Find solution using dual simplex method
MAX Z = -2x1 - x2
subject to
-3x1 - x2 <= -3
-4x1 - 3x2 <= -6
-x1 - 2x2 <= -3
and x1,x2 >= 0;

Solution:

Problem is

Max `Z` `=` ` - ` `2` `x_1` ` - ` `` `x_2`

subject to

` - ` `3` `x_1` ` - ` `` `x_2` ≤ `-3` ` - ` `4` `x_1` ` - ` `3` `x_2` ≤ `-6` ` - ` `` `x_1` ` - ` `2` `x_2` ≤ `-3`

and `x_1,x_2 >= 0; `

`Z` `+` `` `2` `x_1` ` + ` `` `x_2` = `0`

` - ` `3` `x_1` ` - ` `` `x_2` ` + ` `` `S_1` = `-3` ` - ` `4` `x_1` ` - ` `3` `x_2` ` + ` `` `S_2` = `-6` ` - ` `` `x_1` ` - ` `2` `x_2` ` + ` `` `S_3` = `-3`

Tableau-1

`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1` `0`	`Z`	`2`	`1`	`0`	`0`	`0`	`0`
`R_2` `0`	`S_1`	`-3`	`-1`	`1`	`0`	`0`	`-3`
`R_3` `0`	`S_2`	`-4`	`(-3)`	`0`	`1`	`0`	`-6``->`
`R_4` `0`	`S_3`	`-1`	`-2`	`0`	`0`	`1`	`-3`
	`"Ratio"=(Z_j)/(S_2,j)` and `S_2,j<0`	`(2)/(-4)` `=-0.5`	`(1)/(-3)` `=-0.3333``uarr`	`(0)/(0)` `=`---	`(0)/(1)` `=`---	`(0)/(0)` `=`---

Most negative `RHS` is `-6`. So, the leaving basis variable is `S_2`.

Least negative ratio is `-0.3333`. So, the entering variable is `x_2`.

`:.` The pivot element is `-3`.

Entering `=x_2`, Departing `=S_2`, Key Element `=-3`

`R_3`(new)`= R_3`(old) `-: -3`

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_3`(old) =	`-4`	`-3`	`0`	`1`	`0`	`-6`
`R_3`(new)`= R_3`(old) `-: -3`	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`

`R_2`(new)`= R_2`(old) + `R_3`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_2`(old) =	`-3`	`-1`	`1`	`0`	`0`	`-3`
`R_3`(new) =	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`R_2`(new)`= R_2`(old) + `R_3`(new)	`-1.6667`	`0`	`1`	`-0.3333`	`0`	`-1`

`R_4`(new)`= R_4`(old) + `2 R_3`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_4`(old) =	`-1`	`-2`	`0`	`0`	`1`	`-3`
`R_3`(new) =	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`2 xx R_3`(new) =	`2.6667`	`2`	`0`	`-0.6667`	`0`	`4`
`R_4`(new)`= R_4`(old) + `2 R_3`(new)	`1.6667`	`0`	`0`	`-0.6667`	`1`	`1`

`R_1`(new)`= R_1`(old) - `R_3`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1`(old) =	`2`	`1`	`0`	`0`	`0`	`0`
`R_3`(new) =	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`R_1`(new)`= R_1`(old) - `R_3`(new)	`0.6667`	`0`	`0`	`0.3333`	`0`	`-2`

Tableau-2

`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1` `0`	`Z`	`0.6667`	`0`	`0`	`0.3333`	`0`	`-2`
`R_2` `0`	`S_1`	`(-1.6667)`	`0`	`1`	`-0.3333`	`0`	`-1``->`
`R_3` `-1`	`x_2`	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`R_4` `0`	`S_3`	`1.6667`	`0`	`0`	`-0.6667`	`1`	`1`
	`"Ratio"=(Z_j)/(S_1,j)` and `S_1,j<0`	`(0.6667)/(-1.6667)` `=-0.4``uarr`	`(0)/(0)` `=`---	`(0)/(1)` `=`---	`(0.3333)/(-0.3333)` `=-1`	`(0)/(0)` `=`---

Most negative `RHS` is `-1`. So, the leaving basis variable is `S_1`.

Least negative ratio is `-0.4`. So, the entering variable is `x_1`.

`:.` The pivot element is `-1.6667`.

Entering `=x_1`, Departing `=S_1`, Key Element `=-1.6667`

`R_2`(new)`= R_2`(old) `-: -1.6667`

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_2`(old) =	`-1.6667`	`0`	`1`	`-0.3333`	`0`	`-1`
`R_2`(new)`= R_2`(old) `-: -1.6667`	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`

`R_3`(new)`= R_3`(old) - `1.3333 R_2`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_3`(old) =	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`R_2`(new) =	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`
`1.3333 xx R_2`(new) =	`1.3333`	`0`	`-0.8`	`0.2667`	`0`	`0.8`
`R_3`(new)`= R_3`(old) - `1.3333 R_2`(new)	`0`	`1`	`0.8`	`-0.6`	`0`	`1.2`

`R_4`(new)`= R_4`(old) - `1.6667 R_2`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_4`(old) =	`1.6667`	`0`	`0`	`-0.6667`	`1`	`1`
`R_2`(new) =	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`
`1.6667 xx R_2`(new) =	`1.6667`	`0`	`-1`	`0.3333`	`0`	`1`
`R_4`(new)`= R_4`(old) - `1.6667 R_2`(new)	`0`	`0`	`1`	`-1`	`1`	`0`

`R_1`(new)`= R_1`(old) - `0.6667 R_2`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1`(old) =	`0.6667`	`0`	`0`	`0.3333`	`0`	`-2`
`R_2`(new) =	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`
`0.6667 xx R_2`(new) =	`0.6667`	`0`	`-0.4`	`0.1333`	`0`	`0.4`
`R_1`(new)`= R_1`(old) - `0.6667 R_2`(new)	`0`	`0`	`0.4`	`0.2`	`0`	`-2.4`

Tableau-3

`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1` `0`	`Z`	`0`	`0`	`0.4`	`0.2`	`0`	`-2.4`
`R_2` `-2`	`x_1`	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`
`R_3` `-1`	`x_2`	`0`	`1`	`0.8`	`-0.6`	`0`	`1.2`
`R_4` `0`	`S_3`	`0`	`0`	`1`	`-1`	`1`	`0`
	Ratio

Since all `Z_j >= 0` and all `RHS >= 0`, thus the current solution is the optimal solution.

Hence, optimal solution is arrived with value of variables as :
`x_1=0.6,x_2=1.2`

Max `Z=-2.4`