To solve the problem of minimizing and maximizing \( Z = 5x + 2y \) subject to the given constraints, we will follow these steps:
### Step 1: Define the Constraints
The constraints given are:
1. \( x - 2y \leq 2 \)
2. \( 3x + 2y \leq 12 \)
3. \( -3x + 2y \leq 3 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert Inequalities to Equations
To find the boundary lines of the constraints, we convert the inequalities into equalities:
1. \( x - 2y = 2 \)
2. \( 3x + 2y = 12 \)
3. \( -3x + 2y = 3 \)
### Step 3: Find Intercepts
Next, we find the intercepts for each equation.
1. For \( x - 2y = 2 \):
- \( x \)-intercept: Set \( y = 0 \) → \( x = 2 \) → Point: \( (2, 0) \)
- \( y \)-intercept: Set \( x = 0 \) → \( -2y = 2 \) → \( y = -1 \) (not valid since \( y \geq 0 \))
2. For \( 3x + 2y = 12 \):
- \( x \)-intercept: Set \( y = 0 \) → \( 3x = 12 \) → \( x = 4 \) → Point: \( (4, 0) \)
- \( y \)-intercept: Set \( x = 0 \) → \( 2y = 12 \) → \( y = 6 \) → Point: \( (0, 6) \)
3. For \( -3x + 2y = 3 \):
- \( x \)-intercept: Set \( y = 0 \) → \( -3x = 3 \) → \( x = -1 \) (not valid since \( x \geq 0 \))
- \( y \)-intercept: Set \( x = 0 \) → \( 2y = 3 \) → \( y = 1.5 \) → Point: \( (0, 1.5) \)
### Step 4: Find Intersection Points
Now we find the intersection points of the lines:
1. **Intersection of \( x - 2y = 2 \) and \( 3x + 2y = 12 \)**:
- Adding the equations:
\[
x - 2y + 3x + 2y = 2 + 12 \implies 4x = 14 \implies x = \frac{7}{2}
\]
- Substitute \( x = \frac{7}{2} \) into \( x - 2y = 2 \):
\[
\frac{7}{2} - 2y = 2 \implies -2y = 2 - \frac{7}{2} \implies -2y = \frac{-3}{2} \implies y = \frac{3}{4}
\]
- Intersection Point: \( \left( \frac{7}{2}, \frac{3}{4} \right) \)
2. **Intersection of \( 3x + 2y = 12 \) and \( -3x + 2y = 3 \)**:
- Adding the equations:
\[
3x + 2y - 3x + 2y = 12 + 3 \implies 4y = 15 \implies y = \frac{15}{4}
\]
- Substitute \( y = \frac{15}{4} \) into \( 3x + 2y = 12 \):
\[
3x + 2 \cdot \frac{15}{4} = 12 \implies 3x + \frac{30}{4} = 12 \implies 3x = 12 - \frac{30}{4} \implies 3x = \frac{48 - 30}{4} \implies 3x = \frac{18}{4} \implies x = \frac{3}{2}
\]
- Intersection Point: \( \left( \frac{3}{2}, \frac{15}{4} \right) \)
### Step 5: Identify Feasible Region
The feasible region is bounded by the points:
- \( (0, 0) \)
- \( (2, 0) \)
- \( (0, 6) \)
- \( (0, 1.5) \)
- \( \left( \frac{7}{2}, \frac{3}{4} \right) \)
- \( \left( \frac{3}{2}, \frac{15}{4} \right) \)
### Step 6: Evaluate Objective Function at Corner Points
Now we evaluate \( Z = 5x + 2y \) at each corner point:
1. \( (0, 0) \): \( Z = 5(0) + 2(0) = 0 \)
2. \( (2, 0) \): \( Z = 5(2) + 2(0) = 10 \)
3. \( (0, 6) \): \( Z = 5(0) + 2(6) = 12 \)
4. \( (0, 1.5) \): \( Z = 5(0) + 2(1.5) = 3 \)
5. \( \left( \frac{7}{2}, \frac{3}{4} \right) \):
\[
Z = 5 \cdot \frac{7}{2} + 2 \cdot \frac{3}{4} = \frac{35}{2} + \frac{3}{2} = \frac{38}{2} = 19
\]
6. \( \left( \frac{3}{2}, \frac{15}{4} \right) \):
\[
Z = 5 \cdot \frac{3}{2} + 2 \cdot \frac{15}{4} = \frac{15}{2} + \frac{30}{4} = \frac{15}{2} + \frac{15}{2} = 15
\]
### Step 7: Determine Minimum and Maximum Values
- Minimum value of \( Z \) is \( 0 \) at \( (0, 0) \).
- Maximum value of \( Z \) is \( 19 \) at \( \left( \frac{7}{2}, \frac{3}{4} \right) \).
### Conclusion
- The minimum value of \( Z \) is \( 0 \) at the point \( (0, 0) \).
- The maximum value of \( Z \) is \( 19 \) at the point \( \left( \frac{7}{2}, \frac{3}{4} \right) \).
