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Critical Points in 3D

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the critical points of the function f(x, y) = sinx + siny + cos(x+y)
    where 0<=x<=pi/4 and 0<=y<=pi/4


    2. Relevant equations
    First and second order partial derivative of f(x, y)


    3. The attempt at a solution
    To find the critical points, I first find the first partial derivative with respect to x and y.

    fx(x,y) = cosx - sin(x+y)
    fy(x,y) = cosy - sin(x+y)

    Set both of the first partial derivative = 0
    cosx = sin(x+y) x=pi/4 and y=0
    cosy = sin(x+y) x=0 and y=pi/4

    Here is where I got stuck, I noticed two points where the above equations are true, but how do I find all the critical points?

    I checked the two points I found using
    D = fxxfyy - (fxy)^2
    fxx = -sinx - cos(x+y)
    fyy = -siny - cos(x+y)
    fxy = -cos(x+y)

    and I found that D for both points are less than zero, which suggest that they both are saddle points. I then graphed the function f(x, y) and these two points don't appear to be saddle points on f. I assume that I made some mistake somewhere either in the derivative or the graph, if someone can check this for me, it would be greatly appreciated.
     
  2. jcsd
  3. Oct 5, 2009 #2

    LCKurtz

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    Gold Member

    You said:

    fx(x,y) = cosx - sin(x+y)
    fy(x,y) = cosy - sin(x+y)

    Set both of the first partial derivative = 0
    cosx = sin(x+y) x=pi/4 and y=0
    cosy = sin(x+y) x=0 and y=pi/4

    But you must solve these simultaneously. You have picked values that solve them separately, which is irrelevant. But what you can get from those last two equations is that cos(x) = cos(y). And on [0, π/4] that means x = y. Use that in your equation for fx = 0 to see if you can find critical points in your domain.

    Depending on what your problem asks for, you may need to check the boundaries too.:yuck:
     
  4. Oct 5, 2009 #3
    Thanks LCKurtz, I haven't solve it yet but your suggestion is really helpful. It makes a lot more sense when I look at the graph. The maximum of the surface is indeed a point where x and y are equal.
     
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