Maximize Z = −x_{1} + 2x_{2}
Subject to
\[ x_1 + 3 x_2 \leq 10\]
\[ x_1 + x_2 \leq 6\]
\[ x_1  x_2 \leq 2\]
\[ x_1 , x_2 \geq 0\]
Solution
First, we will convert the given inequations into equations, we obtain the following equations:
−x_{1} + 3x_{2} = 10, x_{1} + x_{2} = 6, x_{1} + x_{2} = 2, x_{1} = 0 and x_{2} = 0
Region represented by −x_{1} + 3x_{2} ≤ 10:
The line −x_{1} + 3x_{2}_{ }= 10 meets the coordinate axes at A(−10, 0) and \[B\left( 0, \frac{10}{3} \right)\] respectively. By joining these points we obtain the line −x_{1} + 3x_{2}_{ }= 10.
Clearly (0,0) satisfies the inequation −x_{1} + 3x_{2} ≤ 10 .So,the region in the plane which contain the origin represents the solution set of the inequation
−x_{1} + 3x_{2} ≤ 10.
Region represented by x_{1} + x_{2} ≤ 6:
The line x_{1} + x_{2} = 6 meets the coordinate axes at C(6, 0) and D(0, 6) respectively. By joining these points we obtain the line x_{1} + x_{2} = 6.Clearly (0,0) satisfies the inequation x_{1} + x_{2} ≤ 6. So,the region containing the origin represents the solution set of the inequation x_{1} + x_{2} ≤ 6.
Region represented by x_{1}− x_{2} ≤ 2:
The line x_{1} − x_{2} = 2 meets the coordinate axes at E(2, 0) and F(0, −2) respectively. By joining these points we obtain the line x_{1} − x_{2} = 2.Clearly (0,0) satisfies the inequation x_{1}− x_{2} ≤ 2. So,the region containing the origin represents the solution set of the inequation x_{1}− x_{2} ≤ 2.
Region represented by x_{1} ≥ 0 and x_{2} ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x_{1} ≥ 0 and x_{2} ≥ 0.
The feasible region determined by the system of constraints, −x_{1} + 3x_{2} ≤ 10, x_{1} + x_{2} ≤ 6, x_{1}− x_{2} ≤ 2, x_{1} ≥ 0, and x_{2} ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), E(2, 0), H(4, 2), G(2, 4) and \[B\left( 0, \frac{10}{3} \right)\] .
The values of Z at these corner points are as follows.
Corner point  Z = −x_{1} + 2x_{2} 
O(0, 0)  −1 × 0 + 2 × 0 = 0 
E(2, 0)  −1 × 2 + 2 × 0 = −2 
H(4, 2)  −1 × 4 + 2 × 2 = 0 
G(2, 4)  −1 × 2 + 2 × 4 = 6 
\[B\left( 0, \frac{10}{3} \right)\]

−1 × 0 + 2 × \[\frac{10}{3}\]= \[\frac{20}{3}\]

We see that the maximum value of the objective function Z is \[\frac{20}{3}\]which is at \[B\left( 0, \frac{10}{3} \right)\] .