To find the maximum value of \( Z = \frac{3}{4}x + \frac{1}{2}y \) subject to the given constraints, we will follow these steps:
### Step 1: Identify the constraints
The constraints provided are:
1. \( 2x + y \leq 180 \)
2. \( x \geq 20 \)
3. \( 2x + 3y \geq 120 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert inequalities into equations
To graph the constraints, we convert the inequalities into equations:
1. \( 2x + y = 180 \)
2. \( x = 20 \)
3. \( 2x + 3y = 120 \)
### Step 3: Find intercepts for the equations
- For \( 2x + y = 180 \):
- If \( x = 0 \), then \( y = 180 \) (y-intercept).
- If \( y = 0 \), then \( x = 90 \) (x-intercept).
- For \( 2x + 3y = 120 \):
- If \( x = 0 \), then \( y = 40 \) (y-intercept).
- If \( y = 0 \), then \( x = 60 \) (x-intercept).
### Step 4: Graph the lines
Now, we will plot these lines on a graph:
- The line \( 2x + y = 180 \) connects points (0, 180) and (90, 0).
- The line \( 2x + 3y = 120 \) connects points (0, 40) and (60, 0).
- The line \( x = 20 \) is a vertical line at \( x = 20 \).
### Step 5: Determine the feasible region
The feasible region is where all the constraints overlap. We check the regions defined by the inequalities:
- For \( 2x + y \leq 180 \), we take the area below the line.
- For \( x \geq 20 \), we take the area to the right of the line.
- For \( 2x + 3y \geq 120 \), we take the area above the line.
### Step 6: Find corner points of the feasible region
The corner points of the feasible region can be found by solving the equations:
1. Intersection of \( 2x + y = 180 \) and \( x = 20 \):
- Substitute \( x = 20 \) into \( 2(20) + y = 180 \)
- \( 40 + y = 180 \) → \( y = 140 \)
- Point A: \( (20, 140) \)
2. Intersection of \( 2x + 3y = 120 \) and \( x = 20 \):
- Substitute \( x = 20 \) into \( 2(20) + 3y = 120 \)
- \( 40 + 3y = 120 \) → \( 3y = 80 \) → \( y = \frac{80}{3} \)
- Point B: \( (20, \frac{80}{3}) \)
3. Intersection of \( 2x + 3y = 120 \) and \( 2x + y = 180 \):
- Solve the system:
- From \( 2x + y = 180 \), express \( y = 180 - 2x \).
- Substitute into \( 2x + 3(180 - 2x) = 120 \):
- \( 2x + 540 - 6x = 120 \) → \( -4x = -420 \) → \( x = 105 \) (not feasible since \( x \) must be ≤ 90).
4. Intersection of \( 2x + y = 180 \) and \( y = 0 \):
- \( 2x = 180 \) → \( x = 90 \)
- Point D: \( (90, 0) \)
5. Intersection of \( 2x + 3y = 120 \) and \( y = 0 \):
- \( 2x = 120 \) → \( x = 60 \)
- Point C: \( (60, 0) \)
### Step 7: Evaluate \( Z \) at corner points
Now we evaluate \( Z \) at the corner points:
1. At Point A \( (20, 140) \):
\[
Z_A = \frac{3}{4}(20) + \frac{1}{2}(140) = 15 + 70 = 85
\]
2. At Point B \( (20, \frac{80}{3}) \):
\[
Z_B = \frac{3}{4}(20) + \frac{1}{2}(\frac{80}{3}) = 15 + \frac{40}{3} \approx 15 + 13.33 = 28.33
\]
3. At Point C \( (60, 0) \):
\[
Z_C = \frac{3}{4}(60) + \frac{1}{2}(0) = 45 + 0 = 45
\]
4. At Point D \( (90, 0) \):
\[
Z_D = \frac{3}{4}(90) + \frac{1}{2}(0) = 67.5 + 0 = 67.5
\]
### Step 8: Determine the maximum value
The maximum value of \( Z \) occurs at Point A \( (20, 140) \) where \( Z = 85 \).
### Final Answer
The maximum value of \( Z \) is \( 85 \) at the point \( (20, 140) \).
