To solve the linear programming problem, we need to maximize the objective function \( Z = 45x + 55y \) subject to the given constraints. Let's break down the solution step by step.
### Step 1: Identify the Constraints
The constraints given are:
1. \( x \geq 0 \)
2. \( y \geq 0 \)
3. \( 6x + 4y \leq 120 \)
4. \( 3x + 10y \leq 180 \)
### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( 6x + 4y = 120 \)
2. \( 3x + 10y = 180 \)
### Step 3: Find Intercepts of the Lines
**For the first equation \( 6x + 4y = 120 \):**
- When \( x = 0 \):
\[
4y = 120 \implies y = 30 \quad \text{(y-intercept)}
\]
- When \( y = 0 \):
\[
6x = 120 \implies x = 20 \quad \text{(x-intercept)}
\]
**For the second equation \( 3x + 10y = 180 \):**
- When \( x = 0 \):
\[
10y = 180 \implies y = 18 \quad \text{(y-intercept)}
\]
- When \( y = 0 \):
\[
3x = 180 \implies x = 60 \quad \text{(x-intercept)}
\]
### Step 4: Plot the Lines
Now we plot the lines on a graph:
- The line \( 6x + 4y = 120 \) intersects the axes at \( (20, 0) \) and \( (0, 30) \).
- The line \( 3x + 10y = 180 \) intersects the axes at \( (60, 0) \) and \( (0, 18) \).
### Step 5: Find the Intersection of the Lines
To find the feasible region, we need to find the intersection point of the two lines:
1. From \( 6x + 4y = 120 \) (divide by 2):
\[
3x + 2y = 60 \quad \text{(Equation 1)}
\]
2. From \( 3x + 10y = 180 \) (already in standard form):
\[
3x + 10y = 180 \quad \text{(Equation 2)}
\]
Now, subtract Equation 1 from Equation 2:
\[
(3x + 10y) - (3x + 2y) = 180 - 60
\]
\[
8y = 120 \implies y = 15
\]
Substituting \( y = 15 \) back into Equation 1:
\[
3x + 2(15) = 60
\]
\[
3x + 30 = 60 \implies 3x = 30 \implies x = 10
\]
Thus, the intersection point is \( (10, 15) \).
### Step 6: Identify the Feasible Region
The feasible region is bounded by the points:
- \( (0, 0) \)
- \( (0, 18) \)
- \( (20, 0) \)
- \( (10, 15) \)
### Step 7: Evaluate the Objective Function at Each Vertex
Now we evaluate \( Z = 45x + 55y \) at each vertex:
1. At \( (0, 0) \):
\[
Z = 45(0) + 55(0) = 0
\]
2. At \( (0, 18) \):
\[
Z = 45(0) + 55(18) = 990
\]
3. At \( (20, 0) \):
\[
Z = 45(20) + 55(0) = 900
\]
4. At \( (10, 15) \):
\[
Z = 45(10) + 55(15) = 450 + 825 = 1275
\]
### Step 8: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (10, 15) \) with \( Z = 1275 \).
### Final Answer
The point that provides the solution to the linear programming problem is \( (10, 15) \) with a maximum value of \( Z = 1275 \).
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