Question
solve the following linear programming problem using graphical method?
maximize z=110x1+110x2
subject to
x1+x2</=9
x1>/=2
20x1+50x2</=360
x2>/=3
x1,x2>/=0
maximize z=110x1+110x2
subject to
x1+x2</=9
x1>/=2
20x1+50x2</=360
x2>/=3
x1,x2>/=0
Answers
GPT-4o mini
To solve the linear programming problem using the graphical method, follow these steps:
### Problem Formulation
We need to maximize the objective function:
\[ z = 110x_1 + 110x_2 \]
Subject to the constraints:
1. \( x_1 + x_2 \leq 9 \)
2. \( x_1 \geq 2 \)
3. \( 20x_1 + 50x_2 \leq 360 \)
4. \( x_2 \geq 3 \)
5. \( x_1, x_2 \geq 0 \)
### Step 1: Convert Inequalities to Equalities
To graph the constraints, we convert the inequalities to equalities:
1. \( x_1 + x_2 = 9 \)
2. \( x_1 = 2 \)
3. \( 20x_1 + 50x_2 = 360 \)
4. \( x_2 = 3 \)
### Step 2: Find the Intersection Points
Let's plot each of the lines defined by these equations:
1. **For \( x_1 + x_2 = 9 \):**
- When \( x_1 = 0 \), \( x_2 = 9 \) (point A: (0, 9))
- When \( x_2 = 0 \), \( x_1 = 9 \) (point B: (9, 0))
2. **For \( x_1 = 2 \):**
- This is a vertical line at \( x_1 = 2 \).
3. **For \( 20x_1 + 50x_2 = 360 \):**
- When \( x_1 = 0 \), \( 50x_2 = 360 \) leads to \( x_2 = 7.2 \) (point C: (0, 7.2))
- When \( x_2 = 0 \), \( 20x_1 = 360 \) leads to \( x_1 = 18 \) (point D: (18, 0))
4. **For \( x_2 = 3 \):**
- This is a horizontal line at \( x_2 = 3 \).
### Step 3: Plot the Constraints on the Graph
Next, plot the points and lines on a graph:
- The line from (0, 9) to (9, 0) represents the constraint \( x_1 + x_2 \leq 9 \).
- The vertical line at \( x_1 = 2 \).
- The line from (0, 7.2) to (18, 0) represents the constraint \( 20x_1 + 50x_2 \leq 360 \).
- The horizontal line at \( x_2 = 3 \).
### Step 4: Identify the Feasible Region
The feasible region is the area that satisfies all constraints.
Starting from the inequalities:
- The area below or on the line \( x_1 + x_2 = 9 \).
- The area to the right of the line \( x_1 = 2 \).
- The area below or on the line \( 20x_1 + 50x_2 = 360 \).
- The area above the line \( x_2 = 3 \).
#### Finding Vertices of the Feasible Region
Evaluate the intersections of the constraint lines:
1. Intersection of \( x_1 + x_2 = 9 \) and \( x_2 = 3 \):
\[
x_1 + 3 = 9 \implies x_1 = 6 \quad \Rightarrow (6, 3)
\]
2. Intersection of \( x_1 + x_2 = 9 \) and \( x_1 = 2 \):
\[
2 + x_2 = 9 \implies x_2 = 7 \quad \Rightarrow (2, 7)
\]
3. Intersection of \( 20x_1 + 50x_2 = 360 \) and \( x_2 = 3 \):
\[
20x_1 + 50 \cdot 3 = 360 \implies 20x_1 + 150 = 360 \implies 20x_1 = 210 \implies x_1 = 10.5 \quad \Rightarrow (10.5, 3)
\]
But since \( x_1 \leq 9 \) here, this point is not in the feasible region.
4. Intersection of \( 20x_1 + 50x_2 = 360 \) and \( x_1 = 2 \):
\[
20 \cdot 2 + 50x_2 = 360 \implies 40 + 50x_2 = 360 \implies 50x_2 = 320 \implies x_2 = 6.4 \quad \Rightarrow (2, 6.4)
\]
### Step 5: Evaluate the Objective Function at the Vertices
Now evaluate \( z = 110x_1 + 110x_2 \) at the feasible vertex points:
1. \( (6, 3) \):
\[ z = 110(6) + 110(3) = 660 + 330 = 990 \]
2. \( (2, 7) \):
\[ z = 110(2) + 110(7) = 220 + 770 = 990 \]
3. \( (2, 6.4) \):
\[ z = 110(2) + 110(6.4) = 220 + 704 = 924 \]
### Step 6: Conclusion
The optimal solution occurs at both points \( (6, 3) \) and \( (2, 7) \), yielding the maximum value of:
\[
\text{Max } z = 990
\]
Thus, the maximum value of the objective function, within the feasible region defined by the constraints, is 990 at two points: \( (6, 3) \) and \( (2, 7) \).
