Question

Maximize Z = −x₁ + 2x₂
Subject to
$$\begin{gathered} -x_1+3x_2\le 10 \\ {x}_1+x_2\le 6 \\ {x}_1-x_2\le 2 \\ {x}_1,x_2\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
−x₁ + 3x₂ = 10, x₁ + x₂ = 6, x₁ + x₂ = 2, x₁ = 0 and x₂ = 0

Region represented by −x₁ + 3x₂ ≤ 10:
The line −x₁ + 3x₂ = 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 −x₁ + 3x₂ = 10.
Clearly (0,0) satisfies the inequation −x₁ + 3x₂ ≤ 10 .So,the region in the plane which contain the origin represents the solution set of the inequation
−x₁ + 3x₂ ≤ 10.

Region represented by x₁ + x₂ ≤ 6:
The line x₁ + x₂ = 6 meets the coordinate axes at C(6, 0) and D(0, 6) respectively. By joining these points we obtain the line x₁ + x₂ = 6.Clearly (0,0) satisfies the inequation x₁ + x₂ ≤ 6. So,the region containing the origin represents the solution set of the inequation x₁ + x₂ ≤ 6.

Region represented by x₁− x₂ ≤ 2:
The line x₁ − x₂ = 2 meets the coordinate axes at E(2, 0) and F(0, −2) respectively. By joining these points we obtain the line x₁ − x₂ = 2.Clearly (0,0) satisfies the inequation x₁− x₂ ≤ 2. So,the region containing the origin represents the solution set of the inequation x₁− x₂ ≤ 2.

Region represented by x₁ ≥ 0 and x₂ ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x₁ ≥ 0 and x₂ ≥ 0.

The feasible region determined by the system of constraints, −x₁ + 3x₂ ≤ 10, x₁ + x₂ ≤ 6, x₁− x₂ ≤ 2, x₁ ≥ 0, and x₂ ≥ 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 = −x1 + 2x2
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 $B\left(0,\frac{10}{3}\right)$.