## Calculus with Applications (10th Edition)

$f(x,y)=3x^{2}+7y^{3}-42xy+5$ $f_x(x,y)=6x-42y=0$ $f_y(x,y)=21y^{2}-42x=0 \rightarrow x=\frac{y^{2}}{2}$ Use the substitution method to solve the system of equations $f_x(x,y)=6x-42y=0 \rightarrow 6(\frac{y^{2}}{2})-42y=0$ $3y^{2}-42y=0$ $y=14 \rightarrow x=98$ or $y=0 \rightarrow x=0$ The critical points are (98,14) and (0,0) To identify any extrema: $f_{xx}(x,y)=6$ $f_{yy}(x,y)=42y$ $f_{xy}(x,y)=-42$ For (0,0): $f_{xx}(0,0)=6$ $f_{yy}(0,0)=0$ $f_{xy}(0,0)=-42$ $D=6*0 - (-42)^{2}=-1764\lt0 \rightarrow$ There is a saddle point at (0,0) For (98,14): $f_{xx}(98,14)=6$ $f_{yy}(98,14)=588$ $f_{xy}(98,14)=-42$ $=6*588- (-42)^{2}=1764\gt0$ and $f_{xx}(98,14)=6\gt0 \rightarrow$ There is a relative maximum at (98,14)