Question

Let \( f: \mathbb{R}^2 \to \mathbb{R} \) be defined as \[ f(x, y) = x^2 y^2 + 8x - 4y. \] The number of saddle points of \( f \) is __________ (answer in integer).

Hint:

A saddle point is a point where the discriminant of the second derivative test is negative. This indicates that the point is neither a local maximum nor a local minimum.

Answer 1

Correct Answer: 1

Step 1: Find the first-order partial derivatives of \( f(x, y) \).
To find the critical points, we first compute the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \): \[ f_x(x, y) = \frac{\partial}{\partial x}(x^2 y^2 + 8x - 4y) = 2x y^2 + 8, \] \[ f_y(x, y) = \frac{\partial}{\partial y}(x^2 y^2 + 8x - 4y) = 2x^2 y - 4. \] Step 2: Solve the system of equations \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \).
We need to solve the system of equations: \[ 2x y^2 + 8 = 0 \quad {and} \quad 2x^2 y - 4 = 0. \] - From the second equation, we have: \[ 2x^2 y - 4 = 0 \quad \Rightarrow \quad x^2 y = 2. \] - From the first equation, we have: \[ 2x y^2 + 8 = 0 \quad \Rightarrow \quad x y^2 = -4. \] We now solve the system:
\( x^2 y = 2 \),
\( x y^2 = -4 \).
Dividing the second equation by the first equation, we get: \[ \frac{x y^2}{x^2 y} = \frac{-4}{2} \quad \Rightarrow \quad \frac{y}{x} = -2 \quad \Rightarrow \quad y = -2x. \] Substitute \( y = -2x \) into \( x^2 y = 2 \): \[ x^2 (-2x) = 2 \quad \Rightarrow \quad -2x^3 = 2 \quad \Rightarrow \quad x^3 = -1 \quad \Rightarrow \quad x = -1. \] Substitute \( x = -1 \) into \( y = -2x \): \[ y = -2(-1) = 2. \] Thus, the only critical point is \( (x, y) = (-1, 2) \). Step 3: Determine the nature of the critical point. To classify the critical point, we compute the second-order partial derivatives: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2y^2, \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2x^2, \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 4xy. \] At the critical point \( (x, y) = (-1, 2) \), we compute the second-order partial derivatives: \[ f_{xx}(-1, 2) = 2(2)^2 = 8, \] \[ f_{yy}(-1, 2) = 2(-1)^2 = 2, \] \[ f_{xy}(-1, 2) = 4(-1)(2) = -8. \] Now, compute the discriminant \( D = f_{xx} f_{yy} - (f_{xy})^2 \): \[ D = 8 \times 2 - (-8)^2 = 16 - 64 = -48. \] Since \( D<0 \), the critical point is a saddle point. Thus, the number of saddle points of \( f \) is \( \boxed{1} \).