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Finding absolute minimum and maximum values

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the absolute minimum and maximum values of f on the set D.

    f(x,y)= e-x2-y2(x2+2y2); D is the disk x2+y2 <= 4



    2. Relevant equations

    Second Derivatives test,
    partial derivatives

    3. The attempt at a solution

    fx(x,y) = 0 = (e-x2-y2)(-2x) + (x2+2y2)(-2x e-x2-y2)

    fy(x,y) = 0 = (e-x2-y2)(4y) + (x2+2y2)(-2y e-x2-y2)

    fxy(x,y) = (e-x2-y2)+(-2x)(-2y e-x2-y2) + (x2+2y2)(-2x*-2y e-x2-y2) + (-2x e-x2-y2)(4y)

    fx and fy simplify to:
    fx (x,y) = 1+x2+2y2 = 0
    fy (x,y) = -2y+x2+2y2 = 0

    I'm stymied here because the equation I get for fx seems impossible to solve. Did I make a mistake differentiating?
     
  2. jcsd
  3. Dec 13, 2008 #2

    Dick

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    Homework Helper

    Yes, there's a sign error in f_x. But f_y looks ok, and I don't think (4y)-(2y)*(x^2+2y^2) simplifies to what you got for f_y.
     
  4. Dec 13, 2008 #3
    ah.

    ok.

    so I should have ended up with
    fx= x2+2y2 -1 = 0 and
    fy=x2+2y2-2 = 0

    So to find the critical points I have to solve these two equations.

    Would I be on the right track to say fxthen simplifies to (x-1)(x+1)+2y2 = 0


    If so, I'm embarassed to say I'm still not sure how to proceed from here. I find the examples in my book don't really adapt themselves very well to the exercises, so I'm having trouble getting to the next step. : (
     
  5. Dec 13, 2008 #4
    ok, so it looks like I got my approach wrong.

    The way it's worked out it looks like the e distributes to the other term, and then I should differentiate, which yields


    e-x^2-y^2*2x =- 2x^3*e-x^2-y^2

    and
    e-x^2-y^2 * 4y-4y^3e-x^2-y^2


    which reduces to
    2x(1-x^2) = 0
    4y(1-y^2) = 0

    so we have critical points at x=0,1,-1
    and cp at y=0,1,-1

    we also need to check for extreme values on the boundary.
    now, the boundary function is [tex]x^2+y^2=4[/tex] so [tex]e^(-x^2+y^2)[/tex] is [tex]e^-4[/tex]

    and means that [tex]x^2=4-y^2[/tex]
    so if we plug into our original function, we get

    [tex]f(y)=e^-4 (4-y^2+2y^2)[/tex] for -2<=y<=2

    so
    [tex]f'(y)=e^-4*2y=0[/tex]
    solving for y we get y=0, and plugging that into our boundary equation [tex]x^2+y^2=4[/tex] and solving for x we get x=+/- 2

    There's more, where I have to set something up with Lagrange, but I'm still working on understanding that part. I'll post the rest once I understand it a little better.
     
  6. Dec 13, 2008 #5

    HallsofIvy

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    Of course, it doesn't necessarily follow that there is a critical point inside the circle- and even if there is the max or min are not necessarily there. It is also possible for the max or min (or both) to occur on the boundary of the set which, here, is the circle itself.
     
  7. Dec 13, 2008 #6

    Dick

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    I get f_x=2x-2x(2y^2+x^2)*exp(...) and f_y=4y-2y(2y^2+x^2)*exp(...) and you came pretty close to getting it right the first time. I don't know what you did in your last response. I get five critical points.
     
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