Calculus with Applications (10th Edition)

The critical points are $(126,2646)$ and (0,0). There is a saddle point at (0,0). There is a relative maximum at $(126,2646)$
$f(x,y)=7x^{3}+3y^{2}-126xy-63$ $f_x(x,y)=21x^{2}-126y=0$ $f_y(x,y)=6y-126x=0 \rightarrow y=21x$ Use the substitution method to solve the system of equations $f_x(x,y)=21x^{2}-126y=0 \rightarrow 21x^{2}-126(21x)=0$ $21x^{2}-2646x=0$ $x=126 \rightarrow y=2646$ or $x=0 \rightarrow y=0$ The critical points are $(126,2646)$ and (0,0) To identify any extrema: $f_{xx}(x,y)=42x$ $f_{yy}(x,y)=6$ $f_{xy}(x,y)=-126$ For (0,0): $f_{xx}(0,0)=0$ $f_{yy}(0,0)=6$ $f_{xy}(0,0)=-126$ $D=6\times0 - (-126)^{2}=-15876\lt0 \rightarrow$ There is a saddle point at (0,0) For $(126,2646)$: $f_{xx}(126,2646)=5292$ $f_{yy}(126,2646)=6$ $f_{xy}(126,2646)=-126$ $D=5292\times 6- (-126)^{2}=158766\gt0$ and $f_{xx}(126,2646)=5292\gt0 \rightarrow$ There is a relative maximum at $(126,2646)$