## Calculus with Applications (10th Edition)

There is a relative maximum at $(\frac{-3}{4},\frac{-9}{32})$. The relative maximum is $\frac{539}{128}$. There is a saddle point at (0,0)
We are given $z=f(x,y)=x^{3}-8y^{2}+6xy+4$ $f_{x}(x,y)=3x^{2}+6y$ $f_{y}(x,y)=-16y+6x$ Set the partial derivative equal to 0 $-16y+6x=0$ $x=\frac{8}{3}y$ Now substitute for x in the other equation and solve for y $3(\frac{8}{3}y)^{2}+6y=0$ $\frac{64}{3}y^{2}+6y=0$ $y=0 \rightarrow x=0$ or $y=\frac{-9}{32} \rightarrow x=\frac{-3}{4}$ There are two critical points $(0,0); (\frac{-3}{4},\frac{-9}{32})$ $f_{xx}(x,y)=6x$ $f_{yy}(x,y)=-16$ $f_{xy}(x,y)=6$ $f_{xx}(\frac{-3}{4},\frac{-9}{32}) =\frac{-9}{2}$ $f_{yy}(\frac{-3}{4},\frac{-9}{32})=-16$ $f_{xy}(\frac{-3}{4},\frac{-9}{32})=6$ $D=\frac{-9}{2}\times(-16)-(6)^{2}=36$ Since D>0, and $f_{xx}(\frac{-3}{4},\frac{-9}{32})=\frac{-9}{2}\lt0$, there is a relative maximum at $(\frac{-3}{4},\frac{-9}{32})$. The relative maximum is $\frac{539}{128}$ $f_{xx}(0,0) =0$ $f_{yy}(0,0)=-16$ $f_{xy}(0,0)=6$ $D=0\times(-16)-(6)^{2}=-36$ Since D<0, there is a saddle point at (0,0)