Question

If the function \( f(x, y) \) is defined by \[ f(x, y) = x^3 - \frac{3}{2} x^2 y^2 + y^3, x, y \in \mathbb{R}, \] then

• A. Neither \( (0, 0) \) nor \( (1, 1) \) is a critical point
• B. \( (0, 0) \) is a critical point but \( (1, 1) \) is NOT a critical point
• C. \( (0, 0) \) is NOT a critical point but \( (1, 1) \) is a critical point
• D. \( (0, 0) \) and \( (1, 1) \) are both critical points

Hint:

To find critical points of a function, set the partial derivatives equal to zero and solve for \( x \) and \( y \).

Answer 1

The Correct Option is D

The critical points of a function occur where the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) are both zero.

Step 1: Compute the partial derivatives.
The partial derivatives of \( f(x, y) \) are: \[ \frac{\partial f}{\partial x} = 3x^2 - 3xy^2, \frac{\partial f}{\partial y} = -3x^2 y + 3y^2. \]

Step 2: Evaluate at \( (0, 0) \) and \( (1, 1) \).
At \( (0, 0) \): \[ \frac{\partial f}{\partial x} = 3(0)^2 - 3(0)(0)^2 = 0, \frac{\partial f}{\partial y} = -3(0)^2(0) + 3(0)^2 = 0. \] Thus, \( (0, 0) \) is a critical point. At \( (1, 1) \): \[ \frac{\partial f}{\partial x} = 3(1)^2 - 3(1)(1)^2 = 0, \frac{\partial f}{\partial y} = -3(1)^2(1) + 3(1)^2 = 0. \] Thus, \( (1, 1) \) is also a critical point.

Step 3: Conclusion.
Therefore, both \( (0, 0) \) and \( (1, 1) \) are critical points, and option (A) is incorrect.

Final Answer: \[ \boxed{\text{(A) Neither \( (0, 0) \) nor \( (1, 1) \) is a critical point}}. \]