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Math Help - find absolute extrema & Lagrange Multipliers

  1. #1
    ggw
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    find absolute extrema & Lagrange Multipliers

    1. given: f(x,y) = x^3 - 6xy + y^2 +15x

    find absolute extrema of f over the triangluar region in the xy plane with vertices (0,0), (0,4) & (2,4).


    2. given: T(x,y,z)=100+x^2 + y^2 represent the temperature at each point on the sphere x^2 + y^2 + z^2 =50. Use Lagrange Multipliers to find the Maximum & Minimum temperature on the curve formed by the intersection of the sphere and the plane x - z = 0. (hint: Optimizing a function subject to two constraints)


    thanks for your big help....!
    Last edited by ggw; April 1st 2007 at 06:48 PM.
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  2. #2
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    Quote Originally Posted by ggw View Post
    1. given: f(x,y) = x^3 - 6xy - y^2 +15x

    find absolute extrema of f over the triangluar region in the xy plane with vertices (0,0), (0,4) & (2,4).

    First let us find the equation of each side of the triangle.

    (0,0),(0,4)
    It is the equation of the line in the xy plane which connects those two.
    Hence,
    y=4.

    (0,0),(2,4)
    y=2x

    (0,0),(0,4)
    x=0

    Inside Triangle
    f_x=0
    f_y=0

    Thus,

    3x^2-6y+15=0
    -6x-2y=0

    Solve, (you are big enough to solve the system thyself).

    There exists solutions but not inside the open boundary.

    Hence we check the boundary itself.

    Line y=4
    f(x,4)=x^3-24x-16+15x=x^3-9x-16
    Find extreme values on (0,2) and you find there are none.
    Now check its boundaries x=0 and x=2
    Hence,
    f(0,4)=-16 and f(2,4)=8-18-16=-26

    Line x=0
    f(0,y)=-y^2
    Find extreme values on (0,4) and there are none.
    Now check its boundaries y=0 and y=4.
    f(0,0)=0 and f(0,4)=-16

    Line y=2x
    f(x,y)=x^3-12x^2-4x^2+15x=x^3-16x^2+15x
    Find extreme values on (0,2) and there are none.
    Now check its boundaries x=0 and x=2.
    f(0,0)=0 and f(2,4)=8-64+30=-26

    The smallest values was at,
    (2,4)
    The largest value was at,
    (0,0)
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  3. #3
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    I hope the following attachment helps, because this is only half of the solution.
    Attached Thumbnails Attached Thumbnails find absolute extrema & Lagrange Multipliers-picture2.gif  
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  4. #4
    ggw
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    missing plus sign

    for the f(x,y) = x^3 -6xy + y^2 +15

    the function should be like this, and the critical point should be x =1, y=3. I dont know are there any other criticals point might have in the function.
    Last edited by ggw; April 3rd 2007 at 12:33 PM.
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    Quote Originally Posted by ggw View Post
    for the f(x,y) = x^3 -6xy + y^2 +15

    the function should be like this, and the critical point should be x =1, y=3. I dont know are there any other criticals point might have in the function.
    f_x=3x^2-6y=0
    f_y=-6x+2y=0

    (x,y)={(0,0),(6,18)}
    The only one inside the triangle is,
    (0,0).
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  6. #6
    ggw
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    here is what i did
    Partial Derivatives
    fx =3x^2 -6y + 15
    fy =-6x+2y solve for x & y i got. x = 1, y =3
    => critical points is (1,3)
    f(1,3) = 7
    Line 1: y =4 0<=x<=2
    f(x,4)=x^3 -9x +16
    so, min is f(0,0) = 0, max is f(2,0)=6
    Line 2: x = 0 0<=y<=4
    f(0,y) = y^2
    so, min is f(0,0) = 0, max is f(0,4) =16
    Line 3: y = 2x 0<=x<=2
    f(x,y) = x^3 -8x^2 + 5x
    so, min is f(0,0) =0, f(2,0) = -14 there are none

    what would i conclude this problem, and am I corretct ?


    The second questions, I am not sure if i understand well. Thanks for your help.
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  7. #7
    ggw
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    i am still woring out on problem 2,

    here is what i got

    y = +- sqrt(50)

    and (0,150,0) is relative max. Not really sure about it, need help...!
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