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Maximum and Minimum Values (Trig)

  1. Oct 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the 4 critical points f(x,y) = 5ycos(9x) closest to (0,0)

    2. Relevant equations


    3. The attempt at a solution

    fx = -45ysin(9x)
    fy = 5cos(9x)
    fxx = -45*9ycos(9x)
    fyy = 0
    fxy = -45sin(9x)

    y=0
    x=pi/18

    (0,pi/18) (0,pi/18 + pi/2) (0,pi/18 - pi/2) (0,pi/18 + 3pi/2) Was not correct.

    (All 4 points need to be correct for me to check it )
     
    Last edited: Oct 17, 2014
  2. jcsd
  3. Oct 17, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Usually the x-coordinate comes first and y second in the notation.
    And check the spacing of the solutions in x, it is not pi/2.
     
  4. Oct 17, 2014 #3
    Oops, I tried it correctly first but then swapped to see if that helped with anything, i put it back in x/y

    But, i redid the problem and got..
    y=0, x = (1/18)pi(4n-1)
    y=0, x = (1/18)pi(4n+1)

    Not sure how to translate that to the 4 closest points? Tried this but didn't work..

    ((1/18)pi(4-1),0), ((1/18)pi(0-1),0), ((1/18)pi(4+1),0), ((1/18)pi(0+1),0)
     
  5. Oct 17, 2014 #4

    mfb

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    2016 Award

    Staff: Mentor

    I don't know how powerful the automatic (?) system to check the points is. I would not try to feed it with calculations like that. Maybe it needs decimal numbers, maybe something like 3/18 pi is okay (but where is the point in things like "4-1"?
    ((1/18)pi(4+1),0) is not among the 4 closest points.
     
  6. Oct 17, 2014 #5
    Ah, thanks a bunch!
    Solution:
    ((1/18)pi(3),0), ((1/18)pi(-1),0), ((1/18)pi(-3),0), ((1/18)pi(1),0)
     
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