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Apologies in advance if my terms aren't exact, I'm learning "Mathmatical Economoics" in the hebrew language and some of the terms don't translate well.

I was given the following question: Two sellers are a part of an exchange market with two goods, each has the utility function $u(x,y)=\max(x,y)$. Assume the first seller starts with two units of product $x$ and one unit of $y$.

Find $(a,b)$ s.t seller two starts with $a$ units of $x$ and $b$ units of $y$, for which there is a competetive equilebrium.

Any guidence will be of great help! Thanks in advance

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  • $\begingroup$ What are you confused about specifically? $\endgroup$ – Art Feb 25 at 10:21
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I might be wrong so feel free to point out any mistakes.

Both of them have the same utility function $u(x,y)=max(x,y)$ This means when $Px > Py$ , they'll choose to consume $Y$ instead of $X$ since they get higher satisfaction by selling all units of x and purchasing y.

Take the case of First seller, If he sells all X and purchases Y, His total demand for $Y$ will be $2(Px/Py) + 1$ and since Px/Py >1, total demand will be higher than 3. If he had done the opposite, his total demand will be less than 3. Since the utility function is $max(x,y)$ He will choose the former strategy

Hence, the demand for $X$ will be zero but the supply is $a+2$. Also, there is excess demand of $Y$. Hence, this is not a competitive equilibrium.

Similarly, then $Px <Py$, there will be excess demand for $X$ so that isnt a competitive equilibrium either.

The only case we have left is $Px=Py$ where the seller will be indifferent in consuming either good. Here, each seller consumes all of 1 good each. There will be 2 cases.

Case 1. If seller 1 consumes all $X$, His demand is $2 + (1)Py/Px = 3$ since Px=Py and the supply of X is $a+2$. When we equate this, we get $$a=1$$ Notice that this is the only condition that needs to hold here, B can take any non negative values in this case.

Case 2 is where seller 1 consumes all $Y$ and, His demand is $1+ (2)Px/Py = 3$ and the supply of $Y$ is $b+1$. Since they have to be equal, we get $$b=2$$ And again there's no restriction on $a$ and it can take any non negative values.

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  • $\begingroup$ Thank you! I understand it now, finally :) $\endgroup$ – Gil Bar Koltun Feb 25 at 10:55

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