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Lagrang multipliers to find max and min

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data

    The temperature at a point (x, y) on a metal plate is T(x, y) = 4x^2 − 4xy + y^2 .
    An ant, walking on the plate, traverses a circle of radius 5 centered at the origin.
    Using the method of Lagrange multipliers, find the highest and lowest
    temperatures encountered by the ant.


    2. Relevant equations



    3. The attempt at a solution

    T(x,y) = 4x^2 − 4xy + y^2
    gradient of T = (8x - 4y)i + (2y - 4x)j

    g(x,y) = x^2 + y^2 = 5^2
    gradient of g = (2x)i + (2y)j

    gradient of T = (lambda)gradient of g ----> lambda=#

    8x - 4y = #2x ---->1
    2y - 4x = #2y ---->2

    # = 4-4y = 1-4x

    what am i going to do next?:confused:
     
  2. jcsd
  3. Nov 23, 2009 #2

    lanedance

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    Homework Helper

    haven't checked your work, but notice you also have the original constraint, use that with the equation that you found
     
  4. Nov 23, 2009 #3
    what is the meaning of original constraint?
     
  5. Nov 23, 2009 #4

    lanedance

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    you have a function to optimise T(x, y) = 4x^2 − 4xy + y^2, against a given constraint x^2 + y^2 = 5^2
     
  6. Nov 23, 2009 #5
    if i solved the previous equation, i get, x=(1/4)y
     
  7. Nov 24, 2009 #6
    i've got max= 45 and min=5..
    am i done it right?
     

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  8. Nov 24, 2009 #7

    HallsofIvy

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    Yes, what you have done so far is correct. But that certainly does NOT give "max= 45 and min=5"!
     
  9. Nov 24, 2009 #8
    ok...
    i've checked my calculation already...
    when y = 2x,
    i substitute into x^2+y^2=25
    i got x=+-(5)^1/2

    when x=-2y,
    i substitute into x^2+y^2=25
    i got y= +-(5)^1/2

    so.. i got my critical points..
    ---> [(5^1/2) , (5^1/2)]
    ---> [(5^1/2) , -(5^1/2)]
    ---> [-(5^1/2) , (5^1/2)]
    ---> [-(5^1/2) , -(5^1/2)]

    my critical points are correct?:uhh:
     
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