Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 45

There is a saddle point at $(0,2)$

We are given $z=2x^{2}-3y^{2}+12y$ $f_{x}(x,y)=4x = 0 \rightarrow x=0$ $f_{y}(x,y)=-6y+12=0 \rightarrow y=2$ The critical points is $(0,2)$ $f_{xx}(x,y)=4$ $f_{yy}(x,y)=-6$ $f_{xy}(x,y)=0$ $f_{xx}(0,2)=4$ $f_{yy}(0,2)=-6$ $f_{xy}(0,2)=0$ $D=4\times(-6)-(0)^{2}=-24$ Since D<0, there is a saddle point at $(0,2)$