Question

Suppose f(x,y)=(x−y)(16−xy). Find the saddle points of f.

Answer 1

$$f(x, y)=(x-y)(16-xy)=16x-x^2y-16y+xy^2$$

Find the critical point(s)

$$f_x=16-2xy+y^2, f_y=-16+2xy-x^2$$

$$\begin{cases} f_x=0 \\ f_y=0 \end{cases}=> \begin{cases} 16-2xy+y^2=0 \\ -16+2xy-x^2=0 \end{cases}=>$$ $$=> \begin{cases} (y-x)(y+x)=0 \\ -16+2xy-x^2=0 \end{cases}$$

Critical points: $(-4,-4),\ (-4,4),\ (4,-4),\ (4,4)$

Second Derivatives Test

$$f_{xx}=-2y, f_{xy}=f_{yx}=-2x+2y, f_{yy}=2x$$

$$D=\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}=\begin{vmatrix} -2y & -2x+2y \\ -2x+2y & 2x \end{vmatrix}=$$

$$=-4xy-4y^2 +8xy-4x^2=-4(x^2-xy+y^2)$$

$$x^2-xy+y^2>0, x\in\R, y\in \R=>$$

$$=>D(x,y)<0, x\in\R, t\in \R$$

All critical points are saddle points.

$(-4,-4):$

$$f(-4,-4)=(-4-(-4))(16-(-4)(-4))=0$$

$(-4,4):$

$$f(-4,4)=(-4-4)(16-(-4)(4))=-256$$

$(4,-4):$

$$f(4,-4)=(4-(-4))(16-4(-4))=256$$

$(4,4):$

$$f(4,4)=(4-4)(16-4(4))=0$$

Saddle points: $(-4,-4,0),\ (-4,4,-256),\ (4,-4,256),\ (4,4,0)$