<s>[INST] Absolute Min and Max of $f(x, y)=x^2+4y^2-2x^2y+4$ Using Partial Derivatives Consider this problem:
Find the absolute minimum and absolute maximum of $f(x, y)=x^2+4y^2-2x^2y+4$ on the rectangle given by 
$-1\leq x\leq1$ and $-1\leq y\leq1$
I solved this problem using partial derivatives:
I got $f_x=2x-4xy=0$
And $f_y=8y-2x^2=0$
Solving simultaneously gives the following critical points: $(0,0), (0,\frac{1}{2}), (\sqrt2,\frac{1}{2}) $
And with the restriction: $-1\leq x\leq1$ and $-1\leq y\leq1$
The critical points becomes: $(0,0), (0,\frac{1}{2}) $
$f(0,0)=4 $ and $f(0,\frac{1}{2}) =5$
Now the quagmire is how to use partial derivatives (probably of second order) to determine which is the absolute minimum and maximum. 
Plotting a graph of $z=x^2+4y^2-2x^2y+4$ suggests that:
$(0,0,4) $ is the absolute minimum and
$(0,\frac{1}{2},5) $ the maximum
But I need to be sure I'm on the right path. I also need to know how to use partial derivatives to differentiate a minimum critical point from a maximum critical point. 
 [/INST] Hint:
Since $2x-4xy=0\implies 2x(1-2y)=0\implies x=0$ or $y=\frac{1}{2}$ and
$8y-2x^2=0\implies x^2=4y$,  either $x=0, y=0$ or $y=\frac{1}{2}, x=\pm\sqrt{2}$.
Therefore $(0,0)$ is the only critical point in the interior of the region.

On the left and right edges of the rectangle,  $g(y)=4y^2-2y+5$ and $g^{\prime}(y)=8y-2=0$ if $y=\frac{1}{4}$.
On the top edge, $k(x)=-x^2+8$ so $k^{\prime}(x)=-2x=0$ if $x=0$.
On the bottom edge, $l(x)=3x^2+8$ so $l^{\prime}(x)=6x=0$ if $x=0$.

Now you just need to compare the values of the function at the points
$(0,0)$ and $(1, \frac{1}{4}),\; (-1, \frac{1}{4}),\; (0,1), \;(0,-1), \;(-1, 1), \;(-1, -1),\; (1, -1),\; (1,1)$
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