Maximize $z = 6x_1 – 2x_2+ 3x_3$

Subject to $2x_1 – x_2 + 2x_3 ≤ 2 \\ X_1 + 4x_3 ≤ 4 \\ X_1, x_2, x_3 ≥ 0. $

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2015

The problem in standard form

$Z – bx_1 + 2x_2 – 3x_3 +0s_1 + 0s_2 = 0 \\ 2x_1 –x_2 + 2x_3 + 0s_1 + 0s_2 = 2 \\ X_1 +0x_2 +4x_3 + 0s_1 + s_2 = 4$

Since 2 constraints are there thus introducing two slack variables.

Now $x_1$ column is selected as it was the highest negative No.

Now dividing $x_1$ values with RHS value.

We get $2/2 = 1$

And $4/1 = 4$

So 1 is less and thus first row is selected.

Now $S_1$ will leave and $x_1$ enters.

Dividing $S1/2$ we get $x_1$

In order to make $x_1$ value of z row = 0 we do

$$Z + 6 x 1$$

$$S_2 –X_1$$

We get

Repeating same procedure fill we get all positive in Z row.

$∴x_1=4 x_2=6 x_3=0 Z_{max}=12.$