x 1 + x 2 ≥ 6 x_1+x_2\geq 6 x 1 + x 2 ≥ 6
3 x 1 + 2 x 2 ≤ 30 3x_1+2x_2\leq 30 3 x 1 + 2 x 2 ≤ 30
2 x 1 + x 2 ≤ 5 2x_1+x_2\leq 5 2 x 1 + x 2 ≤ 5
x 1 , x 2 ≥ 0 x_1, x_2\geq 0 x 1 , x 2 ≥ 0
Maximize z = 5 x 1 + 8 x 2 z=5x_1+8x_2 z = 5 x 1 + 8 x 2
Green region shows all solutions of { x 1 + x 2 ≥ 6 3 x 1 + 2 x 2 ≤ 30 2 x 1 + x 2 ≤ 5 \begin{cases}
x_1+x_2\geq 6
\\3x_1+2x_2\leq 30
\\
2x_1+x_2\leq 5
\end{cases} ⎩ ⎨ ⎧ x 1 + x 2 ≥ 6 3 x 1 + 2 x 2 ≤ 30 2 x 1 + x 2 ≤ 5
And red region shows all solutions of x 1 , x 2 ≥ 0 x_1,x_2\geq 0 x 1 , x 2 ≥ 0
We can see that there is no solution of the system { x 1 + x 2 ≥ 6 3 x 1 + 2 x 2 ≤ 30 2 x 1 + x 2 ≤ 5 x 1 , x 2 ≥ 0 \begin{cases}
x_1+x_2\geq 6
\\3x_1+2x_2\leq 30
\\
2x_1+x_2\leq 5
\\ x_1,x_2\geq 0
\end{cases} ⎩ ⎨ ⎧ x 1 + x 2 ≥ 6 3 x 1 + 2 x 2 ≤ 30 2 x 1 + x 2 ≤ 5 x 1 , x 2 ≥ 0 because intersection of these regions is empty.
Therefore, we can’t maximize z = 5 x 1 + 8 x 2 z=5x_1+8x_2 z = 5 x 1 + 8 x 2 because there is no ( x 1 , x 2 ) (x_1,x_2) ( x 1 , x 2 ) , for which all inequalities are satisfied.
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