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Homework Help: Absolute Min/Max, Bounded region

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the abs min/max values of the function

    f(x,y) = e1-2x2-y2

    on the closed and bounded region x2 + y2 <= 1

    3. The attempt at a solution

    First I have to find the critical points

    Dfx = (-4x)e1-2x2-y2
    Dfy = (-2y)e1-2x2-y2

    Clearly e1-2x2-y2 cannot equal 0, therefore

    x=y=0, Critical point is (0,0)

    Now, my professor has done most of these problems by setting
    [tex]\nabla[/tex]f = [tex]\lambda[/tex][tex]\nabla[/tex]g, the gradients of each function

    Even with this method yields x=0 and y=0 to be the only solution.

    I am not too sure how to incorporate the bounded region into this question.

    Do I just look at x=0 and y=0 of x2 + y2 = 1?
    This would result in x2 + 0 = 1
    x=y= +/- 1 giving me two points, (-1, -1), and (1,1), however f(-1,-1) > 1 and

    f(1,1) = e-2 thus a min/max value for the function?

    Is this correct? Any insight would be great, thanks.
  2. jcsd
  3. Nov 9, 2009 #2


    Staff: Mentor

    Neither of the points you found is in your region, which is a circular disk of radius 1, centered at (0, 0).
  4. Nov 9, 2009 #3
    How am I supposed to proceed then using my critical point and incorporating the bound?

    I guess since it is a disk, I can use sin(t) and cos(t) as the coordinates to represent the boundry c(t)= (sint, cost) 0 <= t <= 2[tex]\pi[/tex]

    f(c(t)) = e1-2sin2t-cos2t ?

    At both of the boundaries 0 and 2[tex]\pi[/tex] , f = 1.

    Kinda confused otherwise on computing this
  5. Nov 9, 2009 #4


    Staff: Mentor

    You need to check f at all points on the boundary, not just at t = 0 and t = 2pi. Your exponent on e is 1 - 2sin2 t - cos2 t. I would use an identity to get this exponent in terms of sin or cos alone.
  6. Nov 9, 2009 #5
    Ah ok so now I have


    Now the max and min values of the sin function are [tex]\pi[/tex]/2 and 3[tex]\pi[/tex]/2

    These are the values I look at?
  7. Nov 10, 2009 #6


    Staff: Mentor

    The maximum and minimum values of the sine function are 1 and -1, and they are attained when x is pi/2 and 3pi/2, respectively. The maximum value of -sin2 x is 0, which is attained at x = 0, x = pi, and x = 2pi. The minimum value of this function is -1, attained at x = pi/2 and x = 3pi/2.
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