10.7 Maxima / minima / other

For the function $f(x,y)=2 x^2 + y^2 - 4 x y - 20 y + 3$

1. Compute the following derivatives:
   
   $f_x (x,y)$ =$2*2x+0-4y-0+0=$$4x-4y$
   
   
   
   $f_y(x,y)$ =$0+2y-4x-20+0=$$2y-4x-20$
   
   
   
   $f_{xx}(x,y)$ =$\frac{\del}{\del x}(4x-4y)=$4
   
   
   
   $f_{yy}(x,y)$ =$\frac{\del}{\del y}(2y-4x-20)=$2
   
   
   
   $f_{xy}(x,y)$ =$\frac{\del}{\del y}(f_x)=\frac{\del}{\del y}(4x-4y)=$ -4
2. Give $x$- and $y$-values for any critical points of $f(x,y)$.
   Critical points occur when:
   $$f_x=0=4x-4y\ \Rightarrow\ x=y$$ AND $$f_y=0=2y-4x-20\ \Rightarrow 10=y-2x$$ Substituting $x=y$ in to $10=y-2x=x-2x=-x\ \Rightarrow \ x=-10$ and $y=-10$. There is a critical point at (-10,-10).
3. For each critical point, determine whether $f(x,y)$ has a local maximum at that point, a local minimum, or neither.
   Using $f(x,y)=f(x,y)=2 x^2 + y^2 - 4 x y - 20 y + 3$, calculate some values near the critical point:

   	-11	-10	-9	$x$
   -9	110	104	102
   -10	105	103	105
   -11	102	104	110
   $y$

   Looks like it's a minimum. Even though $f_{xx}=4$ and $f_{yy}=2$ are both positive--concave up, it still appears that in one of the diagonal directions, $f$ is going *down*.

   Using the second derivative test: $$D=f_{xx}f_{yy}-(f_{xy})^2=2\cdot 4-(-4)^2=8-16=-8.$$ If $D\lt 0$ that means we have neither a maximum nor a minimum. In this case we have a saddle point.

4. Make a contour plot (SageMath) and confirm your answer about the critical point.