The minimum value of Z = 10x + By subject to ` 4x +y ge 4, x +3y ge 6, x +y ge 3, xge 0, y ge 0` is

Options

A. 60
B. 27
C. `74/3 `
D. 32

Explanation

To find the minimum value of \( Z = 10x + 8y \) subject to the constraints:

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

we will follow these steps:

### Step 1: Convert Inequalities to Equations
First, we convert the inequalities into equations to find the boundary lines:

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

### Step 2: Find Intersection Points
Next, we will find the intersection points of these lines.

- **Intersection of \( 4x + y = 4 \) and \( x + 3y = 6 \)**:
\[ y = 4 - 4x \quad \text{(from the first equation)} \]
Substitute into the second equation:
\[ x + 3(4 - 4x) = 6 \\ x + 12 - 12x = 6 \\ -11x = -6 \\ x = \frac{6}{11} \]
Substitute \( x \) back to find \( y \):
\[ y = 4 - 4\left(\frac{6}{11}\right) = 4 - \frac{24}{11} = \frac{44 - 24}{11} = \frac{20}{11} \]
So, one intersection point is \( \left(\frac{6}{11}, \frac{20}{11}\right) \).

- **Intersection of \( x + 3y = 6 \) and \( x + y = 3 \)**:
\[ x + 3y = 6 \quad \text{(from the second equation)} \\ y = 3 - x \quad \text{(from the third equation)} \]
Substitute into the second equation:
\[ x + 3(3 - x) = 6 \\ x + 9 - 3x = 6 \\ -2x = -3 \\ x = \frac{3}{2} \]
Substitute \( x \) back to find \( y \):
\[ y = 3 - \frac{3}{2} = \frac{3}{2} \]
So, another intersection point is \( \left(\frac{3}{2}, \frac{3}{2}\right) \).

- **Intersection of \( 4x + y = 4 \) and \( x + y = 3 \)**:
\[ y = 4 - 4x \\ x + (4 - 4x) = 3 \\ x + 4 - 4x = 3 \\ -3x = -1 \\ x = \frac{1}{3} \]
Substitute \( x \) back to find \( y \):
\[ y = 3 - \frac{1}{3} = \frac{8}{3} \]
So, another intersection point is \( \left(\frac{1}{3}, \frac{8}{3}\right) \).

### Step 3: Identify Feasible Region
Now we need to check which of these points satisfy all the original inequalities to determine the feasible region.

### Step 4: Evaluate Z at Each Vertex
Evaluate \( Z = 10x + 8y \) at each of the feasible vertices:

1. For \( \left(\frac{6}{11}, \frac{20}{11}\right) \):
\[ Z = 10\left(\frac{6}{11}\right) + 8\left(\frac{20}{11}\right) = \frac{60}{11} + \frac{160}{11} = \frac{220}{11} = 20 \]

2. For \( \left(\frac{3}{2}, \frac{3}{2}\right) \):
\[ Z = 10\left(\frac{3}{2}\right) + 8\left(\frac{3}{2}\right) = 15 + 12 = 27 \]

3. For \( \left(\frac{1}{3}, \frac{8}{3}\right) \):
\[ Z = 10\left(\frac{1}{3}\right) + 8\left(\frac{8}{3}\right) = \frac{10}{3} + \frac{64}{3} = \frac{74}{3} \approx 24.67 \]

### Step 5: Determine Minimum Value
The minimum value of \( Z \) occurs at the point \( \left(\frac{6}{11}, \frac{20}{11}\right) \) with \( Z = 20 \).

### Final Answer
The minimum value of \( Z \) is **20**.

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