Answer
There is a relative minimum at $(-8,-23)$. The relative minimum is -57
Work Step by Step
We are given $z=f(x,y)=7x^{2}+y^{2}-3x+6y-5xy$
$f_{x}(x,y)=14x-3-5y$
$f_{y}(x,y)=2y+6-5x$
Set the partial derivative equal to 0
$14x-3-5y=0$
$x=\frac{5y+3}{14}$
Now substitute for x in the other equation and solve for y
$2y+6-5(\frac{5y+3}{14})=0$
$28y+84-25y-15=0$
$y=-23 \rightarrow x=-8$
The critical point is $(-8,-23)$
$f_{xx}(x,y)=14$
$f_{yy}(x,y)=2$
$f_{xy}(x,y)=-5$
$f_{xx}(-8,-23) =14$
$f_{yy}(-8,-23)=2$
$f_{xy}(-8,-23)=-5$
$D=14\times2-(-5)^{2}=3$
Since D>0 and $f_{xx}(-8,-23) =14\gt0$, there is a relative minimum at $(-8,-23)$. The relative minimum is -57
