Question

By graphical method, the solution of linear programming problem

\[\text{Maximize}\text{ Z }= 3 x_1 + 5 x_2 \]
\[\text{ Subject }  to \text{ 3 } x_1 + 2 x_2 \leq 18\]
\[ x_1 \leq 4\]
\[ x_2 \leq 6\]
\[ x_1 \geq 0, x_2 \geq 0, \text{ is } \]

Options

• x₁ = 2, x₂ = 0, Z = 6

• x₁ = 2, x₂ = 6, Z = 36

• x₁ = 4, x₂ = 3, Z = 27

• x₁ = 4, x₂ = 6, Z = 42

Answer 1

x₁ = 2, x₂ = 6, Z = 36

We need to maximize the function Z = 3x₁ + 5x₂
First, we will convert the given inequations into equations, we obtain the following equations:  3x₁ + 2x₂ = 18, x₁ = 4, x₂ = 6, x₁ = 0 and x₂ = 0

Region represented by 3x₁ + 2x₂ ≤ 18:
The line 3x₁ + 2x₂ = 18 meets the coordinate axes at A(6, 0) and B(0, 9) respectively. By joining these points we obtain the line 3x₁ + 2x₂ = 18.
Clearly (0,0)  satisfies the inequation 3x₁ + 2x₂ = 18 .So,the region in the plane which contain the origin represents the solution set of the inequation  3x₁ + 2x₂ = 18.

Region represented by x₁ ≤ 4:
The line x₁ = 4 is the line that passes through C(4, 0) and is parallel to the Y axis. The region to the left of the line x₁ = 4 will satisfy the inequation x₁ ≤ 4.
Region represented by x₂ ≤ 6:
The line x₂ = 6 is the line that passes through D(0, 6) and is parallel to the X axis. The region below the line x₂ = 6 will satisfy the inequation x₂ ≤ 6.

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, 3x₁ + 2x₂ ≤ 18, x₁ ≤ 4, x₂ ≤ 6, x₁ ≥ 0, and x₂ ≥ 0, are as followsCorner points are O(0, 0), D(0, 6), F(2, 6), E(4, 3) and C(4, 0).
The values of the objective function at these points are given in the following table
 

Points	Value of Z
O(0, 0)	3(0)+5(0) = 0
D(0, 6)	3(0)+5(6) = 30
F(2, 6)	3(2)+5(6) = 36
E(4, 3)	3(4)+5(3) = 27
C(4, 0)	3(4)+5(0) = 12

We see that the maximum value of the objective function Z is 36 which is at F(2, 6).