Answer
The critical point is $(-8,-23)$.
There is a relative maximum at $(-8,-23)$
Work Step by Step
$f(x,y)=5xy-7x^{2}-y^{2}+3x-6y-4$
$f_x(x,y)=5y-14x+3=0 \rightarrow x=\frac{5y+3}{14}$
$f_y(x,y)=5x-2y-6=0$
Use the substitution method to solve the system of equations
$f_y(x,y)=5x-2y-6=0 \rightarrow 5(\frac{5y+3}{14})-2y-6=0$
$25y+15-28y-84=0 $
$y=-23$
The critical point is: $(-8,-23)$
To identify any extrema:
$f_{xx}(x,y)=-14$
$f_{yy}(x,y)=-2$
$f_{xy}(x,y)=5$
For $(-8,-23)$:
$f_{xx}(-8,-23)=-14$
$f_{yy}(-8,-23)=-2$
$f_{xy}(-8,-23)=5$
$D=(-14)\times(-2) - (5)^{2}=3\gt 0$ and $f_{xx}(-8,-23)=-14 \lt0 \rightarrow$ There is a relative maximum at $(-8,-23)$
