Questions: The function f(x, y)=-x^2 y-y^2-4 y has one critical point, classify it as a relative maximum, relative minimum or a saddle point. A. Relative Maximum B. Relative Minimum C. Saddle point

Solution Steps

To classify the critical point of the function \( f(x, y) = -x^2 y - y^2 - 4y \), we first find the critical points by setting the partial derivatives with respect to \( x \) and \( y \) to zero. Then, we use the second derivative test, which involves computing the Hessian matrix and its determinant at the critical point. The sign of the determinant will help us classify the critical point as a relative maximum, relative minimum, or saddle point.

To find the critical points of the function \( f(x, y) = -x^2 y - y^2 - 4y \), we compute the partial derivatives: \[ f_x = -2xy, \quad f_y = -x^2 - 2y - 4 \] Setting these equal to zero, we solve the system: \[ -2xy = 0 \quad \text{and} \quad -x^2 - 2y - 4 = 0 \] This yields the critical points: \[ (0, -2), \quad (-2i, 0), \quad (2i, 0) \]

Next, we compute the second derivatives: \[ f_{xx} = -2y, \quad f_{yy} = -2, \quad f_{xy} = -2x \]

We evaluate the Hessian matrix at each critical point and determine the classification based on the determinant:

1. At \( (0, -2) \): \[ H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & -2 \end{bmatrix} \] The determinant is \( 4 \cdot (-2) - 0 \cdot 0 = -8 < 0 \), indicating a Saddle Point.

2. At \( (-2i, 0) \): \[ H = \begin{bmatrix} 0 & 4 \\ 4 & -2 \end{bmatrix} \] The determinant is \( 0 \cdot (-2) - 4 \cdot 4 = -16 < 0 \), indicating a Saddle Point.

3. At \( (2i, 0) \): \[ H = \begin{bmatrix} 0 & -4 \\ -4 & -2 \end{bmatrix} \] The determinant is \( 0 \cdot (-2) - (-4) \cdot (-4) = -16 < 0 \), indicating a Saddle Point.

Final Answer