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To solve the linear programming problem of maximizing [tex]\( P = 30x + 40y \)[/tex] subject to the given constraints, we can follow a series of logical steps to find the optimal solution.
### Step 1: Define the Constraints
The problem is subject to the following constraints:
[tex]\[ \begin{aligned} (1) \quad 2x + y & \leq 12, \\ (2) \quad x + y & \leq 8, \\ (3) \quad x + 2y & \leq 14, \\ (4) \quad x & \geq 0, \\ (5) \quad y & \geq 0. \end{aligned} \][/tex]
### Step 2: Graph the Feasible Region
Plotting these constraints on a graph will help us visualize the feasible region.
- For [tex]\(2x + y \leq 12\)[/tex]:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 12\)[/tex]
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 6\)[/tex]
- For [tex]\(x + y \leq 8\)[/tex]:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 8\)[/tex]
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 8\)[/tex]
- For [tex]\(x + 2y \leq 14\)[/tex]:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 7\)[/tex]
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 14\)[/tex]
These lines intersect the axes at the points mentioned.
### Step 3: Find Intersection Points
The intersections of these lines with each other and the axes provide us with the vertices of the feasible region. Let's determine these points.
- Intersection of [tex]\(2x + y = 12\)[/tex] and [tex]\(x + y = 8\)[/tex]:
[tex]\[ \begin{aligned} 2x + y &= 12, \\ x + y &= 8. \end{aligned} \][/tex]
Subtract the second equation from the first:
[tex]\[ (2x + y) - (x + y) = 12 - 8 \implies x = 4. \][/tex]
Substitute [tex]\(x = 4\)[/tex] in [tex]\(x + y = 8\)[/tex]:
[tex]\[ 4 + y = 8 \implies y = 4. \][/tex]
This gives the point [tex]\((4, 4)\)[/tex].
- Intersection of [tex]\(x + y = 8\)[/tex] and [tex]\(x + 2y = 14\)[/tex]:
[tex]\[ \begin{aligned} x + y &= 8, \\ x + 2y &= 14. \end{aligned} \][/tex]
Subtract the first equation from the second:
[tex]\[ (x + 2y) - (x + y) = 14 - 8 \implies y = 6. \][/tex]
Substitute [tex]\(y = 6\)[/tex] in [tex]\(x + y = 8\)[/tex]:
[tex]\[ x + 6 = 8 \implies x = 2. \][/tex]
This gives the point [tex]\((2, 6)\)[/tex].
- Other boundary points come from where the constraints intersect the axes, such as [tex]\((0,0)\)[/tex], [tex]\((0, 7)\)[/tex], and [tex]\((6, 0)\)[/tex].
### Step 4: Evaluate the Objective Function
Evaluate the objective function [tex]\(P = 30x + 40y\)[/tex] at these vertices:
- At [tex]\((0,0)\)[/tex]:
[tex]\[ P = 30 \times 0 + 40 \times 0 = 0 \][/tex]
- At [tex]\((4,4)\)[/tex]:
[tex]\[ P = 30 \times 4 + 40 \times 4 = 120 + 160 = 280 \][/tex]
- At [tex]\((2,6)\)[/tex]:
[tex]\[ P = 30 \times 2 + 40 \times 6 = 60 + 240 = 300 \][/tex]
- At [tex]\((0,7)\)[/tex]:
[tex]\[ P = 30 \times 0 + 40 \times 7 = 280 \][/tex]
- At [tex]\((6,0)\)[/tex]:
[tex]\[ P = 30 \times 6 + 40 \times 0 = 180 \][/tex]
### Step 5: Select the Optimal Solution
The maximum value of [tex]\(P\)[/tex] from the evaluated points is [tex]\(300\)[/tex], which occurs at the point [tex]\((2,6)\)[/tex].
### Conclusion
The optimal solution is:
[tex]\[ x = 2, y = 6 \][/tex]
[tex]\[ \text{Maximum value of } P = 300 \][/tex]
Thus, the maximum value of the objective function [tex]\(P\)[/tex] under the given constraints is [tex]\(300\)[/tex], achieved at the point [tex]\((2,6)\)[/tex].
