Question

Consider the function \( f(x, y) = x^4 + y^4 - 4xy + 1 \). Which of the following is/are correct?

• A. The minimum value of \( f \) occurs at \( (0, 0) \)
• B. The point \( (0, 0) \) is a point of inflection
• C. \( f \) has three critical points
• D. The minimum value of \( f \) is \( -1 \)

Hint:

To analyze critical points of multivariable functions: 1. Compute the first-order partial derivatives and solve for zero to find critical points.
2. Use the Hessian determinant to classify each critical point.
3. Evaluate the function at the critical points to determine extrema.

Answer 1

The Correct Option is B

Step 1: Compute the critical points of \( f(x, y) \).
The critical points are determined by setting the first-order partial derivatives of \( f(x, y) \) to zero: \[ \frac{\partial f}{\partial x} = 4x^3 - 4y = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 4y^3 - 4x = 0. \] From the first equation: \[ x^3 = y. \] Substituting \( x^3 = y \) into the second equation: \[ 4(x^3)^3 - 4x = 0 \quad \Rightarrow \quad 4x^9 - 4x = 0 \quad \Rightarrow \quad 4x(x^8 - 1) = 0. \] This gives: \[ x = 0, \quad x = 1, \quad x = -1. \] For \( x = 0, \, y = 0 \).
For \( x = 1, \, y = 1 \).
For \( x = -1, \, y = -1 \).
Thus, the critical points are \( (0, 0), (1, 1), \) and \( (-1, -1) \). Step 2: Classify the critical points.
The second-order partial derivatives of \( f(x, y) \) are: \[ \frac{\partial^2 f}{\partial x^2} = 12x^2, \quad \frac{\partial^2 f}{\partial y^2} = 12y^2, \quad \text{and} \quad \frac{\partial^2 f}{\partial x \partial y} = -4. \] The Hessian determinant is: \[ H = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 = (12x^2)(12y^2) - (-4)^2 = 144x^2y^2 - 16. \] 1. At \( (0, 0) \), \( H = -16 \) (negative), indicating a saddle point. Thus, \( (0, 0) \) is also a point of inflection. 2. At \( (1, 1) \), \( H = 144(1)(1) - 16 = 128 \) (positive), indicating a local minimum. 3. At \( (-1, -1) \), \( H = 144(1)(1) - 16 = 128 \) (positive), indicating another local minimum. Step 3: Verify the minimum value.
Evaluate \( f(x, y) \) at the critical points: \[ f(0, 0) = 1, \quad f(1, 1) = 1^4 + 1^4 - 4(1)(1) + 1 = -1, \quad f(-1, -1) = (-1)^4 + (-1)^4 - 4(-1)(-1) + 1 = -1. \] The minimum value of \( f \) is \( -1 \), occurring at \( (1, 1) \) and \( (-1, -1) \). Step 4: Analyze the options.
Option (A): Incorrect, as the minimum does not occur at \( (0, 0) \).
Option (B): Correct, as \( (0, 0) \) is a point of inflection (saddle point).
Option (C): Correct, as \( f \) has three critical points: \( (0, 0), (1, 1), (-1, -1) \).
Option (D): Correct, as the minimum value of \( f \) is \( -1 \).
Conclusion: The correct statements are (B), (C), and (D).