91Ó°ÊÓ

Consider the three variables namely${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3$and the two linear inequalities as follows,

$\begin{array}{l}{\mathrm{x}}_1−{\mathrm{x}}_2≤1\\ 2{\mathrm{x}}_2−{\mathrm{x}}_3≤1\end{array}$

with a condition, ${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3â‰\yen 0$

Assume the inequalities be an equation of the form,

$\begin{array}{l}\left[1\right]\text{ }{\mathrm{x}}_1−{\mathrm{x}}_2=1\\ \left[2\right]\text{ }2{\mathrm{x}}_2−{\mathrm{x}}_3=1\\ \left[3\right]\text{ }{\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3=0\end{array}$

Equation’s last line signifies those variables${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3$ can never be negative.

Consider the three cases:

Case (I): ${\mathrm{x}}_1=0$

Case (II):${\mathrm{x}}_2=0$

Case (III):$x_3=0$

In this case, Substitute ${\mathrm{x}}_1=0$in Equation [1],

$\begin{array}{c}\left[1\right]\text{ }{\mathrm{x}}_1−{\mathrm{x}}_2=1\\ 0−{\mathrm{x}}_2=1\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}[âˆ\mu {\mathrm{x}}_1=0]\\ {\mathrm{x}}_2=−1\\ \end{array}$

But no variable can be negative.

Hence, discard the Case (I)

In this case, substitute$x_2=0$in Equation [1] as follows,

$\begin{array}{c}\left[1\right]\text{ }{x}_1−x_2=1\\ x_1−0=1\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}[âˆ\mu x_2=0]\\ x_1=1\\ \end{array}$

Substitute$x_2=0$in Equation [2] as follows,

$\begin{array}{c}\left[2\right]2x_2−x_3=1\\ 0−x_3=1\text{ â¶Ä‰â¶Ä‰}[âˆ\mu x_2=0]\\ x_3=−1\end{array}$

But no variable can be negative.

Hence, discard the Case (II).

In this case, substitute$x_3=0$in Equation as follows,

$\begin{array}{c}\left[2\right]2x_2−x_3=1\\ 2x_2−0=1\text{ â¶Ä‰â¶Ä‰â€‰}[âˆ\mu x_3=0]\\ 2x_2=1\\ x_2=\frac{1}{2}\end{array}$

Substitute${\mathrm{x}}_2=\frac{1}{2}$in Equation [1] as follows,

$\begin{array}{c}\left[1\right]x_1−x_2=1\\ x_1−\frac{1}{2}=1\text{​ â¶Ä‰â¶Ä‰â€‹â€‹â€‰}[âˆ\mu x_2=\frac{1}{2}]\\ x_1=1+\frac{1}{2}\\ x_1=\frac{3}{2}\end{array}$

Therefore, this case is possible.

The maximum of $x_1−2x_3$ can be calculated as follows,

$\begin{array}{c}{x}_1−2x_3=\frac{3}{2}−2×0\\ =\frac{3}{2}\end{array}$

Therefore, the optimal solution at$({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3)$ is $(3/2,1/2,0)$.