Answer
The critical point is: $(10,-3)$.
There is a saddle point at $(10,-3)$
Work Step by Step
$f(x,y)=4y^{2}+2xy+6x+4y-8$
$f_x(x,y)=2y+6=0 \rightarrow y=-3$
$f_y(x,y)=8y+2x+4=0$
Use the substitution method to solve the system of equations
$f_y(x,y)=8y+2x+4=0 \rightarrow 8(-3)+2x+4=0$
$x=10 $
The critical point is: $(10,-3)$
To identify any extrema:
$f_{xx}(x,y)=0$
$f_{yy}(x,y)=8$
$f_{xy}(x,y)=2$
For $(10,-3)$:
$f_{xx}(10,-3)=0$
$f_{yy}(10,-3)=8$
$f_{xy}(10,-3)=2$
$D=0\times8 - (2)^{2}=-4 \lt 0 \rightarrow$ There is a saddle point at $(10,-3)$
