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    This is a question which I always came across in my economics degree course but I'm not sure on how to solve it with the Lagregean multiplier method.

    U=min(2x, 3y), subject to x + 3y = 60

    So we have:

    Z= min(2x, 3y) + \Lambda(x + 3y - 60 )

    But how do I find the partial derivative with min(2x, 3y)?
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    (Original post by SecretDuck)
    This is a question which I always came across in my economics degree course but I'm not sure on how to solve it with the Lagregean multiplier method.

    U=min(2x, 3y), subject to x + 3y = 60

    So we have:

    Z= min(2x, 3y) + \Lambda(x + 3y - 60 )

    But how do I find the partial derivative with min(2x, 3y)?
    I've never actually seen the Lagrangian method used with something like the min function, so I've no idea whether this will work or not, but one thing you could try is to note that min(2x,3y) = either 2x or 3y, depending on which is the smaller, and the partial derivatives of these expressions are just constants! So does the method work if you try to solve the 2 different cases as if they were separate problems?

    This is just a shot in the dark! :confused:
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    (Original post by davros)
    I've never actually seen the Lagrangian method used with something like the min function, so I've no idea whether this will work or not, but one thing you could try is to note that min(2x,3y) = either 2x or 3y, depending on which is the smaller, and the partial derivatives of these expressions are just constants! So does the method work if you try to solve the 2 different cases as if they were separate problems?

    This is just a shot in the dark! :confused:
    I tried it and I get fallacies like 2 = 0 when doing the partial derivatives.
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    Perhaps worth noting that this is pretty easy to solve directly, so I wouldn't use Lagrangians unless forced to.
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    (Original post by SecretDuck)
    I tried it and I get fallacies like 2 = 0 when doing the partial derivatives.
    (Original post by davros)
    ..
    When you think about it, it's fairly obvious(*) that U is going to be maximized when 2x = 3y, which is, unfortunately, the one point where you can't possibly differentiate min(2x, 3y). So I can't see how you can do this with derivatives...

    (*) Not so obvious that I realised it until your post, however...
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    (Original post by DFranklin)
    Perhaps worth noting that this is pretty easy to solve directly, so I wouldn't use Lagrangians unless forced to.
    Langragian is a method that I'm comfortable with but if there's an easier and more efficient way, then I'd love to hear it.
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    (Original post by DFranklin)
    When you think about it, it's fairly obvious(*) that U is going to be maximized when 2x = 3y, which is, unfortunately, the one point where you can't possibly differentiate min(2x, 3y). So I can't see how you can do this with derivatives...

    (*) Not so obvious that I realised it until your post, however...
    and I was secretly hoping there was going to be smart way of solving this problem

    But direct solution is pretty clearly a simple approach here.
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    (Original post by SecretDuck)
    This is a question which I always came across in my economics degree course but I'm not sure on how to solve it with the Lagregean multiplier method.

    U=min(2x, 3y), subject to x + 3y = 60

    So we have:

    Z= min(2x, 3y) + \Lambda(x + 3y - 60 )

    But how do I find the partial derivative with min(2x, 3y)?
    I think for the minimum the partial derivative would be
    2x, and 3y respectively. For f(x,y)
    \frac{\partial f}{\parial x}=2x
    \frac{\partial f}{\partial x}=3y
    which would equal zero when there is no constraint.
    So for z(x,y,\lambda)
    \frac{\partial z}{\partial x}=2x+\lambda =0
    \frac{\partial z}{\partial y}=3y+3\lambda =0
    \frac{\partial z}{\partial \lambda}=x+3y-60 =0
 
 
 
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