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Lagrange Multipliers with Multiple Constraints?

  1. Jun 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Using Lagrange multipliers, find the max and the min values of f:

    f(x,y,z) = x^2 +2y^2+3x^2

    Constraints:

    x + y + z =1
    x - y + 2z = 2


    2. Relevant equations

    ∇f(x) = λ∇g(x) + μ∇h(x)


    3. The attempt at a solution

    Using Lagrange multipliers, I obtained the equations:

    2x = λ + μ
    4y = λ - μ
    6z = λ + 2μ

    With the constraints, I tried to find values for the unknowns, but I can't solve the system of equations. I don't know where to go from here. Can anyone help put? Any input is appreciated!
     
    Last edited: Jun 11, 2014
  2. jcsd
  3. Jun 11, 2014 #2

    Ray Vickson

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    Your equations give you x, y and z in terms of λ and μ. Plug these expressions into the two constraint equations to get two equations for the two unknowns λ, μ.
     
  4. Jun 11, 2014 #3
    Hey, thanks for helping out! Please note the small change in the second constraint equation: it's x -y + 2z = 2. I did exactly that, but it didn't really seem to help solve for the unknowns. By plugging in the expressions obtained into the constraint equations, I got 11λ + 7μ = 12 and 7λ+ 17μ = 24. Solving, I still get an equation with two unknowns.
     
  5. Jun 11, 2014 #4

    Ray Vickson

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    I don't understand what you mean. When you solve the two equations you will get two numbers, one for λ and one for μ. If that is not what you get you had better show the details here.

    BTW: you wrote f(x,y,z) = x^2 +2y^2+3x^2, but I assume you meant 3z^2 in the last term.
     
  6. Jun 11, 2014 #5

    AlephZero

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    Altogether you have 5 equations in 5 unknowns x, y, z, λ, and μ: the two constraint equations, and the three equations you have derived.

    There are lots of ways to solve them (some quicker than others) but you certainly have enough equations to get the solution.

    You don't need the values of λ and μ to get the max and min values of f, but as Ray Vickson said it's probably easiest to find λ and μ first, and then find x y and z.
     
  7. Jun 11, 2014 #6

    HallsofIvy

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    With [itex]2x= \lambda+ \mu[/itex], [itex]2y= \lambda-\mu[/itex], and [itex]2z= \lambda+ 2\mu[/itex] the two constraints become [itex]x+ y+ z= (\lambda + \mu)/2+ (\lambda- \mu)/2+ (\lambda+ 2\mu)/2= 1[/itex], which reduces to [itex]3\lambda+ 2\mu= 2[/itex], and [itex]x- y+ 2z= (\lambda+ \mu)/2- (\lambda- \mu)/2+ 2(\lambda+ 2\mu)/2= 2[/itex], which reduces to [itex]\lambda+ 2\mu= 2[/itex].

    Solve [itex]3\lambda+ 2\mu= 2[/itex] and [itex]\lambda+ 3\mu= 2[/itex].
     
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