Question #235309

Consider the following LPP

Maximize Z = 2x1 + 3x2

Subject to

x1 + 3x2 ≤ 6

3x1 + 2x2 ≤ 6

x1, x2 ≥ 0

a. Determine all the basic solutions of the problem, and classify them as feasible and

infeasible.

b. Carry out the full tableau implementation of the simplex method, starting with the basic

feasible solution (x1, x2) = (0, 0)


1
Expert's answer
2021-09-20T08:54:35-0400

a. Solving the system of linear equations given in the constraints simultaneously, we have that x1=67 and x2=127We can determine the feasible solutions by setting one variable to 0 in each constraint Hence, we have (0,2),(6,0),(0,3),(2,0)(0,2) and (2,0) are feasible solutions, since they satisfy all the constraints and (0,3)and (6,0) are infeasible solutions because they do not satisfy all the constraintsb. Next, we find the optimal solution using simplex methodThe first tableau is given byx1x2x3x42300x3013106x403201623000Applying row reduction techniques to each row and the simplex algorithm, we gettableau 2x1x2x3x42300x231311302x40730231210106The final tableau is given byx1x2x3x42300x23013717127x1210273767005737487Hence the feasible solution which is optimal is x1=127x2=67 and z=487a. \text{ Solving the system of linear equations given in the constraints simultaneously, we }\\ \text{have that $x_1 = \frac{6}{7}$ and $x_2 = \frac{12}{7}$}\\ \text{We can determine the feasible solutions by setting one variable to 0 in each constraint }\\ \text{Hence, we have }\\ (0,2), (6,0),(0,3),(2,0)\\ \text{(0,2) and (2,0) are feasible solutions, since they satisfy all the constraints and (0,3)}\\ \text{and (6,0) are infeasible solutions because they do not satisfy all the constraints}\\ b. \text{ Next, we find the optimal solution using simplex method}\\ \text{The first tableau is given by}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 \\ & & 2 & 3 & 0 & 0\\\hline x_3 & 0 & 1 & 3 & 1 & 0 & 6\\ x_4 & 0 & 3 & 2 & 0 & 1 & 6\\ & & -2 & -3 & 0 & 0 & 0 \end{matrix}\\ \text{Applying row reduction techniques to each row and the simplex algorithm, we get}\\ \text{tableau 2}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 \\ & & 2 & 3 & 0 & 0\\\hline x_2 & 3 & \frac{1}{3} & 1 & \frac{1}{3} & 0 & 2\\ x_4 & 0 & \frac{7}{3} & 0 & -\frac{2}{3} & 1 & 2\\ & & -1 & 0 & 1 & 0 & 6 \end{matrix}\\ \text{The final tableau is given by}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 \\ & & 2 & 3 & 0 & 0\\\hline x_2 & 3 & 0 & 1 & \frac{3}{7} & -\frac{1}{7} & \frac{12}{7}\\ x_1 & 2 & 1 & 0 & -\frac{2}{7} & \frac{3}{7} & \frac{6}{7}\\ & & 0 & 0 & \frac{5}{7} & \frac{3}{7} & \frac{48}{7} \end{matrix}\\ \text{Hence the feasible solution which is optimal is $x_1=\frac{12}{7}$, $x_2=\frac{6}{7}$ and $z=\frac{48}{7}$}


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