To solve the linear programming problem using the graphical method, we need to first plot the constraints and then identify the feasible region. The problem to solve is:
Maximize:
\( z = 30x_1 + 50x_2 \)
Subject to:
- \( 3x_1 + x_2 \geq 9 \)
- \( x_1 + 2x_2 \geq 12 \)
- \( x_1 + 2x_2 \geq 9 \)
- \( x_1, x_2 \geq 0 \)
Step 1: Rearranging the Constraints
We first rearrange the inequalities into slope-intercept form \( y = mx + b \).
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For \( 3x_1 + x_2 \geq 9 \):
Rearranging gives \( x_2 \geq -3x_1 + 9 \)
(The line is \( x_2 = -3x_1 + 9 \)) -
For \( x_1 + 2x_2 \geq 12 \):
Rearranging gives \( 2x_2 \geq -x_1 + 12 \)
or \( x_2 \geq -\frac{1}{2}x_1 + 6 \)
(The line is \( x_2 = -\frac{1}{2}x_1 + 6 \)) -
For \( x_1 + 2x_2 \geq 9 \):
Rearranging gives \( 2x_2 \geq -x_1 + 9 \)
or \( x_2 \geq -\frac{1}{2}x_1 + \frac{9}{2} \)
(The line is \( x_2 = -\frac{1}{2}x_1 + 4.5 \))
Step 2: Plotting the Constraints
Next, we can plot these lines on a graph:
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\( x_2 = -3x_1 + 9 \):
- When \( x_1 = 0 \), \( x_2 = 9 \) (point (0,9))
- When \( x_1 = 3 \), \( x_2 = 0 \) (point (3,0))
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\( x_2 = -\frac{1}{2}x_1 + 6 \):
- When \( x_1 = 0 \), \( x_2 = 6 \) (point (0,6))
- When \( x_1 = 12 \), \( x_2 = 0 \) (point (12,0))
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\( x_2 = -\frac{1}{2}x_1 + 4.5 \):
- When \( x_1 = 0 \), \( x_2 = 4.5 \) (point (0,4.5))
- When \( x_1 = 9 \), \( x_2 = 0 \) (point (9,0))
After plotting these lines, we shade the region that satisfies all constraints (i.e., the feasible region). The inequalities indicate that we shade above each line.
Step 3: Identifying the Feasible Region
The feasible region will be the area where all shaded regions overlap, constrained by the axes where \( x_1 \geq 0 \) and \( x_2 \geq 0 \).
Step 4: Finding the Corner Points
Next, we find the corner points of the feasible region by identifying where the constraint lines intersect.
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Intersection of \( x_2 = -3x_1 + 9 \) and \( x_2 = -\frac{1}{2}x_1 + 6 \): \[ -3x_1 + 9 = -\frac{1}{2}x_1 + 6 \] Solving gives: \[ -3x_1 + \frac{1}{2}x_1 = 6 - 9 \] \[ -\frac{5}{2}x_1 = -3 \implies x_1 = \frac{6}{5}, \quad x_2 = 9 - 3(\frac{6}{5}) = \frac{9 - \frac{18}{5}}{1} = \frac{27}{5} \]
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Intersection of \( x_2 = -\frac{1}{2}x_1 + 6 \) and \( x_2 = -\frac{1}{2}x_1 + 4.5 \): is not useful since they are parallel.
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Others as needed.
Step 5: Evaluating the Objective Function
Calculate \( z \) at each corner point identified in the feasible region:
- \( P1: (0, 6) \) -> \( z = 30(0) + 50(6) = 300 \)
- \( P2: (0, 4.5) \) -> \( z = 30(0) + 50(4.5) = 225 \)
- \( P3: (3, 0) \) -> \( z = 30(3) + 50(0) = 90 \)
- \( P4: (9, 0) \) -> \( z = 30(9) + 50(0) = 270 \)
Step 6: Conclusion
The maximum value occurs at the point \( (0, 6) \) with \( z = 300 \).
Thus, the optimal solution is: \[ \text{Max } z = 300 \text{ at } (x_1, x_2) = (0, 6) \]