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Stationary points

  1. Feb 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the stationary point(s) and also find out whether the stationary point(s) are minimum/maximum
    z = 2x + 5y - 5y2 - 5x2 - 4xy - 1000

    2. Relevant equations


    3. The attempt at a solution
    I'm not 100% sure if I was correct and laid the answer out properly, can someone please help me check it out and correct any mistakes

    fx = 2 - 10x - 4y
    fy = 5 - 10y - 4x
    fxx = -10
    fyy = -10
    fxy = -4

    2 - 10x - 4y = 0, therefore 4y = 2 and y = 0.5
    Possible stationary points (0,0.5) (0,-0.5)

    5 - 10y - 4x = 0, therefore 4x = 5 and x = 1.2
    Possible stationary points (0,0.5) (0,-0.5)
    (1.25,0) (-1.25,0)

    fxxfyy-f2xy = 84 (Greater than 0, and fxx & fyy are less than 0 therefore it's a maximum)
  2. jcsd
  3. Feb 27, 2014 #2


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    Don't understand the "therefore". You solve fx=0 for x=0 . There is only one solution. Where does the (0, -0.5) appear ?

    10x + 4y=2 is the equation for a straight line.

    Idem 5 - 10y - 4x = 0. (0, -0.5) does not satisfy that !

    What is/are the condition(s) for a stationary point ?
  4. Feb 27, 2014 #3


    Staff: Mentor

    4y = 2 only if x = 0, which you don't know. The equation 2 - 10x - 4y = 0 has an infinite number of solutions.
    No, 4x = 5 only if y = 0, which you also don't know.

    Solve the two equations simultaneously to find the point or points that are on both lines. If you're in a calculus class, you should know how to solve a system of two linear equations.
  5. Feb 28, 2014 #4
    Solving both of them simultaneously I got..


    Multiplying the top by 4 and the bottom by 10 gives..


    The x's cancel out, so..


    Subbing y into the original equation makes x=0.12

    I've done the main working out, could you please confirm how I would then present the answer
  6. Feb 28, 2014 #5


    Staff: Mentor

    You really make it hard on yourself. A little work in advance saves time and minimizes the chance for errors. I would write this system as
    10x + 4y = 2
    4x + 10y = 5

    No, but this might have been a transcription error. The first equation should be 8 - 40x - 16y = 0.
    Also, it makes more sense to multiply one equation by a positive number and the other by a negative number. Then you can add the equations rather than subtract one from the other.

    Correct, despite the error I noted.
    No. You're looking for the solution of the system of equations. Use your y value to find the x value for which fx = fy.
    When you have found the critical point, use the other partials to determine whether it's a local maximum, local minimum, or whatever.
  7. Feb 28, 2014 #6
    Thanks for the help I really appreciate it but if you're gonna help me then you might aswell help me set out the answer since I've done the main working outs

    Since I've concluded that a stationary point is at (0,0.5) and it's a maximum point since fxx & fyy are negative, also because fxxfyy-f2xy is positive. How would I actually present the answer?
  8. Feb 28, 2014 #7


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    You found one point where fx=0 AND fy=0 so that is a stationary point.

    Same question for criteria for maximum, minimum, saddle point.

    At (0, 0.5), fxx and fyy are both negative, so it is a maximum.

    You bring in fxy but I don't see much use for that.

    My question: this is homework for a chapter or section in your syllabus or textbook. Don't they summarize these conditions/criteria somewhere ?
  9. Feb 28, 2014 #8


    Staff: Mentor

    Pretty much like what you have above.
  10. Feb 28, 2014 #9

    Ray Vickson

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    The second mixed partials are crucial to overall behavior and cannot be simply dismissed. Having ##f_{xx}, f_{yy} < 0## does not guarantee a maximum; it does eliminate a minimum, but still leaves open the possibility of a saddle point. For example for both functions ## f = -x^2-y^2 \pm 100 xy## the origin is the only stationary point, but is a saddle point rather than a maximum.
  11. Feb 28, 2014 #10


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    Thanks Ray, I learned something. Must have gotten used too much to less pathologic functions! Retro did just fine then.
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