Question:
The function
\[
f(x, y) = x^3 + 2xy + y^3
\]
has a saddle point at
The function
\[
f(x, y) = x^3 + 2xy + y^3
\]
has a saddle point at
Show 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.
Updated On: Dec 11, 2025
- \( (0, 0) \)
- \( \left( -\frac{2}{3}, -\frac{2}{3} \right) \)
- \( \left( -\frac{3}{2}, -\frac{3}{2} \right) \)
- \( (-1, -1) \)
Show Solution
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The Correct Option is A
Solution and Explanation
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 \( \boxed{(A)} (0, 0) \).
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 \( \boxed{(A)} (0, 0) \).
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