Answer
The critical point is: (15,-8).
There is a relative minimum at (15,-8)
Work Step by Step
$f(x,y)=x^{2}+3xy+3y^{2}-6x+3y$
$f_x(x,y)=2x+3y-6=0$
$f_y(x,y)=3x+6y+3=0 \rightarrow x=-2y-1$
Use the substitution method to solve the system of equations:
$f_y(x,y)=2x+3y-6=0 \rightarrow 2(-2y-1)+3y-6=0$
$y=-8$
The critical point is: $(15,-8)$
To identify any extrema:
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=6$
$f_{xy}(x,y)=3$
For $(15,-8)$:
$f_{xx}(15,-8)=2$
$f_{yy}(15,-8)=6$
$f_{xy}(15,-8)=3$
$D=2\times6 - (3)^{2}=3\gt 0$ and $f_{xx}(15,-8)=2\gt0 \rightarrow$ There is a relative minimum at $(15,-8)$
