To minimize the objective function \( Z = 6x + 3y \) subject to the given constraints, we will follow these steps:
### Step 1: Write down the constraints
The constraints given are:
1. \( 4x + y \geq 80 \)
2. \( x + 5y \geq 115 \)
3. \( 3x + 2y \leq 150 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert inequalities to equations
To find the feasible region, we convert the inequalities into equations:
1. \( 4x + y = 80 \) (Equation 1)
2. \( x + 5y = 115 \) (Equation 2)
3. \( 3x + 2y = 150 \) (Equation 3)
### Step 3: Find intercepts for each equation
**For Equation 1:**
- When \( x = 0 \): \( y = 80 \) (Point: \( (0, 80) \))
- When \( y = 0 \): \( x = 20 \) (Point: \( (20, 0) \))
**For Equation 2:**
- When \( x = 0 \): \( 5y = 115 \) → \( y = 23 \) (Point: \( (0, 23) \))
- When \( y = 0 \): \( x = 115 \) (Point: \( (115, 0) \))
**For Equation 3:**
- When \( x = 0 \): \( 2y = 150 \) → \( y = 75 \) (Point: \( (0, 75) \))
- When \( y = 0 \): \( 3x = 150 \) → \( x = 50 \) (Point: \( (50, 0) \))
### Step 4: Plot the lines and find intersection points
Next, we will find the points of intersection of the lines:
**Intersection of Equation 1 and Equation 2:**
1. \( 4x + y = 80 \)
2. \( x + 5y = 115 \)
From Equation 1, express \( y \):
\[ y = 80 - 4x \]
Substituting into Equation 2:
\[ x + 5(80 - 4x) = 115 \]
\[ x + 400 - 20x = 115 \]
\[ -19x = -285 \]
\[ x = 15 \]
Now substitute \( x = 15 \) back into Equation 1:
\[ 4(15) + y = 80 \]
\[ 60 + y = 80 \]
\[ y = 20 \]
So the intersection point is \( (15, 20) \).
**Intersection of Equation 1 and Equation 3:**
1. \( 4x + y = 80 \)
2. \( 3x + 2y = 150 \)
From Equation 1, express \( y \):
\[ y = 80 - 4x \]
Substituting into Equation 3:
\[ 3x + 2(80 - 4x) = 150 \]
\[ 3x + 160 - 8x = 150 \]
\[ -5x = -10 \]
\[ x = 2 \]
Now substitute \( x = 2 \) back into Equation 1:
\[ 4(2) + y = 80 \]
\[ 8 + y = 80 \]
\[ y = 72 \]
So the intersection point is \( (2, 72) \).
**Intersection of Equation 2 and Equation 3:**
1. \( x + 5y = 115 \)
2. \( 3x + 2y = 150 \)
From Equation 2, express \( x \):
\[ x = 115 - 5y \]
Substituting into Equation 3:
\[ 3(115 - 5y) + 2y = 150 \]
\[ 345 - 15y + 2y = 150 \]
\[ -13y = -195 \]
\[ y = 15 \]
Now substitute \( y = 15 \) back into Equation 2:
\[ x + 5(15) = 115 \]
\[ x + 75 = 115 \]
\[ x = 40 \]
So the intersection point is \( (40, 15) \).
### Step 5: Identify the corner points
The corner points of the feasible region are:
1. \( (0, 72) \)
2. \( (15, 20) \)
3. \( (40, 15) \)
4. \( (20, 0) \) (from the first constraint)
### Step 6: Evaluate the objective function at each corner point
Now, we evaluate \( Z = 6x + 3y \) at each corner point:
1. At \( (0, 72) \):
\[ Z = 6(0) + 3(72) = 216 \]
2. At \( (15, 20) \):
\[ Z = 6(15) + 3(20) = 90 + 60 = 150 \]
3. At \( (40, 15) \):
\[ Z = 6(40) + 3(15) = 240 + 45 = 285 \]
4. At \( (20, 0) \):
\[ Z = 6(20) + 3(0) = 120 \]
### Step 7: Determine the minimum value
The minimum value of \( Z \) occurs at the point \( (15, 20) \) with:
\[ Z = 150 \]
### Final Answer
The minimum value of \( Z \) is **150** at the point \( (15, 20) \).
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