Is it possible to solve: If $2x+y=4$ find $\max \min \{x,y\}$?

Is it possible to solve: If $2x+y=4$ find $\max \min \{x,y\}$?

My usual approach to this question is to set WLOG $x\geq y$ and then proceed, but I don't think that works here...

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HINT: What happens when $x = y$? – Srivatsan Dec 31 '11 at 18:11
@Davide Giraudo. I believe your edit changed to original meaning of the question. – Kavka Dec 31 '11 at 19:17

Method 1. You seem to have intuition that $x = y$ is an interesting case to look at, and that is indeed a good idea. We can formalise that as follows: Suppose $z = \min (x,y)$. Then $z \leqslant x$ and $z \leqslant y$. Therefore, $$3z \leqslant 2x + y = 4,$$ which implies that $z = \min (x,y) \leqslant \frac{4}{3}$. On the other hand, it is not hard to construct an example that achieves equality: if $x = y = \frac{4}{3}$, then $2x + y = 4$ and $\min(x,y) = \frac43$. Thus the maximum value of $\min(x,y)$ is indeed $\frac{4}{3}$.
Method 2. It might help visualisation if you eliminate $y$ from the problem. The given problem is equivalent to: maximize $f(x)$ where $$f(x) = \min \{ x, 4 - 2x \}.$$ From the following graph, it is clear that $f(x)$ is maximised when the $y=x$ line meets $y = 4-2x$.