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19. Use simplex method and maximize Z = 20x + 50y subject to the constraints x + y <= 30 and x + 2y <= 40 x and y are positive numbers. 5

To solve the linear programming problem using the Simplex method, we aim to **maximize** the objective function:

\[
Z = 20x + 50y
\]

subject to the following constraints:

\[
x + y \leq 30 \quad \text{(Constraint 1)}
\]
\[
x + 2y \leq 40 \quad \text{(Constraint 2)}
\]
\[
x \geq 0, \quad y \geq 0 \quad \text{(Non-negativity constraints)}
\]

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

\[
x + y + s_1 = 30
\]
\[
x + 2y + s_2 = 40
\]

Here, \( s_1 \) and \( s_2 \) represent the unused resources in each constraint. So, the system becomes:

\[
x + y + s_1 = 30
\]
\[
x + 2y + s_2 = 40
\]

### Step 2: Set up the initial simplex tableau
We rewrite the objective function in standard form for the tableau. The objective function can be written as:

\[
Z - 20x - 50y = 0
\]

Now we set up the initial tableau, including the slack variables:

\[
\begin{array}{c|ccccc|c}
\text{Basic} & x & y & s_1 & s_2 & Z & \text{RHS} \\
\hline
s_1 & 1 & 1 & 1 & 0 & 0 & 30 \\
s_2 & 1 & 2 & 0 & 1 & 0 & 40 \\
Z & -20 & -50 & 0 & 0 & 1 & 0 \\
\end{array}
\]

### Step 3: Apply the Simplex Method

#### Iteration 1: Identify the entering and leaving variables
- **Entering variable:** The most negative coefficient in the \( Z \)-row is \(-50\) (corresponding to \( y \)), so \( y \) is the entering variable.
- **Leaving variable:** We calculate the ratio of the right-hand side to the coefficients of \( y \) in each row:
  - For row 1: \( \frac{30}{1} = 30 \)
  - For row 2: \( \frac{40}{2} = 20 \)

  The smallest ratio is 20, so \( s_2 \) is the leaving variable.

#### Iteration 2: Pivot and update the tableau
We pivot around the element in row 2, column \( y \), and update the tableau. After performing the pivot operation, the new tableau becomes:

\[
\begin{array}{c|ccccc|c}
\text{Basic} & x & y & s_1 & s_2 & Z & \text{RHS} \\
\hline
s_1 & 0 & -1 & 1 & -1 & 0 & 10 \\
y & 0.5 & 1 & 0 & 0.5 & 0 & 20 \\
Z & 5 & 0 & 0 & 25 & 1 & 1000 \\
\end{array}
\]

### Step 4: Check for optimality
There are no more negative coefficients in the \( Z \)-row, so the current solution is optimal.

### Step 5: Interpret the solution
The solution is:
- \( y = 20 \)
- \( x = 10 \) (found by substituting \( y = 20 \) into the equation \( x + 2y = 40 \))
- The maximum value of \( Z = 1000 \)

### Final Answer:
- \( x = 10 \)
- \( y = 20 \)
- Maximum \( Z = 1000 \)

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. What happens if you change the objective function to \( Z = 30x + 40y \)?
2. How does the Simplex method handle problems with more than two constraints?
3. Can you explain the significance of slack variables in the Simplex method?
4. How would you handle a minimization problem using the Simplex method?
5. What is the geometric interpretation of the Simplex method?

**Tip:** Always check for negative coefficients in the objective function row to determine if the current solution is optimal.

Use simplex method and maximize Z = 20x + 50y subject to the constraints x + y <= 30 and x + 2y <= 40. x and y are positive numbers.

Solve the linear programming problem to maximize Z = 20x + 50y using the Simplex method. The solution involves setting up constraints x + y <= 30 and x + 2y <= 40, applying slack variables, and optimizing the objective function using the tableau method. Suitable for Grades 11-12 and college students.

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