Questions: Solve the linear programming problem using the simplex method. Maximize P = -x1 + 2x2 subject to -x1 + x2 ≤ 2 -x1 + 3x2 ≤ 14 x1 - 4x2 ≤ 6 x1, x2 ≥ 0 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The maximum value of P is P= when x1= and x2= . (Simplify your answers.) B. There is no optimal solution.

Solve the linear programming problem using the simplex method.

Maximize P = -x1 + 2x2

subject to -x1 + x2 ≤ 2

-x1 + 3x2 ≤ 14

x1 - 4x2 ≤ 6

x1, x2 ≥ 0

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

A. The maximum value of P is P=  when x1=  and x2= .

(Simplify your answers.)

B. There is no optimal solution.
Transcript text: Solve the linear programming problem using the simplex method. \[ \begin{aligned} \text { Maximize } & P=-x_{1}+2 x_{2} \\ \text { subject to } & -x_{1}+x_{2} \leq 2 \\ & -x_{1}+3 x_{2} \leq 14 \\ & x_{1}-4 x_{2} \leq 6 \\ & x_{1}, x_{2} \geq 0 \end{aligned} \] Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The maximum value of $P$ is $P=$ $\square$ when $\mathrm{x}_{1}=$ $\square$ and $x_{2}=$ $\square$ . (Simplify your answers.) B. There is no optimal solution.
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Solution

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Solution Steps

Step 1: Define the Problem

We are tasked with maximizing the objective function \( P = -x_1 + 2x_2 \) subject to the following constraints: \[ \begin{aligned} -x_1 + x_2 & \leq 2 \\ -x_1 + 3x_2 & \leq 14 \\ x_1 - 4x_2 & \leq 6 \\ x_1, x_2 & \geq 0 \end{aligned} \]

Step 2: Analyze the Constraints

The constraints can be rewritten as:

  1. \( x_2 \leq x_1 + 2 \)
  2. \( x_2 \leq \frac{14 + x_1}{3} \)
  3. \( x_2 \geq \frac{x_1 - 6}{4} \)
Step 3: Determine Feasibility

Upon applying the simplex method, it was determined that the problem is unbounded. This indicates that there is no maximum value for the objective function \( P \) within the given constraints.

Step 4: Conclusion

Since the optimization process indicates that the problem is unbounded, we conclude that there is no optimal solution.

Final Answer

The answer is B. There is no optimal solution.

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