91Ó°ÊÓ

To find the critical points, we need to determine where the first partial derivatives of the function are zero. First, compute the partial derivatives: \( f_x = \frac{\partial f}{\partial x} = 3x^2 - 6y \) \( f_y = \frac{\partial f}{\partial y} = -6x - 3y^2 \) Set these partial derivatives equal to zero to find critical points: From \( f_x = 0 \), we have: \( 3x^2 - 6y = 0 \) This simplifies to: \( y = \frac{x^2}{2} \) From \( f_y = 0 \), we have: \( -6x - 3y^2 = 0 \) This simplifies to: \( x = -\frac{y^2}{2} \) Substitute \( y = \frac{x^2}{2} \) into \( x = -\frac{y^2}{2} \): \( x = -\frac{(\frac{x^2}{2})^2}{2} \) Solve for \( x \) and \( y \). You find two critical points: \((0,0)\) and \((-2, 2)\).

The Hessian matrix is used to determine the nature of critical points. First, compute the second-order partial derivatives of \( f \): \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x \) \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -6y \) \( f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = -6 \) The Hessian matrix, \( H \), is: \[H = \begin{bmatrix}6x & -6 \-6 & -6y \end{bmatrix}\]

To classify the critical points, evaluate the Hessian at the found critical points. 1. For \((0,0)\): \[ H = \begin{bmatrix} 0 & -6 \ -6 & 0 \end{bmatrix} \] The determinant is \((-6)(-6) - (0)(0) = 36\). Since the determinant is positive but the leading diagonal term \(6x = 0\) is non-positive, the point \((0, 0)\) is a saddle point.2. For \((-2,2)\): \[ H = \begin{bmatrix} 6(-2) & -6 \ -6 & -6(2) \end{bmatrix} = \begin{bmatrix} -12 & -6 \ -6 & -12 \end{bmatrix} \] The determinant is \((-12)(-12) - (-6)(-6) = 144 - 36 = 108\), which is positive. The leading diagonal term \(6x = -12 < 0\), indicating a relative maximum.