Given the following problem:
$$max \ x_1 + x_2\\
s.t. \ x_1 \ge 0\\
x_2 \ge 0 \\
x_2 - x_1 \le 1 \\
x_1 + 6x_2 \le 15 \\
4x_1 - x_2 \le 10$$
The result is 5 as shown here: https://i.stack.imgur.com/sLqpd.jpg
Why does the following hold?
$$max \ x_1 + x_2 = min -(x_1 + x_2)$$
This means that the minimization problem result is $-5$, but it $0$. Having problems understanding this.
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Parcly Taxel
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Ran Dom
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Maybe $\text{argmax}(x_1+x_2)=\text{argmin}(-(x_1+x_2))$, which is trivial. $x_1+x_2$ attains its maximum exactly where $-(x_1+x_2)$ attains its minimum. Or $\max(x_1+x_2) = \color{red}{-}\min\left(-(x_1+x_2)\right)$. – Jack D'Aurizio Mar 09 '18 at 16:02
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The maximization problem: $$\text{Max} \ \ z=x_1+x_2 \ \ \text{subject to} \ \ \begin{cases} -x_1+x_2\le 1 \\ x_1+6x_2\le 1 \\ 4x_1-x_2\le 10 \\ x_1,x_2\ge 0\end{cases}$$ is equivalent to the minimization problem: $$\text{Min} \ \ z=-x_1-x_2 \ \ \text{subject to} \ \ \begin{cases} -x_1+x_2\le 1 \\ x_1+6x_2\le 1 \\ 4x_1-x_2\le 10 \\ x_1,x_2\ge 0\end{cases}$$ Indeed, while the feasible region (the pentagon) is the same for both problems, the largest value of $z=x_1+x_2$ occurs for $x_1+x_1=5$ and the smallest value of $z=-x_1-x_2$ occurs for $-x_1-x_2=-5$. Both problems produce the same solution $(x_1,x_2)=(3,2)$.
farruhota
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