The minimum value of `z=2x_(1)+3x_(2)` subjected to the constraints `2x_(1)+7x_(2)ge22,x_(1)+x_(2)ge6,5x_(1)+x_(2)ge10` and `x_(1),x_(2)ge0`, is

Options

A. 14
B. 20
C. 10
D. 16

Explanation

To find the minimum value of \( z = 2x_1 + 3x_2 \) subject to the given constraints, we will follow these steps:

### Step 1: Write down the constraints
The constraints given are:
1. \( 2x_1 + 7x_2 \geq 22 \)
2. \( x_1 + x_2 \geq 6 \)
3. \( 5x_1 + x_2 \geq 10 \)
4. \( x_1 \geq 0 \)
5. \( x_2 \geq 0 \)

### Step 2: Convert inequalities to equations
We can convert the inequalities into equations to find the boundary lines:
1. \( 2x_1 + 7x_2 = 22 \)
2. \( x_1 + x_2 = 6 \)
3. \( 5x_1 + x_2 = 10 \)

### Step 3: Find intercepts for each line
For each equation, we will find the x-intercept and y-intercept.

1. **For \( 2x_1 + 7x_2 = 22 \)**:
- \( x_1 \)-intercept: Set \( x_2 = 0 \) → \( 2x_1 = 22 \) → \( x_1 = 11 \) → Point: \( (11, 0) \)
- \( x_2 \)-intercept: Set \( x_1 = 0 \) → \( 7x_2 = 22 \) → \( x_2 = \frac{22}{7} \approx 3.14 \) → Point: \( (0, \frac{22}{7}) \)

2. **For \( x_1 + x_2 = 6 \)**:
- \( x_1 \)-intercept: Set \( x_2 = 0 \) → \( x_1 = 6 \) → Point: \( (6, 0) \)
- \( x_2 \)-intercept: Set \( x_1 = 0 \) → \( x_2 = 6 \) → Point: \( (0, 6) \)

3. **For \( 5x_1 + x_2 = 10 \)**:
- \( x_1 \)-intercept: Set \( x_2 = 0 \) → \( 5x_1 = 10 \) → \( x_1 = 2 \) → Point: \( (2, 0) \)
- \( x_2 \)-intercept: Set \( x_1 = 0 \) → \( x_2 = 10 \) → Point: \( (0, 10) \)

### Step 4: Graph the constraints
Plot the lines on a graph and identify the feasible region that satisfies all constraints. The feasible region will be bounded by the lines and the axes.

### Step 5: Find the vertices of the feasible region
To find the vertices of the feasible region, we need to solve the equations pairwise:

1. **Intersection of \( 2x_1 + 7x_2 = 22 \) and \( x_1 + x_2 = 6 \)**:
- From \( x_1 + x_2 = 6 \), we have \( x_2 = 6 - x_1 \).
- Substitute into \( 2x_1 + 7(6 - x_1) = 22 \):
\[ 2x_1 + 42 - 7x_1 = 22 \implies -5x_1 = -20 \implies x_1 = 4 \]
\[ x_2 = 6 - 4 = 2 \quad \text{(Point: (4, 2))} \]

2. **Intersection of \( 5x_1 + x_2 = 10 \) and \( x_1 + x_2 = 6 \)**:
- From \( x_1 + x_2 = 6 \), we have \( x_2 = 6 - x_1 \).
- Substitute into \( 5x_1 + (6 - x_1) = 10 \):
\[ 5x_1 + 6 - x_1 = 10 \implies 4x_1 = 4 \implies x_1 = 1 \]
\[ x_2 = 6 - 1 = 5 \quad \text{(Point: (1, 5))} \]

3. **Intersection of \( 2x_1 + 7x_2 = 22 \) and \( 5x_1 + x_2 = 10 \)**:
- From \( 5x_1 + x_2 = 10 \), we have \( x_2 = 10 - 5x_1 \).
- Substitute into \( 2x_1 + 7(10 - 5x_1) = 22 \):
\[ 2x_1 + 70 - 35x_1 = 22 \implies -33x_1 = -48 \implies x_1 = \frac{48}{33} \approx 1.45 \]
\[ x_2 = 10 - 5 \left(\frac{48}{33}\right) = \frac{10 \cdot 33 - 240}{33} = \frac{330 - 240}{33} = \frac{90}{33} \approx 2.73 \quad \text{(Point: (1.45, 2.73))} \]

### Step 6: Evaluate \( z \) at each vertex
Now, we will evaluate \( z = 2x_1 + 3x_2 \) at each vertex:

1. For \( (11, 0) \):
\[ z = 2(11) + 3(0) = 22 \]

2. For \( (6, 0) \):
\[ z = 2(6) + 3(0) = 12 \]

3. For \( (2, 0) \):
\[ z = 2(2) + 3(0) = 4 \]

4. For \( (4, 2) \):
\[ z = 2(4) + 3(2) = 8 + 6 = 14 \]

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

### Step 7: Determine the minimum value
The minimum value of \( z \) occurs at the point \( (4, 2) \) where \( z = 14 \).

### Final Answer
The minimum value of \( z \) is \( \boxed{14} \).