Question

Minimize Z = 3x₁ + 5x₂
Subject to

\[x_1 + 3 x_2 \geq 3\]
\[ x_1 + x_2 \geq 2\]
\[ x_1 , x_2 \geq 0\]

 

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
x₁ + 3x₂ = 3, x₁ + x₂ = 2, x₁ = 0 and x₂ = 0
Region represented by x₁ + 3x₂ ≥ 3 :
The line x₁ + 3x₂ = 3 meets the coordinate axes at A(3, 0) and B(0, 1) respectively. By joining these points we obtain the line x₁ + 3x₂ = 3.
Clearly (0,0) does not satisfies the inequation x₁ + 3x₂ ≥ 3 .So,the region in the plane which does not contain the origin represents the solution set of the inequation x₁ + 3x₂ ≥ 3.
Region represented by x₁ + x₂ ≥ 2:
The line x₁ + x₂ = 2 meets the coordinate axes at C(2, 0) and D(0, 2) respectively. By joining these points we obtain the line x₁ + x₂ = 2.Clearly (0,0) does not 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 ≥ 3 , x₁ + x₂ ≥ 2,x₁ ≥ 0, and x₂ ≥ 0, are as follows

.

The corner points of the feasible region are O(0, 0), B(0, 1), \[E\left( \frac{3}{2}, \frac{1}{2} \right)\]  and C(2, 0).

The values of Z at these corner points are as follows.

Corner point	Z = 3x1 + 5x2
O(0, 0)	3 × 0 + 5 × 0 = 0
B(0, 1)	3 × 0 + 5 × 1 = 5
\[E\left( \frac{3}{2}, \frac{1}{2} \right)\]	\[\frac{3}{2}\] + 5 × \[\frac{1}{2}\]= 7
C(2, 0)	3 × 2 + 5 × 0 = 6

Therefore, the minimum value of Z is 0 at the point O(0, 0). Hence, x₁ = 0 and x₂ = 0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 0.