Solve the linear programming problem using the simplex method. \[ \begin{array}{ll} \text { Maximize } & P=2 x_{1}+3 x_{2}+4 x_{3} \\ \text { subject to } & x_{1}+x_{3} \leq 36 \\ & x_{2}+x_{3} \leq 27 \\ & x_{1}, x_{2}, x_{3} \geq 0 \end{array} \] 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 $\mathrm{P}$ is $\square$, when $\mathrm{x}_{1}=\square, \mathrm{x}_{2}=\square$, and $\mathrm{x}_{3}=\square$. B. There is no optimal solution.

Step 1 ：Convert the inequalities into equalities by introducing slack variables. The problem becomes: Maximize P = 2x1 + 3x2 + 4x3 subject to x1 + x3 + s1 = 36, x2 + x3 + s2 = 27, and x1, x2, x3, s1, s2 ≥ 0.

Step 2 ：Set up the initial simplex tableau. The tableau is a table with rows representing the constraints and columns representing the variables. The last row represents the objective function.

Step 3 ：Perform the simplex iterations. In each iteration, select the pivot column (the column with the most negative coefficient in the last row), and the pivot row (the row with the smallest non-negative ratio of the right-hand side to the pivot column). Swap the pivot row and column, and adjust the other rows to make the pivot column become a unit column.

Step 4 ：Continue the iterations until there are no more negative coefficients in the last row. This means that the optimal solution has been reached.

Step 5 ：The optimal solution is P = 153 when x1 = 36, x2 = 27, and x3 = 0. This means that the maximum value of P is 153, which is achieved when x1 is 36, x2 is 27, and x3 is 0.