Answer to Question #85730 in Calculus for o

Question

Examine the function f (x, y)= xsquare −6xy + 8ysquare −7x + 7y + 14 for extreme values.

Answer 1

First of all, let us obtain the positions of the critical points by solving the following system of equations:

$$\frac{\partial f}{\partial x} =2 x -6 y -7 =0 \\ \frac{\partial f}{\partial y} =-6 x + 16 y +7 =0$$

As a result, one can show that (-35/2, -7) is the only critical point. In order to define its type, one should perform the second derivative test, i.e. calculate the value

$$A = \frac{\partial^2 f}{\partial x^2} = 2$$

and determinant of the Hessian matrix

$$D = \begin{vmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{vmatrix} = \begin{vmatrix} 2 & -6 \\ -6 & 16 \end{vmatrix} = -4$$

at this point. As long as we got D < 0 (the sign of A in this case does not matter), one can conclude that the point (-35/2, -7) is a saddle point.

Answer: there is only one extreme point (-35/2, -7) which is a saddle point.