Question

For the function \(F: \R^2 → \R\) specified as F(x, y) = x³ - y³ + 9xy, which of the following options is/are correct

• A. one saddle point
• B. one strict local minimum
• C. one strict local maximum
• D. one global maximum

Answer 1

The Correct Option is A, B

To determine the nature of the critical points for the function \(F(x, y) = x^3 - y^3 + 9xy\), we'll follow these steps:

• Find Critical Points:
  • Compute the partial derivatives: 
  • \(\frac{\partial F}{\partial x} = 3x^2 + 9y\)
  • \(\frac{\partial F}{\partial y} = -3y^2 + 9x\)
  • Set these derivatives to zero:
  • System of equations:
    • \(3x^2 + 9y = 0\)
    • \(-3y^2 + 9x = 0\)
  • Solve the system to find critical points:
    • From \(3x^2 + 9y = 0\), we have \(y = -\frac{1}{3}x^2\)
    • Substitute into \(-3y^2 + 9x = 0\):
    • \(-3\left(-\frac{1}{3}x^2\right)^2 + 9x = 0\)
    • \(-\frac{1}{3}x^4 + 9x = 0\)
    • Factor: \(x(-\frac{1}{3}x^3 + 9) = 0\)
    • So, \(x = 0\) or \(x^3 = 27\) \(\Rightarrow x = 3\), as negative roots are not feasible for this power
    • If \(x = 0\), \(y = -\frac{1}{3}(0)^2 = 0\)
    • If \(x = 3\), \(y = -\frac{1}{3}(3)^2 = -3\)
    • Critical points: \((0, 0)\), \((3, -3)\)
• Determine Nature of Critical Points Using Second Derivative Test:
  • Compute second partial derivatives:
  • \(\frac{\partial^2 F}{\partial x^2} = 6x\)
  • \(\frac{\partial^2 F}{\partial y^2} = -6y\)
  • \(\frac{\partial^2 F}{\partial x \partial y} = 9\)
  • Evaluate Determinant: \(D = \frac{\partial^2 F}{\partial x^2}\frac{\partial^2 F}{\partial y^2} - \left(\frac{\partial^2 F}{\partial x \partial y}\right)^2\)
  • At \((0, 0)\):
    • \(D = (0)(0) - (9)^2 = -81 < 0\) (saddle point)
  • At \((3, -3)\):
    • \(D = (18)(18) - (9)^2 = 243\) (positive)
    • \(\frac{\partial^2 F}{\partial x^2}(3, -3) = 18 > 0\) (local minimum)

Conclusion: The function \(F(x, y)\) has one saddle point at \( (0, 0) \) and one strict local minimum at \( (3, -3) \).