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:

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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; `

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` `=` ` - ` `2` `x_1` ` - ` `` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2` ` + ` `0` `S_3`

subject to

` - ` `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`

and `x_1,x_2,S_1,S_2,S_3 >= 0`

Tableau-1	`C_j`	`-2`	`-1`	`0`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1` `0`	`S_1`	`-3`	`-1`	`1`	`0`	`0`	`-3`
`R_2` `0`	`S_2`	`-4`	`(-3)`	`0`	`1`	`0`	`-6``->`
`R_3` `0`	`S_3`	`-1`	`-2`	`0`	`0`	`1`	`-3`
	`Z_j`	`0`	`0`	`0`	`0`	`0`	`Z=0`
	`C_j-Z_j`	`-2`	`-1`	`0`	`0`	`0`
	`"Ratio"=(C_j-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` and its row index is `2`. So, the leaving basis variable is `S_2`.

Least positive ratio is `0.3333` and its column index is `2`. So, the entering variable is `x_2`.

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

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

`R_2`(new)`= R_2`(old) `-: -3`

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_2`(old) =	`-4`	`-3`	`0`	`1`	`0`	`-6`
`R_2`(new)`= R_2`(old) `-: -3`	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`

`R_1`(new)`= R_1`(old) + `R_2`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1`(old) =	`-3`	`-1`	`1`	`0`	`0`	`-3`
`R_2`(new) =	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`R_1`(new)`= R_1`(old) + `R_2`(new)	`-1.6667`	`0`	`1`	`-0.3333`	`0`	`-1`

`R_3`(new)`= R_3`(old) + `2 R_2`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_3`(old) =	`-1`	`-2`	`0`	`0`	`1`	`-3`
`R_2`(new) =	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`2 xx R_2`(new) =	`2.6667`	`2`	`0`	`-0.6667`	`0`	`4`
`R_3`(new)`= R_3`(old) + `2 R_2`(new)	`1.6667`	`0`	`0`	`-0.6667`	`1`	`1`

Tableau-2	`C_j`	`-2`	`-1`	`0`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1` `0`	`S_1`	`(-1.6667)`	`0`	`1`	`-0.3333`	`0`	`-1``->`
`R_2` `-1`	`x_2`	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`R_3` `0`	`S_3`	`1.6667`	`0`	`0`	`-0.6667`	`1`	`1`
	`Z_j`	`-1.3333`	`-1`	`0`	`0.3333`	`0`	`Z=-2`
	`C_j-Z_j`	`-0.6667`	`0`	`0`	`-0.3333`	`0`
	`"Ratio"=(C_j-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` and its row index is `1`. So, the leaving basis variable is `S_1`.

Least positive ratio is `0.4` and its column index is `1`. So, the entering variable is `x_1`.

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

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

`R_1`(new)`= R_1`(old) `-: -1.6667`

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1`(old) =	`-1.6667`	`0`	`1`	`-0.3333`	`0`	`-1`
`R_1`(new)`= R_1`(old) `-: -1.6667`	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`

`R_2`(new)`= R_2`(old) - `1.3333 R_1`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_2`(old) =	`1.3333`	`1`	`0`	`-0.3333`	`0`	`2`
`R_1`(new) =	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`
`1.3333 xx R_1`(new) =	`1.3333`	`0`	`-0.8`	`0.2667`	`0`	`0.8`
`R_2`(new)`= R_2`(old) - `1.3333 R_1`(new)	`0`	`1`	`0.8`	`-0.6`	`0`	`1.2`

`R_3`(new)`= R_3`(old) - `1.6667 R_1`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_3`(old) =	`1.6667`	`0`	`0`	`-0.6667`	`1`	`1`
`R_1`(new) =	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`
`1.6667 xx R_1`(new) =	`1.6667`	`0`	`-1`	`0.3333`	`0`	`1`
`R_3`(new)`= R_3`(old) - `1.6667 R_1`(new)	`0`	`0`	`1`	`-1`	`1`	`0`

Tableau-3	`C_j`	`-2`	`-1`	`0`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`RHS`
`R_1` `-2`	`x_1`	`1`	`0`	`-0.6`	`0.2`	`0`	`0.6`
`R_2` `-1`	`x_2`	`0`	`1`	`0.8`	`-0.6`	`0`	`1.2`
`R_3` `0`	`S_3`	`0`	`0`	`1`	`-1`	`1`	`0`
	`Z_j`	`-2`	`-1`	`0.4`	`0.2`	`0`	`Z=-2.4`
	`C_j-Z_j`	`0`	`0`	`-0.4`	`-0.2`	`0`
	Ratio

Since all `C_j-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`

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