Question
SOLVE the following linear programming problem using graphic method?
maximize z=30x1+50x2
subject to
3x1+x2>/=9
x1+2x2>/=12
x1+2x2>/=9
x1,x2>/=0
maximize z=30x1+50x2
subject to
3x1+x2>/=9
x1+2x2>/=12
x1+2x2>/=9
x1,x2>/=0
Answers
GPT-4o mini
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:**
1. \( 3x_1 + x_2 \geq 9 \)
2. \( x_1 + 2x_2 \geq 12 \)
3. \( x_1 + 2x_2 \geq 9 \)
4. \( x_1, x_2 \geq 0 \)
### Step 1: Rearranging the Constraints
We first rearrange the inequalities into slope-intercept form \( y = mx + b \).
1. **For \( 3x_1 + x_2 \geq 9 \)**:
Rearranging gives \( x_2 \geq -3x_1 + 9 \)
(The line is \( x_2 = -3x_1 + 9 \))
2. **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 \( x_2 = -\frac{1}{2}x_1 + 6 \))
3. **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:
1. **\( x_2 = -3x_1 + 9 \)**:
- When \( x_1 = 0 \), \( x_2 = 9 \) (point (0,9))
- When \( x_1 = 3 \), \( x_2 = 0 \) (point (3,0))
2. **\( x_2 = -\frac{1}{2}x_1 + 6 \)**:
- When \( x_1 = 0 \), \( x_2 = 6 \) (point (0,6))
- When \( x_1 = 12 \), \( x_2 = 0 \) (point (12,0))
3. **\( x_2 = -\frac{1}{2}x_1 + 4.5 \)**:
- 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.
1. 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:
\[
-3x_1 + \frac{1}{2}x_1 = 6 - 9
\]
\[
-\frac{5}{2}x_1 = -3 \implies x_1 = \frac{6}{5}, \quad x_2 = 9 - 3(\frac{6}{5}) = \frac{9 - \frac{18}{5}}{1} = \frac{27}{5}
\]
2. 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.
3. 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:
\[
\text{Max } z = 300 \text{ at } (x_1, x_2) = (0, 6)
\]
**Maximize:**
\( z = 30x_1 + 50x_2 \)
**Subject to:**
1. \( 3x_1 + x_2 \geq 9 \)
2. \( x_1 + 2x_2 \geq 12 \)
3. \( x_1 + 2x_2 \geq 9 \)
4. \( x_1, x_2 \geq 0 \)
### Step 1: Rearranging the Constraints
We first rearrange the inequalities into slope-intercept form \( y = mx + b \).
1. **For \( 3x_1 + x_2 \geq 9 \)**:
Rearranging gives \( x_2 \geq -3x_1 + 9 \)
(The line is \( x_2 = -3x_1 + 9 \))
2. **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 \( x_2 = -\frac{1}{2}x_1 + 6 \))
3. **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:
1. **\( x_2 = -3x_1 + 9 \)**:
- When \( x_1 = 0 \), \( x_2 = 9 \) (point (0,9))
- When \( x_1 = 3 \), \( x_2 = 0 \) (point (3,0))
2. **\( x_2 = -\frac{1}{2}x_1 + 6 \)**:
- When \( x_1 = 0 \), \( x_2 = 6 \) (point (0,6))
- When \( x_1 = 12 \), \( x_2 = 0 \) (point (12,0))
3. **\( x_2 = -\frac{1}{2}x_1 + 4.5 \)**:
- 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.
1. 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:
\[
-3x_1 + \frac{1}{2}x_1 = 6 - 9
\]
\[
-\frac{5}{2}x_1 = -3 \implies x_1 = \frac{6}{5}, \quad x_2 = 9 - 3(\frac{6}{5}) = \frac{9 - \frac{18}{5}}{1} = \frac{27}{5}
\]
2. 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.
3. 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:
\[
\text{Max } z = 300 \text{ at } (x_1, x_2) = (0, 6)
\]