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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.

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    $\begingroup$ Do not neglect to check the boundary square. You could have an edge or corner extreme. $\endgroup$ Aug 3 '15 at 20:36
  • $\begingroup$ You will need the second derivatives $f_{xx}, f_{yy}, f_{xy}$. Then apply the mathworld.wolfram.com/SecondDerivativeTest.html. Definitely check the boundaries - absolute minima and maxima are not always local min/max. $\endgroup$
    – Marconius
    Aug 3 '15 at 20:45
  • $\begingroup$ @Marconius thanks for the link. Just experimented with the point $f(1,1) $ since it appear to be extremes and got 7! So is that the absolute maximum? $\endgroup$
    – Obinoscopy
    Aug 3 '15 at 21:00
  • $\begingroup$ No, you only apply the second derivative test to the candidate min/max points that you found previously: $(0,0), (0,\frac{1}{2})$. Convexity doesn't tell you whether a point is a local extrema or not. $\endgroup$
    – Marconius
    Aug 3 '15 at 21:57
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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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  • $\begingroup$ The values are $4,4.75, 4.75, 8, 8, 7, 11, 11, 7$ respectively $\endgroup$
    – Obinoscopy
    Aug 3 '15 at 21:21
  • $\begingroup$ Right -- so the largest is the absolute max, and the smallest is the absolute min. $\endgroup$
    – user84413
    Aug 3 '15 at 21:25

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