## Calculus with Applications (10th Edition)

There is a relative minimum at $(\frac{-9}{2},4)$. The relative minimum is $\frac{-141}{4}$
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 $\frac{-141}{4}$.