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Find all critical points of $f(x,y) = x^3 - 12xy + 8y^3$ and state whether the function has a relative minimum, relative maximum, or a saddle at the critical points.

So I have:

$f_x = 3x^2 -12 y$

$f_y = -12x + 24y^2$


$f_{xx} = 6x$

$f_{yy} = 48y$

$f_{xy} = -12$


I found that my critical points were: $(0,0)$ and $(2,1)$, but I still need to classify them.

My question is, how do I check if $f_{xx}$ is positive or negative? This seems like a silly question (I'm sure it is), but do I plug in $(0,0)$ for $f_{xx}$? Wouldn't that then be $6\cdot 0 > 0$? That seems wrong.

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    $\begingroup$ $6 \cdot 0 = 0$, so $f_{xx}(0, 0)$ is neither positive or negative. $\endgroup$ – user61527 Aug 22 '13 at 7:04
  • $\begingroup$ Yes, you do plug in. $\endgroup$ – André Nicolas Aug 22 '13 at 7:07
  • $\begingroup$ @T.Bongers So, plugging in $(0,0)$ for $f_{xx}$ is what I do? I just take the x coordinate from $(0,0)$? Wait, if it's neither positive nor negative, does that mean this is a saddle point? $\endgroup$ – Ozera Aug 22 '13 at 7:08
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    $\begingroup$ @Ozera I assume you're trying to use the second partial derivative test. Note that if we evaluate $f_{xx} f_{yy} - f_{xy}^2$ at $(0, 0)$, we get $0 \cdot 0 - (-12)^2 = -144 < 0$, so the sign of $f_{xx}$ is irrelevant. $\endgroup$ – user61527 Aug 22 '13 at 7:12
  • $\begingroup$ O, that's true. I understand now. Thanks. $\endgroup$ – Ozera Aug 22 '13 at 7:12
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Given:

$$\tag 1 f(x,y) = x^3 - 12xy + 8y^3$$

Find and classify all critical points.

The critical points are: $(0,0)$ and $(2,1)$.

The partial derivatives are:

  • $f_x = 3x^2 -12 y$
  • $f_y = -12x + 24y^2$
  • $f_{xx} = 6x$
  • $f_{yy} = 48y$
  • $f_{xy} = f_{yx} = -12$

The Hessian determinant is given by:

$$\det(H) = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{vmatrix} \ $$

If you are using the Hessian, there are four conditions you need to test:

  • $(01)$ $f_{xx} \gt 0$ and $\det H \gt 0 \rightarrow$ local minimum
  • $(02)$ $f_{xx} \lt 0$ and $\det H \gt 0 \rightarrow$ local maximum
  • $(03)$ $\det(H) \lt 0 \rightarrow$ saddle point
  • $(04)$ $\det(H) = 0 \rightarrow$ no statement can be made using this approach

For $(0,0)$, condition $(4)$ tells us that $\det H = 0$, so nothing can be said about this critical point, neither a min or max. Of course, we could have also looked at $$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2.$$

Since $g(x,y) \lt 0$, this is not an extremum.

For $(2,1)$, we have: $f_{2,1} = 12 > 0$ and $\det H = 432 >0 \rightarrow$ a local minimum. The value of $f(x,y)$ at this minimum is $f(2,1) = -8$.

Plots are given by:

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  • $\begingroup$ Talk about thorough! You nailed it! +1 $\endgroup$ – Namaste Aug 23 '13 at 0:35

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