Problem 14 - College Algebra and Calculus: An Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=x^{2}-3 x y-y^{2} $$

Step-by-Step Solution

Answer

The function \(f(x, y)=x^{2}-3 x y-y^{2}\) has a saddle point at the origin (0,0).

Step 1: Find the first partial derivatives

Compute the first partial derivative with respect to \( x, f_x \), and with respect to \( y, f_y \): \n\( f_x = 2x -3y \) and \( f_y = -3x -2y \).

Step 2: Set \( f_x \) and \( f_y \) to zero

In order to find the critical points we have: \n \(2x -3y = 0\) (i) \n and \(-3x -2y=0\) (ii). Solving this system of equations we find \(x = 0\) and \(y = 0\). Therefore, we have a critical point at (0,0).

Step 3: Find the second partial derivatives

Compute \( f_{xx} , f_{xy} , f_{yy} \) \n \( f_{xx} = 2, f_{xy} = -3, f_{yy} = -2 \).

Step 4: Form the Hessian matrix

The Hessian Matrix H is the square matrix of second partial derivatives. \n Hessian H = \( \[ [f_{xx} \ f_{xy}] , [f_{yx} \ f_{yy}] \] \) = \( \[ [2 \ -3] , [-3 \ -2] \] \)

Step 5: Evaluate the determinant of the Hessian Matrix

The determinant of the Hessian is: D = \( f_{xx}f_{yy} - f_{xy}^2 = 2*(-2) - (-3)^2 = -4 - 9 = -13\)

Step 6: Second derivative test

Given that D < 0, we conclude that we have a saddle point at (0,0)