To solve the given linear programming problem graphically, we will follow these steps:
### Step 1: Define the Objective Function and Constraints
We need to minimize the objective function:
\[ Z = 200x + 500y \]
subject to the constraints:
1. \( x + 2y \geq 10 \) (Constraint 1)
2. \( 3x + 4y \leq 24 \) (Constraint 2)
3. \( x \geq 0 \) (Constraint 3)
4. \( y \geq 0 \) (Constraint 4)
### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( x + 2y = 10 \)
2. \( 3x + 4y = 24 \)
### Step 3: Find Intercepts for Each Constraint
**For Constraint 1: \( x + 2y = 10 \)**
- To find the x-intercept, set \( y = 0 \):
\[ x + 2(0) = 10 \implies x = 10 \]
(Point: (10, 0))
- To find the y-intercept, set \( x = 0 \):
\[ 0 + 2y = 10 \implies y = 5 \]
(Point: (0, 5))
**For Constraint 2: \( 3x + 4y = 24 \)**
- To find the x-intercept, set \( y = 0 \):
\[ 3x + 4(0) = 24 \implies x = 8 \]
(Point: (8, 0))
- To find the y-intercept, set \( x = 0 \):
\[ 3(0) + 4y = 24 \implies y = 6 \]
(Point: (0, 6))
### Step 4: Graph the Constraints
Now, we will plot the lines on a graph:
- Draw the line for \( x + 2y = 10 \) connecting points (10, 0) and (0, 5).
- Draw the line for \( 3x + 4y = 24 \) connecting points (8, 0) and (0, 6).
### Step 5: Determine Feasible Region
- For \( x + 2y \geq 10 \), shade the region above the line.
- For \( 3x + 4y \leq 24 \), shade the region below the line.
- Since \( x \geq 0 \) and \( y \geq 0 \), we only consider the first quadrant.
### Step 6: Identify the Vertices of the Feasible Region
The feasible region is bounded by the lines and the axes. The vertices of the feasible region are:
1. (0, 5)
2. (0, 6)
3. (8, 0)
4. The intersection point of the two lines.
### Step 7: Find the Intersection Point
To find the intersection of the lines \( x + 2y = 10 \) and \( 3x + 4y = 24 \), we can solve the equations simultaneously.
From \( x + 2y = 10 \):
\[ x = 10 - 2y \]
Substituting into \( 3x + 4y = 24 \):
\[ 3(10 - 2y) + 4y = 24 \]
\[ 30 - 6y + 4y = 24 \]
\[ -2y = -6 \implies y = 3 \]
Substituting \( y = 3 \) back into \( x + 2y = 10 \):
\[ x + 2(3) = 10 \implies x = 4 \]
Thus, the intersection point is (4, 3).
### Step 8: Evaluate the Objective Function at Each Vertex
Now we evaluate \( Z = 200x + 500y \) at each vertex:
1. At (0, 5):
\[ Z = 200(0) + 500(5) = 2500 \]
2. At (0, 6):
\[ Z = 200(0) + 500(6) = 3000 \]
3. At (8, 0):
\[ Z = 200(8) + 500(0) = 1600 \]
4. At (4, 3):
\[ Z = 200(4) + 500(3) = 800 + 1500 = 2300 \]
### Step 9: Determine the Minimum Value
The minimum value of \( Z \) occurs at the point (4, 3):
\[ \text{Minimum } Z = 2300 \]
### Final Answer
The minimum value of \( Z \) is \( 2300 \) at the point \( (4, 3) \).
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