Minimize `Z = 2x + 3y ` subject to constraints `x + y ge 6 , 2x + y ge 7, x + 4y ge 8 , x ge 0 , y ge 0`

Answer

To minimize the function \( Z = 2x + 3y \) subject to the constraints:

1. \( x + y \geq 6 \)
2. \( 2x + y \geq 7 \)
3. \( x + 4y \geq 8 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the Constraints
We start by rewriting the inequalities as equations to find the boundary lines:

1. \( x + y = 6 \)
2. \( 2x + y = 7 \)
3. \( x + 4y = 8 \)

### Step 2: Find Intercepts for Each Line
We will find the intercepts for each line to plot them:

- For \( x + y = 6 \):
- When \( x = 0 \), \( y = 6 \) → Point (0, 6)
- When \( y = 0 \), \( x = 6 \) → Point (6, 0)

- For \( 2x + y = 7 \):
- When \( x = 0 \), \( y = 7 \) → Point (0, 7)
- When \( y = 0 \), \( x = 3.5 \) → Point (3.5, 0)

- For \( x + 4y = 8 \):
- When \( x = 0 \), \( y = 2 \) → Point (0, 2)
- When \( y = 0 \), \( x = 8 \) → Point (8, 0)

### Step 3: Find Points of Intersection
Next, we find the points of intersection of the lines:

1. **Intersection of \( x + y = 6 \) and \( 2x + y = 7 \)**:
- Subtract the first equation from the second:
\[ (2x + y) - (x + y) = 7 - 6 \implies x = 1 \]
- Substitute \( x = 1 \) back into \( x + y = 6 \):
\[ 1 + y = 6 \implies y = 5 \]
- Intersection Point: \( (1, 5) \)

2. **Intersection of \( 2x + y = 7 \) and \( x + 4y = 8 \)**:
- Multiply the first equation by 4:
\[ 8x + 4y = 28 \]
- Now subtract \( x + 4y = 8 \):
\[ 8x + 4y - (x + 4y) = 28 - 8 \implies 7x = 20 \implies x = \frac{20}{7} \]
- Substitute \( x = \frac{20}{7} \) back into \( 2x + y = 7 \):
\[ 2\left(\frac{20}{7}\right) + y = 7 \implies \frac{40}{7} + y = 7 \implies y = 7 - \frac{40}{7} = \frac{9}{7} \]
- Intersection Point: \( \left(\frac{20}{7}, \frac{9}{7}\right) \)

3. **Intersection of \( x + y = 6 \) and \( x + 4y = 8 \)**:
- From \( x + y = 6 \), we have \( y = 6 - x \).
- Substitute into \( x + 4y = 8 \):
\[ x + 4(6 - x) = 8 \implies x + 24 - 4x = 8 \implies -3x = -16 \implies x = \frac{16}{3} \]
- Substitute \( x = \frac{16}{3} \) back into \( y = 6 - x \):
\[ y = 6 - \frac{16}{3} = \frac{18 - 16}{3} = \frac{2}{3} \]
- Intersection Point: \( \left(\frac{16}{3}, \frac{2}{3}\right) \)

### Step 4: Identify the Feasible Region
The feasible region is determined by the intersection points and the constraints. The points we have are:

1. \( (0, 7) \)
2. \( (1, 5) \)
3. \( \left(\frac{20}{7}, \frac{9}{7}\right) \)
4. \( \left(\frac{16}{3}, \frac{2}{3}\right) \)
5. \( (6, 0) \)

### Step 5: Evaluate the Objective Function at Each Vertex
Now we evaluate \( Z = 2x + 3y \) at each of the vertices:

1. At \( (0, 7) \):
\[ Z = 2(0) + 3(7) = 21 \]

2. At \( (1, 5) \):
\[ Z = 2(1) + 3(5) = 2 + 15 = 17 \]

3. At \( \left(\frac{20}{7}, \frac{9}{7}\right) \):
\[ Z = 2\left(\frac{20}{7}\right) + 3\left(\frac{9}{7}\right) = \frac{40}{7} + \frac{27}{7} = \frac{67}{7} \approx 9.57 \]

4. At \( \left(\frac{16}{3}, \frac{2}{3}\right) \):
\[ Z = 2\left(\frac{16}{3}\right) + 3\left(\frac{2}{3}\right) = \frac{32}{3} + \frac{6}{3} = \frac{38}{3} \approx 12.67 \]

5. At \( (6, 0) \):
\[ Z = 2(6) + 3(0) = 12 \]

### Step 6: Determine the Minimum Value
Comparing all the values of \( Z \):
- \( Z(0, 7) = 21 \)
- \( Z(1, 5) = 17 \)
- \( Z\left(\frac{20}{7}, \frac{9}{7}\right) \approx 9.57 \)
- \( Z\left(\frac{16}{3}, \frac{2}{3}\right) \approx 12.67 \)
- \( Z(6, 0) = 12 \)

The minimum value occurs at the point \( \left(\frac{20}{7}, \frac{9}{7}\right) \) with \( Z \approx 9.57 \).

### Final Answer
The minimum value of \( Z \) is approximately \( 9.57 \) at the point \( \left(\frac{20}{7}, \frac{9}{7}\right) \).

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