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Problem 11

# Solve each linear programming problem by the simplex method. $$\begin{array}{ll} \text { Maximize } & P=3 x+4 y \\ \text { subject to } & x+y \leq 4 \\ & 2 x+y \leq 5 \\ & x \geq 0, y \geq 0 \end{array}$$

Expert verified
The optimal solution is $$x = 0$$, $$y = 1$$, yielding the maximum value for the objective function $$P = 4$$.
See the step by step solution

## Step 1: Convert inequalities to equations by adding slack variables

Introduce slack variables s1 and s2 to convert the inequality constraints into equations: 1. $$x + y + s_1 = 4$$ 2. $$2x + y + s_2 = 5$$ Now, we have the following linear system with non-negative variables: $$\begin{cases} x + y +s_1 = 4 \\ 2x + y +s_2 = 5\\ x \geq 0, y \geq 0, s_1 \geq 0, s_2 \geq 0 \\ \text{Maximize } P = 3x + 4y \\ \end{cases}$$

## Step 2: Setup the initial tableau

Convert the linear system into an initial simplex tableau: $$\begin{array}{c|cccc|c} & x & y & s_1 & s_2 & \text{RHS (Right Hand Side)} \\ \hline \text{P} & -3 & -4 & 0 & 0 & 0\\ s_1 & 1 & 1 & 1 & 0 & 4\\ s_2 & 2 & 1 & 0 & 1 & 5 \end{array}$$

## Step 3: Perform pivot operations to obtain an optimal solution

Identify the entering variable: Since we need to maximize the objective function P, we choose the variable with the most negative coefficient in the objective function row, which is y. Identify the leaving variable: Divide the RHS by the positive coefficients of y in each constraint to find the minimum non-negative ratio: $$s_1: \frac{4}{1} = 4 \\ s_2: \frac{5}{1} = 5 \\$$ y will enter, and s1 will leave. Perform the pivot operation on element (2,2): $$\begin{array}{c|cccc|c} & x & y & s_1 & s_2 & \text{RHS} \\ \hline \text{P} & 1 & 0 & 4 & 0 & 16 \\ s_1 & 1 & 1 & 1 & 0 & 4 \\ s_2 & 0 & 1 & -2 & 1 & 1 \end{array}$$ Now, we must check if there is a more negative coefficient in the objective function row. There are no more negative coefficients, indicating that the optimal solution has been found.

## Step 4: Interpret the tableau to find the optimal solution

To obtain the final solution, we'll take the values of the variables from the tableau. We'll assign the values of the non-basic variables (s1 and s2) as 0: s_1 = 4 (basic variable - leaving variable) \\ s_2 = 0 (non-basic variable) \\ x = 0 (basic variable) \\ y = 1 (non-basic variable - entering variable) So, the optimal solution is $$x = 0$$, $$y = 1$$, which yields the maximum value for P, $$P = 3(0) + 4(1) = 4$$.

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