To solve the problem of maximizing and minimizing the objective function \( Z = 3x + 4y \) subject to the constraints:
1. \( 3x + 4y \leq 80 \)
2. \( x + 3y \leq 30 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
we will follow these steps:
### Step 1: Identify the Constraints
We have the following constraints:
- \( 3x + 4y \leq 80 \)
- \( x + 3y \leq 30 \)
- \( x \geq 0 \)
- \( y \geq 0 \)
### Step 2: Find the Boundary Lines
To find the boundary lines of the inequalities, we convert them into equations:
1. From \( 3x + 4y = 80 \):
- When \( x = 0 \): \( 4y = 80 \) → \( y = 20 \) (Point A: \( (0, 20) \))
- When \( y = 0 \): \( 3x = 80 \) → \( x = \frac{80}{3} \approx 26.67 \) (Point B: \( \left(\frac{80}{3}, 0\right) \))
2. From \( x + 3y = 30 \):
- When \( x = 0 \): \( 3y = 30 \) → \( y = 10 \) (Point C: \( (0, 10) \))
- When \( y = 0 \): \( x = 30 \) (Point D: \( (30, 0) \))
### Step 3: Plot the Lines
Plot the points A, B, C, and D on a graph and draw the lines for the equations \( 3x + 4y = 80 \) and \( x + 3y = 30 \).
### Step 4: Identify the Feasible Region
The feasible region is where all constraints overlap. This region will be bounded by the lines and the axes.
### Step 5: Find the Intersection of the Lines
To find the intersection point of the two lines \( 3x + 4y = 80 \) and \( x + 3y = 30 \), we can solve the equations simultaneously.
1. Multiply the second equation by 3:
\[
3(x + 3y) = 3(30) \implies 3x + 9y = 90
\]
2. Now subtract the first equation from this result:
\[
(3x + 9y) - (3x + 4y) = 90 - 80 \implies 5y = 10 \implies y = 2
\]
3. Substitute \( y = 2 \) back into the second equation:
\[
x + 3(2) = 30 \implies x + 6 = 30 \implies x = 24
\]
Thus, the intersection point P is \( (24, 2) \).
### Step 6: Evaluate the Objective Function at the Corner Points
Evaluate \( Z = 3x + 4y \) at the corner points:
1. At \( (0, 0) \):
\[
Z = 3(0) + 4(0) = 0
\]
2. At \( (0, 10) \):
\[
Z = 3(0) + 4(10) = 40
\]
3. At \( \left(\frac{80}{3}, 0\right) \):
\[
Z = 3\left(\frac{80}{3}\right) + 4(0) = 80
\]
4. At \( (30, 0) \):
\[
Z = 3(30) + 4(0) = 90 \quad \text{(not feasible since it violates the first constraint)}
\]
5. At \( (24, 2) \):
\[
Z = 3(24) + 4(2) = 72 + 8 = 80
\]
### Step 7: Determine Maximum and Minimum Values
From the evaluations:
- Minimum \( Z = 0 \) at \( (0, 0) \)
- Maximum \( Z = 80 \) at \( (24, 2) \) and \( \left(\frac{80}{3}, 0\right) \)
### Final Answer
The maximum value of \( Z \) is \( 80 \) and the minimum value of \( Z \) is \( 0 \).
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