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I am working with this problem and I've come up with these inequalities. I have the next two equalities:

$y_1=\frac{1}{2}(x_1+x_2) $

$y_2=\frac{1}{2}(x_1-x_2)$

where the limits are:

$\qquad l_1\le x_1 \le u_1,$ $\qquad l_2\le x_2 \le u_2$

How can I prove that the limits of the variables $y_1,y_2$ are :

$\frac{1}{2}(l_1+l_2)\le y_1 \le \frac{1}{2}(u_1+u_2)$

$\frac{1}{2}(l_1-u_2)\le y_2 \le \frac{1}{2}(u_1-l_2)$

and $x_1,x_2\in {\rm I\!R}$,

$x_1,x_2 \ge 0$

I've done it graphically and I am certain that the feasible region of the two dimensional problem of having $x_1,x_2$ is a square or rectangle, depending on the upper and lower bounds. When I map these points into another Cartesian plane involving now $y_1,y_2$ this square or rectangle is rotated. I would like to know how this can be proven besides graphically.

Thank you so much in advance!

frrndn
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3 Answers3

1

How can we make $y_1=\frac12(x_1+x_2)$ as low as possible? We see that it is in a linear relationship with $x_1$ and $x_2$, such that if we lower $x_1$ or $x_2$ we lower $y_1$. It follows that the minimum is attained when $x_1,x_2$ are as low as possible – i.e. when we have $y_1=\frac12(l_1+l_2)$.

Similar reasoning applies to finding the upper bound of $y_1$ and the bounds of $y_2$. For $y_2$ also note that increasing $x_2$ will decrease (not increase) $y_2$ and vice versa.

Parcly Taxel
  • 103,344
1

Your claim can be proven using known properties of $\le$.

If $ l_1\le x_1 \le u_1$ and $ l_2\le x_2 \le u_2,$

then we can add these inequalities to get $ l_1+l_2\le x_1+x_2 \le u_1+u_2$

and divide by $2$ to get $\frac12( l_1+l_2)\le y_1 \le \frac12(u_1+u_2).$

Also $-u_2\le -x_2\le -l_{\,2},$ so, adding that to $l_1\le x_1\le u_1$, and dividing by $2,$

we have $\frac12(l_1-u_2)\le y_2\le \frac12 (u_1-l_2)$.

J. W. Tanner
  • 60,406
0

Just see the following steps:

  • $ x_1\le u_1 \mbox{ and } x_2\le u_2 \Rightarrow x_1+x_2\le u_1+u_2$,
  • but $x_1+x_2=2y_1$, then $2y_1\le u_1+u_2$.
  • $ x_2\le u_2 \Rightarrow -u_2\le -x_2 $ and $l_1\le x_1 \mbox{ so } l_1-u_2\le x_1-x_2$ ,
  • but $x_1-x_2=2y_1\Rightarrow l_1-u_2\le 2y_1$.

So you can conclude the rest as the same form

L F
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