To solve the given Linear Programming Problem (LPP) graphically, we will follow these steps:
### Step 1: Write the objective function and constraints
We need to minimize the objective function:
\[ Z = 4x + 5y \]
Subject to the constraints:
1. \( 5x + y \geq 10 \)
2. \( x + y \geq 6 \)
3. \( x + 4y \geq 12 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert inequalities to equations
To graph the constraints, we convert the inequalities into equations:
1. \( 5x + y = 10 \)
2. \( x + y = 6 \)
3. \( x + 4y = 12 \)
### Step 3: Find intercepts of each line
**For \( 5x + y = 10 \):**
- When \( x = 0 \): \( y = 10 \) (Point: \( (0, 10) \))
- When \( y = 0 \): \( 5x = 10 \) → \( x = 2 \) (Point: \( (2, 0) \))
**For \( x + y = 6 \):**
- When \( x = 0 \): \( y = 6 \) (Point: \( (0, 6) \))
- When \( y = 0 \): \( x = 6 \) (Point: \( (6, 0) \))
**For \( x + 4y = 12 \):**
- When \( x = 0 \): \( 4y = 12 \) → \( y = 3 \) (Point: \( (0, 3) \))
- When \( y = 0 \): \( x = 12 \) (Point: \( (12, 0) \))
### Step 4: Plot the lines on a graph
Plot the points obtained from the intercepts on a graph and draw the lines corresponding to each equation.
### Step 5: Determine the feasible region
The feasible region is where all the constraints overlap. Since all inequalities are of the form \( \geq \), the feasible region will be above the lines.
### Step 6: Identify corner points of the feasible region
Find the intersection points of the lines:
1. **Intersection of \( 5x + y = 10 \) and \( x + y = 6 \):**
\[
\begin{align*}
5x + y &= 10 \\
x + y &= 6 \\
\text{Subtract the second from the first:} \\
5x + y - (x + y) &= 10 - 6 \\
4x &= 4 \\
x &= 1 \\
y &= 6 - 1 = 5 \\
\text{Point: } (1, 5)
\end{align*}
\]
2. **Intersection of \( x + y = 6 \) and \( x + 4y = 12 \):**
\[
\begin{align*}
x + y &= 6 \\
x + 4y &= 12 \\
\text{Subtract the first from the second:} \\
(x + 4y) - (x + y) &= 12 - 6 \\
3y &= 6 \\
y &= 2 \\
x &= 6 - 2 = 4 \\
\text{Point: } (4, 2)
\end{align*}
\]
3. **Intersection of \( 5x + y = 10 \) and \( x + 4y = 12 \):**
\[
\begin{align*}
5x + y &= 10 \\
x + 4y &= 12 \\
\text{Multiply the second by 5:} \\
5x + 20y &= 60 \\
\text{Subtract the first from this:} \\
20y - y &= 60 - 10 \\
19y &= 50 \\
y &= \frac{50}{19} \approx 2.63 \\
x &= 12 - 4y = 12 - 4 \cdot \frac{50}{19} = \frac{28}{19} \approx 1.47 \\
\text{Point: } \left(\frac{28}{19}, \frac{50}{19}\right)
\end{align*}
\]
### Step 7: Evaluate the objective function at each corner point
1. At \( (0, 10) \):
\[ Z = 4(0) + 5(10) = 50 \]
2. At \( (1, 5) \):
\[ Z = 4(1) + 5(5) = 4 + 25 = 29 \]
3. At \( (4, 2) \):
\[ Z = 4(4) + 5(2) = 16 + 10 = 26 \]
4. At \( (12, 0) \):
\[ Z = 4(12) + 5(0) = 48 \]
5. At \( \left(\frac{28}{19}, \frac{50}{19}\right) \):
\[ Z = 4\left(\frac{28}{19}\right) + 5\left(\frac{50}{19}\right) = \frac{112}{19} + \frac{250}{19} = \frac{362}{19} \approx 19.05 \]
### Step 8: Determine the minimum value
The minimum value of \( Z \) occurs at the point \( (4, 2) \) with:
\[ \text{Minimum } Z = 26 \]
### Final Answer
The minimum value of \( Z \) is \( 26 \) at the point \( (4, 2) \).
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