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Calculus 3

  1. Jul 23, 2009 #1
    ]I have to find the local/absolute extrema for the following function:
    f(x,y)=x^2+y^2

    bounded by the triangle x=0,y=0,y+2x=2

    So far i have:
    fx(x,y)=2x
    fy(x,y)=2y
    fxx(x,y)=2
    fyy(x,y)=2
    fxy&fyx(x,y)=0

    critical pts at (0,0,0)
    domain: 0<=x<=1, 0<=y<=2

    i dont know what i should do next?
     
  2. jcsd
  3. Jul 23, 2009 #2

    Office_Shredder

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    Since it's only a function of two dimensions, the critical point you get is actually at (0,0). Notice that there are precisely two points where a maximum or a minimum can occur: At a critical point, or on the boundary of the domain. So now you have to check the boundary for any maxima (that you found a critical point on the boundary is pure coincidence; the rest of the boundary still needs to be checked)
     
  4. Jul 23, 2009 #3
    So this is what i get:

    local max: (0,2,4)
    abs. min:(1,0,1)
    abs. max:(1,2,5)
     
  5. Jul 24, 2009 #4

    Office_Shredder

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    What happened to the critical point you found earlier?
     
  6. Jul 24, 2009 #5
    (0,0,0) is an absolute min.

    how does that look now?
     
  7. Jul 24, 2009 #6
    with (1,0,1) being a local min.
     
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