Chapter 9 - Multiveriable Calculus - 9.3 Maxima and Minima - 9.3 Exercises - Page 488: 1

There is a saddle point at (-1,2)

We are given $f(x,y)=xy+y-2x$ $f_{x}(x,y)=y -2 = 0 \rightarrow y=2$ $f_{y}(x,y)=x+1=0 \rightarrow x=-1$ The critical point is (-1,2) $f_{xx}(x,y)=0$ $f_{yy}(x,y)=0$ $f_{xy}(x,y)=1$ $D=0.0-1^{2}=-1$ Since D<0, there is a saddle point at (-1,2).