### Step 1: Define the Constraints
The problem is subject to the following constraints:
[tex]\[ \begin{aligned} (1) \quad 2x + y & \leq 12, \\ (2) \quad x + y & \leq 8, \\ (3) \quad x + 2y & \leq 14, \\ (4) \quad x & \geq 0, \\ (5) \quad y & \geq 0. \end{aligned} \][/tex]
### Step 2: Graph the Feasible Region
Plotting these constraints on a graph will help us visualize the feasible region.
- For [tex]\(2x + y \leq 12\)[/tex]:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 12\)[/tex]
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 6\)[/tex]
- For [tex]\(x + y \leq 8\)[/tex]:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 8\)[/tex]
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 8\)[/tex]
- For [tex]\(x + 2y \leq 14\)[/tex]:
- When [tex]\(x = 0\)[/tex]: [tex]\(y = 7\)[/tex]
- When [tex]\(y = 0\)[/tex]: [tex]\(x = 14\)[/tex]
These lines intersect the axes at the points mentioned.
### Step 3: Find Intersection Points
The intersections of these lines with each other and the axes provide us with the vertices of the feasible region. Let's determine these points.
- Intersection of [tex]\(2x + y = 12\)[/tex] and [tex]\(x + y = 8\)[/tex]:
[tex]\[ \begin{aligned} 2x + y &= 12, \\ x + y &= 8. \end{aligned} \][/tex]
Subtract the second equation from the first:
[tex]\[ (2x + y) - (x + y) = 12 - 8 \implies x = 4. \][/tex]
Substitute [tex]\(x = 4\)[/tex] in [tex]\(x + y = 8\)[/tex]:
[tex]\[ 4 + y = 8 \implies y = 4. \][/tex]
This gives the point [tex]\((4, 4)\)[/tex].
- Intersection of [tex]\(x + y = 8\)[/tex] and [tex]\(x + 2y = 14\)[/tex]:
[tex]\[ \begin{aligned} x + y &= 8, \\ x + 2y &= 14. \end{aligned} \][/tex]
Subtract the first equation from the second:
[tex]\[ (x + 2y) - (x + y) = 14 - 8 \implies y = 6. \][/tex]
Substitute [tex]\(y = 6\)[/tex] in [tex]\(x + y = 8\)[/tex]:
[tex]\[ x + 6 = 8 \implies x = 2. \][/tex]
This gives the point [tex]\((2, 6)\)[/tex].
- Other boundary points come from where the constraints intersect the axes, such as [tex]\((0,0)\)[/tex], [tex]\((0, 7)\)[/tex], and [tex]\((6, 0)\)[/tex].
### Step 4: Evaluate the Objective Function
Evaluate the objective function [tex]\(P = 30x + 40y\)[/tex] at these vertices:
- At [tex]\((0,0)\)[/tex]:
[tex]\[ P = 30 \times 0 + 40 \times 0 = 0 \][/tex]
- At [tex]\((4,4)\)[/tex]:
[tex]\[ P = 30 \times 4 + 40 \times 4 = 120 + 160 = 280 \][/tex]
- At [tex]\((2,6)\)[/tex]:
[tex]\[ P = 30 \times 2 + 40 \times 6 = 60 + 240 = 300 \][/tex]
- At [tex]\((0,7)\)[/tex]:
[tex]\[ P = 30 \times 0 + 40 \times 7 = 280 \][/tex]
- At [tex]\((6,0)\)[/tex]:
[tex]\[ P = 30 \times 6 + 40 \times 0 = 180 \][/tex]
### Step 5: Select the Optimal Solution
The maximum value of [tex]\(P\)[/tex] from the evaluated points is [tex]\(300\)[/tex], which occurs at the point [tex]\((2,6)\)[/tex].
### Conclusion
The optimal solution is:
[tex]\[ x = 2, y = 6 \][/tex]
[tex]\[ \text{Maximum value of } P = 300 \][/tex]
Thus, the maximum value of the objective function [tex]\(P\)[/tex] under the given constraints is [tex]\(300\)[/tex], achieved at the point [tex]\((2,6)\)[/tex].
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