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Second derivatives and local max/min

  1. Nov 8, 2008 #1
    If f(x,y)= xy+(1/x)+(1/y)

    I find that fx= y-(1/x2)=0 and fy=x-(1/y2)=0.

    Solving for coordinates x and y by substituting the equations, I find that y=0 or 1. However, if i try to solve for x-coordinates with y=0, I get an infinity. So does that mean I ignore the possibility of y=0?

    Another question is with second-derivative test where D=fxx*fyy-fxy2.

    I know that if D>0 and fxx<0, then it is a local max. But what if I was told D>0 and given only fyy<0 and was to determine if it's a local max or min. Could I make the same argument with only fyy<0 that it is a local max?

    Thanks a lot
  2. jcsd
  3. Nov 9, 2008 #2


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    Yes. Any (x,y) point with y= 0 is not in the domain of the function.

    -fxy2 is always negative. In order that D be positive it is necessary that fxxfyy be positive (and greater than fxy2). That means that fxx and fyy must have the same sign. Knowing that fyy< 0 immediately tells you that fxx< 0.

  4. Nov 9, 2008 #3
    Oh right. Thanks for clearing that up HallsofIvy :smile:
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