Linear programming - Model formulation, Graphical Method | PPT - Free Printable

Problem Description:

The task is to solve a linear programming problem using the graphical method. The objective is to minimize the function:

\[
Z = 6x_1 + 3x_2
\]

subject to the constraints:

1. \( 2x_1 + 4x_2 \geq 16 \)
2. \( 4x_1 + 3x_2 \geq 24 \)
3. \( x_1, x_2 \geq 0 \)

The provided image shows the graph of the constraints, and we need to determine the feasible region and the point where the objective function is minimized.

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Step-by-Step Solution:

#### 1. Graph the Constraints:
The constraints are inequalities, so we first convert them into equations to plot the lines:

- Constraint 1: \( 2x_1 + 4x_2 = 16 \)
- Simplify: \( x_1 + 2x_2 = 8 \)
- Find intercepts:
- When \( x_1 = 0 \): \( 2x_2 = 8 \implies x_2 = 4 \)
- When \( x_2 = 0 \): \( x_1 = 8 \)
- Plot the line passing through points \( (0, 4) \) and \( (8, 0) \).

- Constraint 2: \( 4x_1 + 3x_2 = 24 \)
- Find intercepts:
- When \( x_1 = 0 \): \( 3x_2 = 24 \implies x_2 = 8 \)
- When \( x_2 = 0 \): \( 4x_1 = 24 \implies x_1 = 6 \)
- Plot the line passing through points \( (0, 8) \) and \( (6, 0) \).

- Non-negativity Constraints: \( x_1 \geq 0 \) and \( x_2 \geq 0 \)
- These restrict the feasible region to the first quadrant.

#### 2. Identify the Feasible Region:
The feasible region is the area where all constraints are satisfied simultaneously. From the graph:
- The feasible region is the area above both constraint lines and within the first quadrant.
- The intersection points of the constraint lines and the axes define the vertices of the feasible region.

#### 3. Find the Intersection Points:
To find the exact coordinates of the intersection points of the constraint lines, solve the system of equations:

- Intersection of \( x_1 + 2x_2 = 8 \) and \( 4x_1 + 3x_2 = 24 \):
- From \( x_1 + 2x_2 = 8 \): \( x_1 = 8 - 2x_2 \)
- Substitute into \( 4x_1 + 3x_2 = 24 \):
\[
4(8 - 2x_2) + 3x_2 = 24
\]
\[
32 - 8x_2 + 3x_2 = 24
\]
\[
32 - 5x_2 = 24
\]
\[
-5x_2 = -8 \implies x_2 = \frac{8}{5}
\]
- Substitute \( x_2 = \frac{8}{5} \) back into \( x_1 = 8 - 2x_2 \):
\[
x_1 = 8 - 2\left(\frac{8}{5}\right) = 8 - \frac{16}{5} = \frac{40}{5} - \frac{16}{5} = \frac{24}{5}
\]
- Intersection point: \( \left( \frac{24}{5}, \frac{8}{5} \right) \)

- Intersection of \( x_1 + 2x_2 = 8 \) and \( x_1 = 0 \):
- From \( x_1 + 2x_2 = 8 \):
\[
0 + 2x_2 = 8 \implies x_2 = 4
\]
- Intersection point: \( (0, 4) \)

- Intersection of \( 4x_1 + 3x_2 = 24 \) and \( x_2 = 0 \):
- From \( 4x_1 + 3x_2 = 24 \):
\[
4x_1 + 0 = 24 \implies x_1 = 6
\]
- Intersection point: \( (6, 0) \)

#### 4. Evaluate the Objective Function at the Vertices:
The vertices of the feasible region are:
1. \( \left( \frac{24}{5}, \frac{8}{5} \right) \)
2. \( (0, 4) \)
3. \( (6, 0) \)

Evaluate the objective function \( Z = 6x_1 + 3x_2 \) at each vertex:

- At \( \left( \frac{24}{5}, \frac{8}{5} \right) \):
\[
Z = 6\left(\frac{24}{5}\right) + 3\left(\frac{8}{5}\right) = \frac{144}{5} + \frac{24}{5} = \frac{168}{5} = 33.6
\]

- At \( (0, 4) \):
\[
Z = 6(0) + 3(4) = 0 + 12 = 12
\]

- At \( (6, 0) \):
\[
Z = 6(6) + 3(0) = 36 + 0 = 36
\]

#### 5. Determine the Minimum Value:
The minimum value of \( Z \) occurs at the vertex \( (0, 4) \), where \( Z = 12 \).

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Final Answer:

\[
\boxed{(0, 4)}
\]