Answer
The critical points are (98,14) and (0,0).
There is a saddle point at (0,0).
There is a relative maximum at (98,14)
Work Step by Step
$f(x,y)=3x^{2}+7y^{3}-42xy+5$
$f_x(x,y)=6x-42y=0$
$f_y(x,y)=21y^{2}-42x=0 \rightarrow x=\frac{y^{2}}{2}$
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)
