Solve the following linear programming problem algebraically

Solve the following linear programming problem algebraically using the simplex method (not the tableau). At each \"update\" (which must be clearly shown), - state the basic feasible solution. - identify which variable which increases \'z\' the most by entering the basis (if not optimal) - identify which variable leaves the basis (if not optimal) maximize z = 2x_1 + 3x_2 + x_3 subject to X_1 + x_2 + 4x_3 lessthanorequalto 100 x_1 + 2x_2 + x_3 lessthanorequalto 150 3x_1 + 2x_2 + x_3 lessthanorequalto 320 with X_1, x_2, x_3 Greaterthanorequalto 0 * your solution should contain the following information at the end: Maximum z = when x_1 =, x_2 =, x_3 =

Solution

MAZIMIZE: Z = 2 X1 + 3 X2 + X3

X1 + X2 + 4 X3 100

X1 + 2 X2 + X3 150

3 X1 + 2 X2 + X3 320

X1, X2, X3 0

The problem is converted to standard form by adding slack, surplus and artificial variables as appropiate. As the constraint is of type \'\' we should add the slack variables X4, X5 & X6.

MAZIMIZE: Z = 2 X1 + 3 X2 + 1 X3 + 0 X4 + 0 X5 + 0 X6

1 X1 + 1 X2 + 4 X3 + 1 X4 = 100

1 X1 + 2 X2 + 1 X3 + 1 X5 = 150

3 X1 + 2 X2 + 1 X3 + 1 X6 = 320

X1, X2, X3, X4, X5, X6 0

The leaving variable is X5 and entering variable is X2.

The leaving variable is X4 and entering variable is X1.

The optimal solution value is Z = 250
X1 = 50, X2 = 50, X3 = 0

Tableau 1			2	3	1	0	0	0
Base	Cb	X0	X1	X2	X3	X4	X5	X6
X4	0	100	1	1	4	1	0	0
X5	0	150	1	2	1	0	1	0
X6	0	320	3	2	1	0	0	1
Z		0	-2	-3	-1	0	0	0