Question

Use the simplex method to obtain the optimal solution of the following linear programming model

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 35𝑥1 + 50𝑥2

𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

3𝑥1 + 𝑥2 ≤ 30

𝑥1 + 2𝑥2 ≤ 15

4𝑥1 + 4𝑥2 ≤ 40 𝑥1,

𝑥2 ≥ 0

Answer 1

Problem is

Max  $Z=35x_1+50x_2$ subject to:

$3x_1+x_2≤30$

$x_1+2x_2≤15$

$4x_1+4x_2≤40$

and $x_1,x_2≥0$ ;

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '≤' we should add slack variable $S_1$

2. As the constraint-2 is of type '≤' we should add slack variable $S_2$

3. As the constraint-3 is of type '≤' we should add slack variable $S_3$

After introducing slack variables

$Max \ Z=35x_1+50x_2+0S_1+0S_2+0S_3$

subject to:

$3x_1+x_2+S_1=30\\ x_1+2x_2+S_2=15\\ 4x_1+4x_2+S_3=40\\$

and $x_1,x_2,S_1,S_2,S_3≥0$

Negative minimum Zj-Cj is -50 and its column index is 2. So, the entering variable is $x_2$ .

Minimum ratio is 7.5 and its row index is 2. So, the leaving basis variable is $S_2$ .

∴ The pivot element is 2.

Entering =$x_2$ , Departing =$S_2$ , Key Element =2

Negative minimum Zj-Cj is -10 and its column index is 1. So, the entering variable is $x_1$ .

Minimum ratio is 5 and its row index is 3. So, the leaving basis variable is $S_3$ .

∴ The pivot element is 2.

Entering =$x_1$ , Departing =$S_3$ , Key Element =2

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :

$x_1$ =5,$x_2$ =5

Max Z=425