Lagragean with a minimum?

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!

I tried it and I get fallacies like 2 = 0 when doing the partial derivatives.

..

Perhaps worth noting that this is pretty easy to solve directly, so I wouldn't use Lagrangians unless forced to.

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...

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)?