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Math Help - Lagrante multipliers check + general question

  1. #1
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    Lagrante multipliers check + general question

    Hello! First of all I'd like somebody to check if what I'm doing is correct. Second question: see end

    I need to find the extrema of 6 - 4x - 3y subject to x + y = 1.

    ∂f/∂x = -4 and ∂g/∂x = 2x and ∂f/∂y = -3 and ∂g/∂y = 2y

    So we've got:
    { -4 = 2
    λx
    { -3 = 2λy
    { x + y = 1

    <=>

    { -2/x =
    λ
    { -3/2y = λ
    { x + y = 1

    If -2/x = -3/2y then 3x = 4y, this means 3x/4 = y

    Substitute this into contraint:
    x + (3x/4) = 1 <=> 25x - 16 = 0 <=> x =
    4/5

    If y = 3x/4, then 3(
    4/5) / 4 = 3/5

    Now here comes my second question. Now that I've got
    x = 4/5 and y = 3/5, how do I know that the points are
    (4/5, 3/5) and (-4/5, -3/5)
    OR
    (4/5, 3/5) , (-4/5; -3/5) , (4/5; -3/5) , (-4/5, 3/5) ?

    Thanks!
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  2. #2
    Newbie russo's Avatar
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    Re: Lagrante multipliers check + general question

    Hi, I've redone the exercise and yes the critical points are right.

    Regarding to your second question, as I've done it before, I think you use the points wich satisfies your \lambda identity (  y = \frac{3}{4} x), so if x=\frac{4}{5} then y=\frac{3}{5} and so on. So you actually check the first pair. I think is that way but still if you can check it, better.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    St. Augustine, FL.
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    Re: Lagrante multipliers check + general question

    Yes, you have only two critical values for the reason given by russo.
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