{xy})^2 \] After calculating the second-order derivatives and evaluating at the critical points, we find that the point \( (0, 0) \) is a saddle point, and the points \( (1, 1) \) and \( (-1, -1) \) correspond to local maxima. Thus, the number of critical points where \( f \) has a local maximum is: \[ \boxed{2} \]