The maximum and minimum values of `Z= 5x+2y`, subject to the constraints `2x+3y ge 3, x-2y ge 2, 6x+4y le 24, -3x+2y le 3" and "x, y ge 0`, are respectively

Options

A. `(18)/(7), (2)/(7)`
B. `(19, (63)/(13)`
C. `19, 63`
D. `19, 13`

Answer

To solve the problem of finding the maximum and minimum values of \( Z = 5x + 2y \) subject to the given constraints, we will follow these steps:

### Step 1: Identify the constraints
The constraints given are:
1. \( 2x + 3y \geq 3 \)
2. \( x - 2y \geq 2 \)
3. \( 6x + 4y \leq 24 \)
4. \( -3x + 2y \leq 3 \)
5. \( x \geq 0 \)
6. \( y \geq 0 \)

### Step 2: Convert inequalities into equations
To find the feasible region, we convert the inequalities into equations:
1. \( 2x + 3y = 3 \)
2. \( x - 2y = 2 \)
3. \( 6x + 4y = 24 \)
4. \( -3x + 2y = 3 \)

### Step 3: Find intersection points of the lines
We will find the intersection points by solving pairs of equations:

1. **Intersection of \( 2x + 3y = 3 \) and \( x - 2y = 2 \)**:
- From \( x - 2y = 2 \), we can express \( x = 2 + 2y \).
- Substitute into \( 2(2 + 2y) + 3y = 3 \):
\[ 4 + 4y + 3y = 3 \implies 7y = -1 \implies y = -\frac{1}{7} \quad (\text{not feasible since } y \geq 0) \]

2. **Intersection of \( 2x + 3y = 3 \) and \( 6x + 4y = 24 \)**:
- From \( 6x + 4y = 24 \), simplify to \( 3x + 2y = 12 \).
- Multiply \( 2x + 3y = 3 \) by 2:
\[ 4x + 6y = 6 \]
- Now solve:
\[ 3x + 2y = 12 \quad \text{and} \quad 4x + 6y = 6 \]
- Using elimination or substitution, we find:
\[ y = 0 \implies x = 4 \quad (4, 0) \]

3. **Intersection of \( x - 2y = 2 \) and \( 6x + 4y = 24 \)**:
- Substitute \( x = 2 + 2y \) into \( 6(2 + 2y) + 4y = 24 \):
\[ 12 + 12y + 4y = 24 \implies 16y = 12 \implies y = \frac{3}{4} \]
- Substitute \( y \) back to find \( x \):
\[ x = 2 + 2(\frac{3}{4}) = 2 + \frac{3}{2} = \frac{7}{2} \]
- So the intersection point is \( \left(\frac{7}{2}, \frac{3}{4}\right) \).

4. **Intersection of \( -3x + 2y = 3 \) and \( 6x + 4y = 24 \)**:
- Solve \( 6x + 4y = 24 \) for \( y \):
\[ 4y = 24 - 6x \implies y = 6 - \frac{3}{2}x \]
- Substitute into \( -3x + 2(6 - \frac{3}{2}x) = 3 \):
\[ -3x + 12 - 3x = 3 \implies -6x = -9 \implies x = \frac{3}{2} \]
- Substitute \( x \) back to find \( y \):
\[ y = 6 - \frac{3}{2}(\frac{3}{2}) = 6 - \frac{9}{4} = \frac{15}{4} \]

### Step 4: Identify feasible region
Plot the points \( (4, 0) \), \( \left(\frac{7}{2}, \frac{3}{4}\right) \), and \( \left(\frac{3}{2}, \frac{15}{4}\right) \) on the graph to identify the feasible region.

### Step 5: Evaluate the objective function at the vertices
1. At \( (4, 0) \):
\[ Z = 5(4) + 2(0) = 20 \]
2. At \( \left(\frac{7}{2}, \frac{3}{4}\right) \):
\[ Z = 5\left(\frac{7}{2}\right) + 2\left(\frac{3}{4}\right) = \frac{35}{2} + \frac{3}{2} = \frac{38}{2} = 19 \]
3. At \( \left(\frac{3}{2}, \frac{15}{4}\right) \):
\[ Z = 5\left(\frac{3}{2}\right) + 2\left(\frac{15}{4}\right) = \frac{15}{2} + \frac{30}{4} = \frac{15}{2} + \frac{15}{2} = 15 \]

### Step 6: Determine maximum and minimum values
- Maximum value of \( Z \) is \( 20 \) at \( (4, 0) \).
- Minimum value of \( Z \) is \( 15 \) at \( \left(\frac{3}{2}, \frac{15}{4}\right) \).

### Final Answer:
- Maximum value: \( 20 \)
- Minimum value: \( 15 \)