Given the following linear programming model minimize cos

Question

t 50x1 +40x2
6x1+4 x2=24
X1+
2x2=6 find an optimal solution for the problem graphically?

Answer 1

First, let us identify the constraints and objective function in the given problem.

Objective function: Minimize cost = 50x1 + 40x2
Constraints:
1. 6x1 + 4x2 <= 24
2. x1 + 2x2 <= 6

Now let's transform the constraints into the equations for lines:
1. x2 = (24 - 6x1) / 4
2. x2 = (6 - x1) / 2

Next, let's graph the constraint equations and indicate their feasible regions:

1. For constraint 1:
x2 = (24 - 6x1) / 4
When x1 = 0, x2 = 6; and when x2 = 0, x1 = 4
So we have the line passing through points (0, 6) and (4, 0), and the feasible region lies below this line, including the line itself.

2. For constraint 2:
x2 = (6 - x1) / 2
When x1 = 0, x2 = 3; and when x2 = 0, x1 = 6
So we have the line passing through points (0, 3) and (6, 0), and the feasible region lies below this line, including the line itself.

The feasible region for the given problem is the intersection of the feasible regions of both constraints. In this case, it is the quadrilateral with vertices A(0,0), B(0, 3), C(2, 2), and D(4, 0).

Now we need to evaluate the objective function at each of these vertices to find the minimum value and the corresponding values of x1 and x2.

1. Vertex A (0,0): Cost = 50(0) + 40(0) = 0
2. Vertex B (0,3): Cost = 50(0) + 40(3) = 120
3. Vertex C (2,2): Cost = 50(2) + 40(2) = 180
4. Vertex D (4,0): Cost = 50(4) + 40(0) = 200

The minimum cost occurs for vertex A with the value 0, so the optimal solution is x1 = 0, x2 = 0, and the minimum cost is 0.