Question

+ Use the simplex method to solve the linear programming problem. Maximize z= 2X2 + 3x2 subject to: 5x1 + x2 = 70 3x4 + 2x2 5 90 X1 + X2 580 X1, X220. with Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The maximum is when X1 = and X2 (Simplify your answers.) OB. There is no maximum.

Answer 1

Max Z = 2X1 + 3X2 subject to 5X1 + X2 570 3X1 + 2x2 5 90 *1 + *2580 and x1, x2 = 0;

The problem is then converted to standard form by adding slack variables

1. As the constraint-1 is of type'<' we should add slack variable Sı 2. As the constraint-2 is of type'' we should add slack variable S2 3. As the constraint-3 is of type '<' we should add slack variable S3 After introducing slack variables Max Z = 2x1 + 3x2 + 0S1 + 052 +0S3 subject to X2 + S = 70 3X1 + 2x2 + S2 = 90 x1 + x2 + S3 = 80 and xq, X2, S1, S2, S320 5X1 +

Now we make the simplex table:

Iteration-1 C 2 3 0 0 0 MinRatio Хв B CB X2 *1 X2 Si S2 S₃ 70 S1 0 70 5 1 1 0 0 = 70 1 S2 0 90 3 (2) 0 1 0 90 2 45 80 Sz 0 80 1 0 0 1 = 80 1 N Z= 0 0 0 0 0 0 C; - Z; 2 31 0 0 0

In row we write the coefficient of the objective function.

CA 3

Positive maximum C;- Z; is 3 and its column index is 2. So, the entering variable is X2. Minimum ratio is 45 and its row index is 2. So, the leaving basis variable is S2. : The pivot element is 2. Entering = x2, Departing = S2, Key Element = 2

the key element= pivot element= 2

Now we prepare the second iteration table:

firs we prepare row 2. Which obtained by diving the old row by 2

every new element in the table is prepared using

(corresponding no. of key row) x (corresponding no. of key column) New number= old number- pivot element

Iteration-2 C 2 3 0 0 0 B CB Хв * X2 Si S2 S3 MinRatio Si 0 25 3.5 0 1 -0.5 0 X2 3 45 1.5 1 0 0.5 0 S3 0 35 -0.5 0 0 -0.5 Z= 135 Z; 4.5 3 0 1.5 0 C; - ; -2.5 0 0 -1.5 0

Since all C;-Z; <0 Hence, optimal solution is arrived with value of variables as : x1 = 0, x2 = 45 Max Z= 135

Option A

The maximum is 135 and