Proving the solution to linear programming problem is always the boundaries

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Show that the solution $\mathbf{y}$ to the following linear problem:

$$\min c_1x_1 + c_2x_2 + \cdot \cdot \cdot + c_nx_n$$ $$\text{subject to} \ \ a_i\leq x_i\leq b_i \ \ \ \ i = 1,2,....n$$

is:

$$ \mathbf{y}_i = \left\{ \begin{array}{ll} a_i & \quad c_i > 0 \\ b_i & \quad c_i < 0 \end{array} \right. $$

Any idea or suggestion on how to go about this probelm?

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Rewrite the constraints as $x_i \ge a_i$ and $-x_i \ge -b_i$, construct the dual linear program with dual variables $\alpha_i \ge 0$ and $\beta_i \ge 0$, and note that the dual solution $\alpha^*_i = \max\{c_i,0\}$ and $\beta^*_i = \max\{-c_i,0\}$ is dual feasible and has the same objective value as your proposed primal feasible solution.

If you multiply both sides of $x_i \ge a_i$ by $\alpha^*_i$ and both sides of $-x_i \ge -b_i$ by $\beta^*_i$ and then add these up, you obtain $$\sum_i (\alpha^*_i - \beta^*_i) x_i \ge \sum_i (\alpha^*_i a_i - \beta^*_i b_i). \tag1$$ By the choice of $\alpha^*_i$ and $\beta^*_i$, we have $\alpha^*_i - \beta^*_i = c_i$, and $(1)$ becomes $$\sum_i c_i x_i \ge \sum_{i:c_i > 0} c_i a_i + \sum_{i:c_i < 0} c_i b_i,$$ and this lower bound on the objective function certifies the optimality of your proposed solution $\mathbf{y}$.