Find solution using Simplex method (BigM method)
MIN Z = 2000x1 + 1500x2
subject to
6x1 + 2x2 >= 8
2x1 + 4x2 >= 12
4x1 + 12x2 >= 24
and x1,x2 >= 0

Solution:
Problem is

Min `Z` `=` `` `2000` `x_1` ` + ` `1500` `x_2`

subject to

`` `6` `x_1` ` + ` `2` `x_2` ≥ `8` `` `2` `x_1` ` + ` `4` `x_2` ≥ `12` `` `4` `x_1` ` + ` `12` `x_2` ≥ `24`

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 subtract surplus variable `S_1` and add artificial variable `A_1`

2. As the constraint-2 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_2`

3. As the constraint-3 is of type '`>=`' we should subtract surplus variable `S_3` and add artificial variable `A_3`

After introducing surplus,artificial variables

Min `Z` `=` `` `2000` `x_1` ` + ` `1500` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2` ` + ` `0` `S_3` ` + ` `M` `A_1` ` + ` `M` `A_2` ` + ` `M` `A_3`

subject to

`` `6` `x_1` ` + ` `2` `x_2` ` - ` `` `S_1` ` + ` `` `A_1` = `8` `` `2` `x_1` ` + ` `4` `x_2` ` - ` `` `S_2` ` + ` `` `A_2` = `12` `` `4` `x_1` ` + ` `12` `x_2` ` - ` `` `S_3` ` + ` `` `A_3` = `24`

and `x_1,x_2,S_1,S_2,S_3,A_1,A_2,A_3 >= 0`

Iteration-1		`C_j`	`2000`	`1500`	`0`	`0`	`0`	`M`	`M`	`M`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_1`	`A_2`	`A_3`	MinRatio `(X_B)/(x_2)`
`A_1`	`M`	`8`	`6`	`2`	`-1`	`0`	`0`	`1`	`0`	`0`	`(8)/(2)=4`
`A_2`	`M`	`12`	`2`	`4`	`0`	`-1`	`0`	`0`	`1`	`0`	`(12)/(4)=3`
`A_3`	`M`	`24`	`4`	`(12)`	`0`	`0`	`-1`	`0`	`0`	`1`	`(24)/(12)=2``->`
`Z=44M`		`Z_j`	`12M`	`18M`	`-M`	`-M`	`-M`	`M`	`M`	`M`
		`Z_j-C_j`	`12M-2000`	`18M-1500``uarr`	`-M`	`-M`	`-M`	`0`	`0`	`0`

Positive maximum `Z_j-C_j` is `18M-1500` and its column index is `2`. So, the entering variable is `x_2`.

Minimum ratio is `2` and its row index is `3`. So, the leaving basis variable is `A_3`.

`:.` The pivot element is `12`.

Entering `=x_2`, Departing `=A_3`, Key Element `=12`

`R_3`(new)`= R_3`(old) `-: 12`

`R_3`(old) =	`24`	`4`	`12`	`0`	`0`	`-1`	`0`	`0`
`R_3`(new)`= R_3`(old) `-: 12`	`2`	`1/3`	`1`	`0`	`0`	`-1/12`	`0`	`0`

`R_1`(new)`= R_1`(old) - `2 R_3`(new)

`R_1`(old) =	`8`	`6`	`2`	`-1`	`0`	`0`	`1`	`0`
`R_3`(new) =	`2`	`1/3`	`1`	`0`	`0`	`-1/12`	`0`	`0`
`2 xx R_3`(new) =	`4`	`2/3`	`2`	`0`	`0`	`-1/6`	`0`	`0`
`R_1`(new)`= R_1`(old) - `2 R_3`(new)	`4`	`16/3`	`0`	`-1`	`0`	`1/6`	`1`	`0`

`R_2`(new)`= R_2`(old) - `4 R_3`(new)

`R_2`(old) =	`12`	`2`	`4`	`0`	`-1`	`0`	`0`	`1`
`R_3`(new) =	`2`	`1/3`	`1`	`0`	`0`	`-1/12`	`0`	`0`
`4 xx R_3`(new) =	`8`	`4/3`	`4`	`0`	`0`	`-1/3`	`0`	`0`
`R_2`(new)`= R_2`(old) - `4 R_3`(new)	`4`	`2/3`	`0`	`0`	`-1`	`1/3`	`0`	`1`

