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

  1. Oct 24, 2010 #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) = 1 + 2xy - x^2 - y^2

    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

    I tried to use the Second Derivative Test to find the local mins, maxes, and saddle points but it's inconclusive, and I don't know how else to find them. My textbook says the answer is "f has a local maximum value of 1 at all points of the form (x, x)"

    This is my work for the Second Derivative Test:

    fx = 2y - 2x = 0 --> 2y = 2x --> y = x

    fy = 2x - 2y = 0 --> 2x - 2(x) = 0 --> 0 = 0

    so i guess there are critical points at every value where y = x... which matches the textbook's answer.
    and then:

    fxx = -2

    fyy = -2

    fxy = 2

    so D = fxx * fyy - (fxy)^2 = (-2)*(-2) - 2^2 = 4 - 4 = 0 so the test is inconclusive

    Is there a different way to find the local mins, maxes, and saddle points?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 24, 2010 #2
    got the answer, for anyone else who looks this up.

    Rewriting the function as f(x, y) = (x - y)^2 + 1,
    we see that the minimum value must be 1 (since 0 is the smallest value of a square),
    and this is attained whenever y = x (i.e., points of the form (x, x)).
     
  4. Oct 24, 2010 #3

    Mark44

    Staff: Mentor

    fx = 2y - 2x
    fx = 0 ==> y = x

    fy = 2x - 2y
    fy = 0 ==> x = y

    fx and fy are both zero along the line y = x.
    The function can be written as f(x, y) = 1 - (x2 - 2xy + y2), and the right side can be written in factored form, which should give you some ideas for finding the global maxima and minima.
     
  5. Oct 24, 2010 #4
    that does make a lot of sense... i never thought to take a more direct approach. thank you for your help!
     
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