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Math Help - Lagrange multipliers

  1. #1
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    Lagrange multipliers

    Q1) Find the ABSOLUTE MAXIMUM AND MINIMUM of the function f(x,y) = 4x^3 +y^2 subject to the constraint 2x^2 +y^2 =1 .

    Q2) Find the MAXIMUM AND MINIMUM of the function f(x,y) = 4x^3 +y^2 subject to the constraint 2x^2 +y^2 =1 .

    Are they asking to solve for the same thing? Do they have the same answer?
    Please advise me how to complete them. Thank you very much.
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  2. #2
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    Quote Originally Posted by littlemu View Post
    Q1) Find the ABSOLUTE MAXIMUM AND MINIMUM of the function f(x,y) = 4x^3 +y^2 subject to the constraint 2x^2 +y^2 =1 .
    We have,
    f(x,y)=4x^3+y^2
    c(x,y)=2x^2+y^2 with condition c(x,y)=1.

    Thus,
    grad (f) = k*grad (c)

    Thus,
    <12x^2,2y> = k * <4x,2y>

    Thus,
    12x^2 = 4kx
    2y = 2yk
    2x^2+y^2=1

    Simplify them by canceling the constants,
    3x^2 = kx (1)
    y=yk (2)
    2x^2+y^2 = 1 (3)

    Look at equation (2),
    It tells us:
    y=0 or k=1.

    y=0
    Then, (3) tells us that,
    2x^2 = 1
    Thus, x=+/- sqrt(2)/2

    k=1
    Then, (1) tells us that,
    3x^2 = x
    Thus,
    x=0 or x=1/3.

    If x=0 then (3) tells us that y=+/- 1.
    If x=1/3 then (3) tells us that y=+/- sqrt(7)/3

    Thus, all the critical points are:
    (-sqrt(2)/2,0)
    (sqrt(2)/2,0)
    (0,1)
    (0,-1)
    (1/3, -sqrt(7)/3)
    (1/3), sqrt(7)/3)

    Check each one for function:
    f(x,y) = 4x^3 +y^2
    To see the maximum and minimum.
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  3. #3
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    Thank you very much
    Last edited by littlemu; April 28th 2007 at 09:34 PM.
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