We start by finding the first and secondpartial derivatives of f: \[f_x=2x+4y\] \[f_{xx}=2\] \[f_{xy}=4\] \[f_y=4x+6y-6\] \[f_{yy}=6\] \[f_{yx}=4\]
We find the critical points by finding where \(f_x\) and \(f_y\) are simultaneously 0. Setting \(f_x=0\), we have: \[f_x=0 \quad \Rightarrow \quad 2x+4y=0\] \[f_y=0 \quad \Rightarrow \quad 4x+6y-6=0\] Using system of equations to find x and y: \[\text{Isolate } x \text{ for } 2x+4y=0; \quad x=-2y\] \[\text{Subsitute }x=-2y; \quad [4(-2y)+6y-6=0]\] \[\text{Isolate } y \text{ for } [4(-2y)+6y-6=0]; \quad y=-3\] \[\text{For }x=-2y;\qquad \text{Substitute } y=-3\] \[\therefore x=6, y=-3\] One critical point at \((6,-3)\)
Second Derivative Test, \[D=f_{xx}(x,y)f_{yy}(x,y)-f_{xy}(x,y)^2\] \[D=2*6-(4)^2=-4\]
\(D=−4\), so this point corresponds to a saddle point.