Solve the linear programming problem using the simplex method.
Maximize
Upper P equals 30 x 1 plus 40 x 2
subject to
2 x 1 plus x 2 less than or equals 60
x 1 plus x 2 less than or equals 42
x 1 plus 2 x 2 less than or equals 72
x 1 comma x 2 greater than or equals 0
Question content area bottom
Part 1
Use the simplex method to solve the problem. Select the correct choice below​ and, if​ necessary, fill in the answer boxes to complete your choice.
A.
The maximum value of P is 
select: 1980 
enter your response here when x 1equals
  
enter your response here and x 2equals
  
enter your response here.
B.
There is no optimal solution.

We are asked to solve a linear programming problem using the simplex method to maximize \( P = 30x_1 + 40x_2 \) subject to the constraints:

\[
\begin{aligned}
2x_1 + x_2 &\leq 60 \\
x_1 + x_2 &\leq 42 \\
x_1 + 2x_2 &\leq 72 \\
x_1, x_2 &\geq 0
\end{aligned}
\]

### Step 1: Convert inequalities to equalities by introducing slack variables
We introduce slack variables \( s_1, s_2, \) and \( s_3 \) to convert the inequalities into equalities:

\[
\begin{aligned}
2x_1 + x_2 + s_1 &= 60 \\
x_1 + x_2 + s_2 &= 42 \\
x_1 + 2x_2 + s_3 &= 72 \\
x_1, x_2, s_1, s_2, s_3 &\geq 0
\end{aligned}
\]

The objective function becomes:

\[
P = 30x_1 + 40x_2
\]

### Step 2: Set up the initial simplex tableau

| Basic Variables | \( x_1 \) | \( x_2 \) | \( s_1 \) | \( s_2 \) | \( s_3 \) | RHS |
|-----------------|-----------|-----------|-----------|-----------|-----------|------|
| \( s_1 \)       | 2         | 1         | 1         | 0         | 0         | 60   |
| \( s_2 \)       | 1         | 1         | 0         | 1         | 0         | 42   |
| \( s_3 \)       | 1         | 2         | 0         | 0         | 1         | 72   |
| \( P \)         | -30       | -40       | 0         | 0         | 0         | 0    |

### Step 3: Perform the simplex algorithm

- **Step 3.1: Identify the entering variable**: The most negative value in the bottom row (objective function row) is \( -40 \) under \( x_2 \), so \( x_2 \) is the entering variable.

- **Step 3.2: Identify the departing variable**: To determine the departing variable, compute the ratios of RHS to the corresponding values in the \( x_2 \) column:
  \[
  \frac{60}{1} = 60, \quad \frac{42}{1} = 42, \quad \frac{72}{2} = 36
  \]
  The smallest positive ratio is \( 36 \), so the row corresponding to \( s_3 \) will leave the basis, and \( s_3 \) will be replaced by \( x_2 \).

- **Step 3.3: Pivot operation**: We perform the necessary row operations to make \( x_2 \) the basic variable in the third row. Let's now update the tableau.

### Step 4: Updated tableau after pivot

After performing the pivot operation, we get the following updated tableau:

| Basic Variables | \( x_1 \) | \( x_2 \) | \( s_1 \) | \( s_2 \) | \( s_3 \) | RHS |
|-----------------|-----------|-----------|-----------|-----------|-----------|------|
| \( s_1 \)       | 0         | -0.5      | 1         | 0         | -0.5      | 42   |
| \( s_2 \)       | 0         | 0.5       | 0         | 1         | -0.5      | 6    |
| \( x_2 \)       | 0.5       | 1         | 0         | 0         | 0.5       | 36   |
| \( P \)         | -10       | 0         | 0         | 0         | 20        | 1440 |

### Step 5: Check for optimality

There are no more negative values in the bottom row, so the current solution is optimal.

### Step 6: Interpret the solution

- The maximum value of \( P \) is 1440.
- \( x_1 = 36 \) and \( x_2 = 36 \).

Thus, the correct choice is:

**A**. The maximum value of \( P \) is 1440 when \( x_1 = 36 \) and \( x_2 = 36 \).

---

Would you like more details on any of the steps? Here are some related questions to deepen understanding:

1. How does the simplex method determine the entering and departing variables?
2. What is the significance of the slack variables in linear programming?
3. How are pivot operations performed in the simplex method?
4. Can the simplex method handle constraints with equality signs instead of inequalities?
5. What happens if there is no optimal solution in a linear programming problem?

**Tip:** When applying the simplex method, always ensure that the objective function row has no negative values before concluding optimality.

Solve the linear programming problem using the simplex method. Maximize P = 30x1 + 40x2 subject to 2x1 + x2 ≤ 60, x1 + x2 ≤ 42, x1 + 2x2 ≤ 72, x1, x2 ≥ 0.

Learn how to solve the linear programming problem using the simplex method with a step-by-step guide. Maximize P = 30x1 + 40x2 subject to constraints. This solution involves converting inequalities to equalities, setting up the initial simplex tableau, and performing pivot operations to find the optimal solution.

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