Solve the linear programming problem.

Maximize
[tex]\[ P = 30x + 40y \][/tex]

Subject to
[tex]\[
\begin{aligned}
2x + y & \leq 12 \\
x + y & \leq 8 \\
x + 2y & \leq 14 \\
x, y & \geq 0
\end{aligned}
\][/tex]



Answer :

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].