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Extrema of function

  1. Jun 12, 2010 #1
    given this function

    f(x) = x^3+y^3+3x^2-18y^2+81y+5

    i should i find the extrema(i hope this is how it is called in english)

    so we should have gradf = 0

    hence
    3x^2+6x = 0 and 3y^2-36y+81 = 0

    here i get x = 0 or x = -2 and y = 54/9 or y = 2

    now what i want to ask is, which points will i have to take in order to check of extrema?

    it will be
    P1 = (0,54/9) and P2 = (-2,2)?
    P1 = (0,2) and P2 = (-2,54/9)?

    which one is the correct?

    Thanks in advance
     
  2. jcsd
  3. Jun 12, 2010 #2

    lanedance

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    Homework Helper

    i don't think your y values are correct... try subtituting them back in
    3y^2-36y+81 = 3(y^2 -12 +27) = 0

    you will need to check all of the points, either by subtituting in or a 2nd derivative check
     
  4. Jun 12, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    First, as lanedance said, your solution for y is incorrect (it is easy to see that y= 2 does not satify the equation: 3(4)- 36(2)+ 81= -72+ 93= 21, not 0).

    Second, once you have correct solutions for x and y, since the two equations are completely separate, any value of x can be used with any value of y- there are four critical points.
     
  5. Jun 12, 2010 #4
    thanks a lot for your help :)
     
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