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.
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.
Solution
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:
\( x_2 \leq x_1 + 2 \)
\( x_2 \leq \frac{14 + x_1}{3} \)
\( x_2 \geq \frac{x_1 - 6}{4} \)
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.