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Lagrange multipliers open constraint

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the maximum and minimum values of the function f(x, y) =49 − x^2 − y^2
    subject to the constraint x + 3y = 10.

    3. The attempt at a solution
    ∇f = <2x,2y>
    ∇g = <1,3>

    ∇f =λ∇g

    2x = λ
    2y = 3λ
    2x = 2y/3
    x = y/3

    y/3 + 3y = 10
    y = 3
    x = 1

    f(1,3) = 39

    Now that is the only point I got, how should I find out whether it is a maximum/minimum/neither? I understand for closed constaints, like x^2 + y^2 =1, but here I don't understand what I am supposed to do. I know that it is a maximum, by looking at the answer key, but I want to understand the process of figuring it out.
     
  2. jcsd
  3. Nov 15, 2014 #2

    Ray Vickson

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    IF there is finite max or min, the Lagrange conditions hold there. If you find only a single point that satisfies Lagrange, that is telling you something.
     
  4. Nov 15, 2014 #3
    In class, my professor found out if it was a max/min looking at the end behaviour, but I have no idea why it was like that. He said as
    x -> inf, y -> -inf
    x -> -inf, y-> inf
    So f(1,3) is a maximum, and it has no minimum. This is what I don't really understand.
     
  5. Nov 15, 2014 #4

    Ray Vickson

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    Well, can you find feasible (x,y) values giving f < -1,000? f < -10,000? f < -1,000,000? etc.
     
  6. Nov 15, 2014 #5
    Ohhhhhh, so because f(x,y) when x gets really big will make f < 0, and the same for y getting really big f < 0, so f(1,3) has to be a maximum and there is no minimum because you can keep making f smaller, I understand, would this method work for all cases like this? Or is there a better way, like some kind of "official" check?
     
    Last edited: Nov 15, 2014
  7. Nov 15, 2014 #6

    Ray Vickson

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    I outlined one way: the Lagrange conditions have a single solution, so there is only one point that could be a max or a min (and might not even be that---it could be a s addle point). In your example it is easy enough to check that you actually have a maximum, and since that is the only optimum, there cannot be a minimum. However, that might not mean in general that you get something unbounded below. If you had something like ##\max / \min f = 10 - e^{-x^2 - y^2}## and a constraint like the one in your question, you would have a finite minimum, but no maximum; nevertheless, the "supremum" would be at 10, so you have ##f \leq 10##, and you could come as close as you want to 10 without ever reaching it.
     
  8. Nov 15, 2014 #7
    So you can use the second derivative test to check whether it is a saddle point or not correct?
    D = fxxfyy - fyx^2
    D < 0, saddle point
     
  9. Nov 16, 2014 #8

    Ray Vickson

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    In an unconstrained problem, the second-order necessary conditions for a (local) min of ##f(x,y)## at the stationary point ##(x,y) = (x^*,y^*)## are:
    [tex] f_{xx} \geq 0, \:f_{yy} \geq 0 \; \text{at} \; (x,y) = (x^*,y^*)\\
    f_{xx} f_{yy} - f_{xy}^2 \geq 0 \; \text{at} \; (x,y) = (x^*,y^*) [/tex]
    The second-order sufficient conditions for a strict (local) min at ##(x,y) = (x^*,y^*)## are as above, but with all inequalities being strict.

    In a constrained problem with a linear constraint, necessary second-order conditions for a (local) min at ##(x^*,y^*)## positive semi-definiteness of the Hessian matrix
    [tex] H_f (x^*,y^*) =\left. \pmatrix{ f_{xx} & f_{xy}\\f_{xy} & f_{yy}} \right|_{(x,y) = (x*^,y^*)} [/tex]
    projected down into the tangent subspace of the constraint. A sufficient condition for a strict constrained local min is that we have positive-definiteness instead of semi-definiteness in the above, provided that the Lagrange multiplier is nonzero. (If the Lagrange multiplier is zero it is trickier).

    For a constrained problem with a nonlinear constraint, you must replace the function f by the Lagrangian
    [tex] L(x,y,\lambda^*) = f(x,y) - \lambda^* g(x,y) [/tex]
    in the tests listed above. (That is, we look at the Hessian of the Lagrangian instead of the function f). Here, ##\lambda^*## is the value of the Lagrange multiplier ##\lambda## at the solution ##(x^*,y^*)##. Note that we fix ##\lambda## at the value ##\lambda^*##, but let ##(x,y)## vary around the point ##(x^*,y^*)##.

    All this is a very lengthy way of saying that your suggested test above is not really correct in all its details.
     
  10. Nov 16, 2014 #9
    I didn't understand three quarters of what you posted, but I understand the result. I don't think I will need to know it in this much detail for the midterm though. Thanks for the explanation!
     
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