Find solution using dual simplex method
MAX Z = -15x1 - 10x2
subject to
-3x1 - 5x2 <= -5
-5x1 - 2x2 <= -3
and x1,x2 >= 0;

Solution:

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Problem is

Max `Z` `=` ` - ` `15` `x_1` ` - ` `10` `x_2`

subject to

` - ` `3` `x_1` ` - ` `5` `x_2` ≤ `-5` ` - ` `5` `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`

After introducing slack variables

Max `Z` `=` ` - ` `15` `x_1` ` - ` `10` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2`

subject to

` - ` `3` `x_1` ` - ` `5` `x_2` ` + ` `` `S_1` = `-5` ` - ` `5` `x_1` ` - ` `2` `x_2` ` + ` `` `S_2` = `-3`

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

Tableau-1	`C_j`	`-15`	`-10`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1` `0`	`S_1`	`-3`	`(-5)`	`1`	`0`	`-5``->`
`R_2` `0`	`S_2`	`-5`	`-2`	`0`	`1`	`-3`
	`Z_j`	`0`	`0`	`0`	`0`	`Z=0`
	`Z_j-C_j`	`15`	`10`	`0`	`0`
	`"Ratio"=(Z_j-C_j)/(S_1,j)` and `S_1,j<0`	`(15)/(-3)` `=-5`	`(10)/(-5)` `=-2``uarr`	`(0)/(1)` `=`---	`(0)/(0)` `=`---

Most negative `RHS` is `-5` and its row index is `1`. So, the leaving basis variable is `S_1`.

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

`:.` The pivot element is `-5`.

Entering `=x_2`, Departing `=S_1`, Key Element `=-5`

`R_1`(new)`= R_1`(old) `-: -5`

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1`(old) =	`-3`	`-5`	`1`	`0`	`-5`
`R_1`(new)`= R_1`(old) `-: -5`	`0.6`	`1`	`-0.2`	`0`	`1`

`R_2`(new)`= R_2`(old) + `2 R_1`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_2`(old) =	`-5`	`-2`	`0`	`1`	`-3`
`R_1`(new) =	`0.6`	`1`	`-0.2`	`0`	`1`
`2 xx R_1`(new) =	`1.2`	`2`	`-0.4`	`0`	`2`
`R_2`(new)`= R_2`(old) + `2 R_1`(new)	`-3.8`	`0`	`-0.4`	`1`	`-1`

Tableau-2	`C_j`	`-15`	`-10`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1` `-10`	`x_2`	`0.6`	`1`	`-0.2`	`0`	`1`
`R_2` `0`	`S_2`	`(-3.8)`	`0`	`-0.4`	`1`	`-1``->`
	`Z_j`	`-6`	`-10`	`2`	`0`	`Z=-10`
	`Z_j-C_j`	`9`	`0`	`2`	`0`
	`"Ratio"=(Z_j-C_j)/(S_2,j)` and `S_2,j<0`	`(9)/(-3.8)` `=-2.3684``uarr`	`(0)/(0)` `=`---	`(2)/(-0.4)` `=-5`	`(0)/(1)` `=`---

Most negative `RHS` is `-1` and its row index is `2`. So, the leaving basis variable is `S_2`.

Least negative ratio is `-2.3684` and its column index is `1`. So, the entering variable is `x_1`.

`:.` The pivot element is `-3.8`.

Entering `=x_1`, Departing `=S_2`, Key Element `=-3.8`

`R_2`(new)`= R_2`(old) `-: -3.8`

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_2`(old) =	`-3.8`	`0`	`-0.4`	`1`	`-1`
`R_2`(new)`= R_2`(old) `-: -3.8`	`1`	`0`	`0.1053`	`-0.2632`	`0.2632`

`R_1`(new)`= R_1`(old) - `0.6 R_2`(new)

	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1`(old) =	`0.6`	`1`	`-0.2`	`0`	`1`
`R_2`(new) =	`1`	`0`	`0.1053`	`-0.2632`	`0.2632`
`0.6 xx R_2`(new) =	`0.6`	`0`	`0.0632`	`-0.1579`	`0.1579`
`R_1`(new)`= R_1`(old) - `0.6 R_2`(new)	`0`	`1`	`-0.2632`	`0.1579`	`0.8421`

Tableau-3	`C_j`	`-15`	`-10`	`0`	`0`
`C_B`	`"Basis"`	`x_1`	`x_2`	`S_1`	`S_2`	`RHS`
`R_1` `-10`	`x_2`	`0`	`1`	`-0.2632`	`0.1579`	`0.8421`
`R_2` `-15`	`x_1`	`1`	`0`	`0.1053`	`-0.2632`	`0.2632`
	`Z_j`	`-15`	`-10`	`1.0526`	`2.3684`	`Z=-12.3684`
	`Z_j-C_j`	`0`	`0`	`1.0526`	`2.3684`
	Ratio

Since all `Z_j-C_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=0.2632,x_2=0.8421`

Max `Z=-12.3684`

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