Answer
The critical points are $(27,9)$ and (0,0).
There is a saddle point at (0,0).
There is a relative maximum at $(27,9)$
Work Step by Step
$f(x,y)=3x^{2}+2y^{3}-18xy+42$
$f_x(x,y)=6x-18y=0 \rightarrow x=3y$
$f_y(x,y)=6y^{2}-18x=0$
Use the substitution method to solve the system of equations
$f_y(x,y)=6y^{2}-18x=0 \rightarrow 6y^{2}-18(3y)=0$
$6y^{2}-54y=0$
$y=9 \rightarrow x=27$ or $x=0 \rightarrow y=0$
The critical points are $(27,9)$ and (0,0)
To identify any extrema:
$f_{xx}(x,y)=6$
$f_{yy}(x,y)=12y$
$f_{xy}(x,y)=-18$
For (0,0):
$f_{xx}(0,0)=6$
$f_{yy}(0,0)=0$
$f_{xy}(0,0)=-18$
$D=6\times0 - (-18)^{2}=-324\lt0 \rightarrow$ There is a saddle point at (0,0)
For $(27,9)$:
$f_{xx}(27,9)=6$
$f_{yy}(27,9)=108$
$f_{xy}(27,9)=-18$
$D=6\times 108 - (-18)^{2}=324\gt0$ and $f_{xx}(27,9)=6\gt0 \rightarrow$ There is a relative maximum at $(27,9)$
