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Math Help - Lagrange Multiplier (difficult?)

  1. #1
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    Lagrange Multiplier (difficult?)

    f(x,y) = d = x + y
    g(x,y) = 3x-4xy+6y = 140

    Then by my calculations,
    grad(f) = 2xi + 2yj
    grad(g) = (6x-4y)i + (-4x+12y)j

    Then grad(f) = λgrad(g) produces these equations
    2x = λ(6x-4y) (eq.1)
    2y = λ(-4x+12y) (eq.2)

    rearranging 1 gives:
    λ = 2x/(6x-4y) (eq.3)
    rearranging 2 gives:
    λ = 2y/(-2x+6y) (eq.4)

    then by cross multiplying,
    -2x + 6xy = 3xy - 2y

    taking all terms to left hand side,
    -2x +3xy + 2y = 0
    (-2x - y)(x - 2y)=0
    therefore x = y/-2 or x = 2y

    This is where I am stuck, now I have a relationship between x and y, what do I do next??
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  2. #2
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    Quote Originally Posted by sk1001 View Post
    f(x,y) = d = x + y
    g(x,y) = 3x-4xy+6y = 140

    Then by my calculations,
    grad(f) = 2xi + 2yj
    grad(g) = (6x-4y)i + (-4x+12y)j

    Then grad(f) = λgrad(g) produces these equations
    2x = λ(6x-4y) (eq.1)
    2y = λ(-4x+12y) (eq.2)

    rearranging 1 gives:
    λ = 2x/(6x-4y) (eq.3)
    rearranging 2 gives:
    λ = 2y/(-2x+6y) (eq.4)

    then by cross multiplying,
    -2x + 6xy = 3xy - 2y

    taking all terms to left hand side,
    -2x +3xy + 2y = 0
    (-2x - y)(x - 2y)=0
    therefore x = y/-2 or x = 2y

    This is where I am stuck, now I have a relationship between x and y, what do I do next??
    I'm assuming your algebra and arithemetic is correct.

    Substitute x = y/-2 or x = 2y into 3x-4xy+6y = 140 to get an equation with a single unknown ......
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  3. #3
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    I'm fairly sure of my working to that point
    Going on from your advice, say I taken the case x=y/-2 and sub into the constraint equation g(x) = 3x^2 - 4xy....

    I return y=4. Now I have a value of x and y to I find the corresponding lambda? I am not that familiar with lagrange multipliers and my textbook doesnt cover the topic greatly so if you could guide me through the method step by step it would be greatly appreciated!
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  4. #4
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    Quote Originally Posted by sk1001 View Post
    I'm fairly sure of my working to that point
    Going on from your advice, say I taken the case x=y/-2 and sub into the constraint equation g(x) = 3x^2 - 4xy....

    I return y=4. Now I have a value of x and y to I find the corresponding lambda? I am not that familiar with lagrange multipliers and my textbook doesnt cover the topic greatly so if you could guide me through the method step by step it would be greatly appreciated!
    Let me guess ...... This is a question on a university multivariable calculus assignment, it's asking you to find the maximum and minimum distance of the surface 3x-4xy+6y = 140 from the origin, and it's due in the next 24 hours or so.

    Why do you need to calculate the value of lambda? Get the values of x and y and hence the points (x, y) that give the extremum. Then test the nature of these solutions.
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