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Let $a=(a_{1},a_{2},\ldots,a_{n})\in\mathbb{R}^{n}$, $b=(b_{1},b_{2},\ldots,b_{n})\in\mathbb{R}^{n}$, $c=(c_{1},c_{2},\ldots,c_{n}) \in\mathbb{R}^{n}$ be given.

Let us firstly consider the problem of finding the global minimum of the function in two real variables $$ F(x,y)=\sum_{k=1}^{n} (a_{k}x+b_{k}y-c_{k})^{2}. $$ This problem has its origin, for example, from the least-squares method.

It is easy to obtain the explicit formula for the local (and global at the same time) minimum point $(x^{\ast},y^{\ast})$ of the function $F(x,y)$. I have calculated and I have got this result: $$ x^{\ast}=\frac{\langle a,c \rangle |b|^{2} - \langle a,b\rangle \langle b,c\rangle}{|a|^{2}|b|^{2} - \langle a,b\rangle^{2}}, $$ $$ y^{\ast}=\frac{|a|^{2} \langle b,c \rangle - \langle a,b\rangle \langle a,c\rangle}{|a|^{2}|b|^{2} - \langle a,b\rangle^{2}}. $$ Here is, as usual, $$ |a|^{2}=\sum_{k=1}^{n}a_{k}^{2}, \quad \langle a,b\rangle=\sum_{k=1}^{n}a_{k}b_{k}. $$

Let us now consider the problem of finding the global minimum of the "analogous" function in two real variables (Euclidean distance in $\mathbb{R}^{n}$ is replaced with the Chebyshev distance) $$ \Phi(x,y)=\max_{k=\overline{1,n}} |a_{k}x+b_{k}y-c_{k}|, $$ or, in expanded form: $$ \Phi(x,y)=\max \Big\{|a_{1}x+b_{1}y-c_{1}|, |a_{2}x+b_{2}y-c_{2}|, \ldots, |a_{n}x+b_{n}y-c_{n}| \Big\}. $$ Is it possible to write the explicit form for the global minimum point of the function $\Phi(x,y)$? Please give me some links (references) to the solution of this problem.

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