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Given a function $z$, with constraints $|x|\le 1$, $|y|\leq2$, find its minimum and maximum $$ z = x + |x-y|$$ ... $$\frac{\partial f(x, y)}{\partial x} = \frac{x-y}{|x-y|} + 1$$ $$\frac{\partial f(x, y)}{\partial y} = \frac{y-x}{|x-y|}$$ Thus, it yields a system of equations, and by solving it we're to find function's extrema points: \begin{cases} \frac{x-y}{|x-y|} + 1 = 0 \\ \frac{y-x}{|x-y|} = 0 \end{cases} I was unable to derive any solutions by brute force, and it is not quite clear to me if it is possible to do that with analytic methods. How to proceed?

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Using calculus here is unnecessary and, given the use of absolute values, arguably inappropriate.

What you'd like to do for the maximum is simultaneously make $x$ as large as possible and $x$ and $y$ as different as possible. The constraints $|x|\le1, |y|\le2$ allow you to this by choosing $x=1$ and $y=-2$, for a maximum value $z=1+|1-(-2)|=4$.

For the minimum, you want to simultaneously make $x$ as small as possible and $x$ and $y$ as close as possible. This happens with $x=y=-1$, for a minimum value of $z=-1+|-1-(-1)|=-1$.

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$$\begin{align} z&=x+|x-y|\\ &\geqslant x+0\\ &\geqslant-1 \end{align}$$ And $$\begin{align} z&=x+|x-y|\\ &=x+|x+(-y)|\\ &\leqslant x+|x|+|-y|\\ &=x+|x|+|y|\\ &\leqslant |x|+|x|+|y|\\ &= 2|x|+|y|\\ &\leqslant 2+2\\ &=4 \end{align}$$ Can you find $(x,y)$ pairs wich for $z=-1$ and $z=4$?

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