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Finding min and max

  1. Oct 20, 2011 #1
    So we first find where fx(x,y) = 0 and fy(x,y) = 0, where fx and fy are the partial derivatives of z = f(x,y). Once we find those critical points, we use D = (fxx)(fyy) - (fxy)^2.
    If D > 0 and fxx > 0, we have a local min at that point.
    If D > 0 and fxx < 0, we have a local max at that point.
    If D < 0, we have a saddle point.
    If D = 0, no information can be found using the second derivative test.

    My question is:
    1. How do we deal with the D = 0 situation? How would we find if that point's a max or min?
    2. What if fxx = 0?
  2. jcsd
  3. Oct 23, 2011 #2
  4. Oct 23, 2011 #3


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    Homework Helper

    The D = 0 case has to be examined more closely, but the means for doing that are not usually discussed in a first multivariate course.

    It is analogous to what happens in single-variable calculus. If both f'(a) and f''(a) equal zero, it could mean that there is a point of inflection at x = a [e.g., x = 0 for f(x) = x3 ], but there could also be a very flat maximum or minimum there [as with x = 0 for f(x) = x4 ] .

    What do you check? If there is a change of concavity at x = a (that is, the sign of f''(x) changes as x passes through a ) , then there is a point of inflection there. If the concavity does not change, x = a is an extremum.

    You would need to do something comparable for z = f(x,y) , but now you have to make checks in two dimensions. (Imagine the fun you can have with functions of even more variables...)

    [Oh, and please don't 'bump' posts in PF; it doesn't get attention any faster, it just confuses the process of having Helpers check for threads needing help. (It raises the reply count and makes it look like the thread has already been getting help...)]
  5. Oct 24, 2011 #4


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    Science Advisor

    If [itex]f_{xx}= 0[/itex] then [itex]D\le 0[/itex].
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