1. PF Insights is off to a great start! Fresh and interesting articles on all things science and math. Here: PF Insights

Finding local min, max, and saddle points in multivariable calculus

  1. 1. The problem statement, all variables and given/known data

    Find the local maximum and minimum values and saddle point(s) of the function.
    f(x,y) = (e^x) * cosy

    The textbook answer is "none." I sort of reached that conclusion but I don't know if my reasoning is correct...

    2. Relevant equations

    The Second Derivative Test: let D = D(a,b) = fxx(a,b)*fyy(a,b) - [fxy(a,b)]^2
    if D > 0 and fxx(a,b) > 0, then f(a,b) is a local minimum
    if D > 0 and fxx(a,b) < 0, then f(a,b) is a local maximum
    if D < 0 then f(a,b) is a saddle point
    if D = 0 then the test is inconclusive

    3. The attempt at a solution

    Since the textbook answer is none, I suppose in my answer I should demonstrate how I came to the conclusion that there are no mins, maxes, or saddle points. I worked something out that seems to lead to that conclusion, but I'm not sure if my reasoning makes sense. That's the thing I really would like help with--does this line of reasoning lead to the correct conclusion?

    I've took the first partial derivatives and set them equal to zero:

    fx = (e^x) * cosy = 0
    fy = -(e^x) * siny = 0

    and since e^x never reaches 0, the only points where these statements can be true is where

    cosy = 0 and/or siny = 0
    which occurs at pi/2, -pi/2 and 0

    Next, take the relevant second partials:

    fxx = (e^x)*cosy
    fyy = -(e^x)*cosy
    fxy = -(e^x)*siny

    D = (e^x)cosy*(-e^x)cosy - [(-e^x)siny]^2
    = -(e^2x)*cos(y)^2 - (e^x)*sin(y)^2
    = -(e^2x) * [ cos(y)^2 + sin(y)^2 ]
    = -(e^2x)

    Since D is always < 0 (since e^2x is always positive), this analysis suggests there is a continuous line of saddle points, which is not possible/doesn't make sense. So, there must not be any local mins, maxes, or saddle points.

    Is it correct to reason that way?

    Any help, advice, or criticism would be much appreciated.
  2. jcsd
  3. I think that here:
    fx = (e^x) * cosy = 0
    fy = -(e^x) * siny = 0
    Is the key. Because those partial derivatives are never zero simultaneously. Which means that there are no critical points.
  4. ah, that makes sense. thanks for your help.
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?