Question

The function \[ f(x, y) = x^3 + 2xy + y^3 \] has a saddle point at

• A. \( (0, 0) \)
• B. \( \left( -\frac{2}{3}, -\frac{2}{3} \right) \)
• C. \( \left( -\frac{3}{2}, -\frac{3}{2} \right) \)
• D. \( (-1, -1) \)

Hint:

To identify saddle points, compute the first and second partial derivatives, and use the second derivative test to analyze the nature of the critical points.

Answer 1

The Correct Option is A

Step 1: Calculate the first and second partial derivatives.
The first partial derivatives of \( f(x, y) = x^3 + 2xy + y^3 \) are: \[ f_x = 3x^2 + 2y, \quad f_y = 2x + 3y^2. \] At the saddle point, \( f_x = 0 \) and \( f_y = 0 \). Setting these equal to zero gives: \[ 3x^2 + 2y = 0, \quad 2x + 3y^2 = 0. \]
Step 2: Solve for the critical points.
Solving these equations gives the critical point \( (0, 0) \).
Step 3: Use the second derivative test.
The second partial derivatives are: \[ f_{xx} = 6x, \quad f_{yy} = 6y, \quad f_{xy} = 2. \] At \( (0, 0) \), we calculate the determinant of the Hessian matrix: \[ D = f_{xx} f_{yy} - (f_{xy})^2 = (0)(0) - (2)^2 = -4. \] Since \( D<0 \), \( (0, 0) \) is a saddle point.

Step 4: Conclusion.
Thus, the correct answer is \( \boxed{(A)} (0, 0) \).