### Problem Formulation
We need to maximize the objective function:
\[ z = 110x_1 + 110x_2 \]
Subject to the constraints:
1. \( x_1 + x_2 \leq 9 \)
2. \( x_1 \geq 2 \)
3. \( 20x_1 + 50x_2 \leq 360 \)
4. \( x_2 \geq 3 \)
5. \( x_1, x_2 \geq 0 \)
### Step 1: Convert Inequalities to Equalities
To graph the constraints, we convert the inequalities to equalities:
1. \( x_1 + x_2 = 9 \)
2. \( x_1 = 2 \)
3. \( 20x_1 + 50x_2 = 360 \)
4. \( x_2 = 3 \)
### Step 2: Find the Intersection Points
Let's plot each of the lines defined by these equations:
1. **For \( x_1 + x_2 = 9 \):**
- When \( x_1 = 0 \), \( x_2 = 9 \) (point A: (0, 9))
- When \( x_2 = 0 \), \( x_1 = 9 \) (point B: (9, 0))
2. **For \( x_1 = 2 \):**
- This is a vertical line at \( x_1 = 2 \).
3. **For \( 20x_1 + 50x_2 = 360 \):**
- When \( x_1 = 0 \), \( 50x_2 = 360 \) leads to \( x_2 = 7.2 \) (point C: (0, 7.2))
- When \( x_2 = 0 \), \( 20x_1 = 360 \) leads to \( x_1 = 18 \) (point D: (18, 0))
4. **For \( x_2 = 3 \):**
- This is a horizontal line at \( x_2 = 3 \).
### Step 3: Plot the Constraints on the Graph
Next, plot the points and lines on a graph:
- The line from (0, 9) to (9, 0) represents the constraint \( x_1 + x_2 \leq 9 \).
- The vertical line at \( x_1 = 2 \).
- The line from (0, 7.2) to (18, 0) represents the constraint \( 20x_1 + 50x_2 \leq 360 \).
- The horizontal line at \( x_2 = 3 \).
### Step 4: Identify the Feasible Region
The feasible region is the area that satisfies all constraints.
Starting from the inequalities:
- The area below or on the line \( x_1 + x_2 = 9 \).
- The area to the right of the line \( x_1 = 2 \).
- The area below or on the line \( 20x_1 + 50x_2 = 360 \).
- The area above the line \( x_2 = 3 \).
#### Finding Vertices of the Feasible Region
Evaluate the intersections of the constraint lines:
1. Intersection of \( x_1 + x_2 = 9 \) and \( x_2 = 3 \):
\[
x_1 + 3 = 9 \implies x_1 = 6 \quad \Rightarrow (6, 3)
\]
2. Intersection of \( x_1 + x_2 = 9 \) and \( x_1 = 2 \):
\[
2 + x_2 = 9 \implies x_2 = 7 \quad \Rightarrow (2, 7)
\]
3. Intersection of \( 20x_1 + 50x_2 = 360 \) and \( x_2 = 3 \):
\[
20x_1 + 50 \cdot 3 = 360 \implies 20x_1 + 150 = 360 \implies 20x_1 = 210 \implies x_1 = 10.5 \quad \Rightarrow (10.5, 3)
\]
But since \( x_1 \leq 9 \) here, this point is not in the feasible region.
4. Intersection of \( 20x_1 + 50x_2 = 360 \) and \( x_1 = 2 \):
\[
20 \cdot 2 + 50x_2 = 360 \implies 40 + 50x_2 = 360 \implies 50x_2 = 320 \implies x_2 = 6.4 \quad \Rightarrow (2, 6.4)
\]
### Step 5: Evaluate the Objective Function at the Vertices
Now evaluate \( z = 110x_1 + 110x_2 \) at the feasible vertex points:
1. \( (6, 3) \):
\[ z = 110(6) + 110(3) = 660 + 330 = 990 \]
2. \( (2, 7) \):
\[ z = 110(2) + 110(7) = 220 + 770 = 990 \]
3. \( (2, 6.4) \):
\[ z = 110(2) + 110(6.4) = 220 + 704 = 924 \]
### Step 6: Conclusion
The optimal solution occurs at both points \( (6, 3) \) and \( (2, 7) \), yielding the maximum value of:
\[
\text{Max } z = 990
\]
Thus, the maximum value of the objective function, within the feasible region defined by the constraints, is 990 at two points: \( (6, 3) \) and \( (2, 7) \).