91Ó°ÊÓ

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: $$ \begin{cases} x + y +s_1 = 4 \\ 2x + y +s_2 = 5\\ x \geq 0, y \geq 0, s_1 \geq 0, s_2 \geq 0 \\ \text{Maximize } P = 3x + 4y \\ \end{cases} $$

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

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.

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, \(P = 3(0) + 4(1) = 4\).