Answer
There is a relative minimum at $(\frac{-9}{2},4)$. The relative minimum is $\frac{-141}{4}$
Work Step by Step
We are given $z=f(x,y)=x^{2}+y^{2}+9x-8y+1$
$f_{x}(x,y)=2x+9 = 0 \rightarrow x=\frac{-9}{2}$
$f_{y}(x,y)=2y-8=0 \rightarrow y=4$
The critical point is $(\frac{-9}{2},4)$
$f_{xx}(x,y)=2$
$f_{yy}(x,y)=2$
$f_{xy}(x,y)=0$
$f_{xx}(\frac{-9}{2},4) =2$
$f_{yy}(\frac{-9}{2},4)=2$
$f_{xy}(\frac{-9}{2},4)=0$
$D=2\times2-(0)^{2}=4$
Since D>0, and $f_{xx}(\frac{-9}{2},4) =2\gt0$, there is a relative minimum at $(\frac{-9}{2},4)$. The relative minimum is $\frac{-141}{4}$.