Iteration-2		`C_j`	`2000`	`1500`	`0`	`0`	`0`	`M`	`M`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_1`	`A_2`	MinRatio `(X_B)/(x_1)`
`A_1`	`M`	`4`	`(16/3)`	`0`	`-1`	`0`	`1/6`	`1`	`0`	`(4)/(16/3)=3/4=0.75``->`
`A_2`	`M`	`4`	`2/3`	`0`	`0`	`-1`	`1/3`	`0`	`1`	`(4)/(2/3)=6`
`x_2`	`1500`	`2`	`1/3`	`1`	`0`	`0`	`-1/12`	`0`	`0`	`(2)/(1/3)=6`
`Z=8M+3000`		`Z_j`	`6M+500`	`1500`	`-M`	`-M`	`(M)/(2)-125`	`M`	`M`
		`Z_j-C_j`	`6M-1500``uarr`	`0`	`-M`	`-M`	`(M)/(2)-125`	`0`	`0`

Positive maximum `Z_j-C_j` is `6M-1500` and its column index is `1`. So, the entering variable is `x_1`.

Minimum ratio is `0.75` and its row index is `1`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `16/3`.

Entering `=x_1`, Departing `=A_1`, Key Element `=16/3`

`R_1`(new)`= R_1`(old) `xx3/16`

`R_1`(old) =	`4`	`16/3`	`0`	`-1`	`0`	`1/6`	`0`
`R_1`(new)`= R_1`(old) `xx3/16`	`3/4`	`1`	`0`	`-3/16`	`0`	`1/32`	`0`

`R_2`(new)`= R_2`(old) - `2/3 R_1`(new)

`R_2`(old) =	`4`	`2/3`	`0`	`0`	`-1`	`1/3`	`1`
`R_1`(new) =	`3/4`	`1`	`0`	`-3/16`	`0`	`1/32`	`0`
`2/3 xx R_1`(new) =	`1/2`	`2/3`	`0`	`-1/8`	`0`	`1/48`	`0`
`R_2`(new)`= R_2`(old) - `2/3 R_1`(new)	`7/2`	`0`	`0`	`1/8`	`-1`	`5/16`	`1`

`R_3`(new)`= R_3`(old) - `1/3 R_1`(new)

`R_3`(old) =	`2`	`1/3`	`1`	`0`	`0`	`-1/12`	`0`
`R_1`(new) =	`3/4`	`1`	`0`	`-3/16`	`0`	`1/32`	`0`
`1/3 xx R_1`(new) =	`1/4`	`1/3`	`0`	`-1/16`	`0`	`1/96`	`0`
`R_3`(new)`= R_3`(old) - `1/3 R_1`(new)	`7/4`	`0`	`1`	`1/16`	`0`	`-3/32`	`0`

Iteration-3		`C_j`	`2000`	`1500`	`0`	`0`	`0`	`M`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	`A_2`	MinRatio `(X_B)/(S_3)`
`x_1`	`2000`	`3/4`	`1`	`0`	`-3/16`	`0`	`1/32`	`0`	`(3/4)/(1/32)=24`
`A_2`	`M`	`7/2`	`0`	`0`	`1/8`	`-1`	`(5/16)`	`1`	`(7/2)/(5/16)=56/5=11.2``->`
`x_2`	`1500`	`7/4`	`0`	`1`	`1/16`	`0`	`-3/32`	`0`	---
`Z=(7M)/(2)+4125`		`Z_j`	`2000`	`1500`	`(M)/(8)-1125/4`	`-M`	`(5M)/(16)-625/8`	`M`
		`Z_j-C_j`	`0`	`0`	`(M)/(8)-1125/4`	`-M`	`(5M)/(16)-625/8``uarr`	`0`

Positive maximum `Z_j-C_j` is `(5M)/(16)-625/8` and its column index is `5`. So, the entering variable is `S_3`.

Minimum ratio is `11.2` and its row index is `2`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `5/16`.

Entering `=S_3`, Departing `=A_2`, Key Element `=5/16`

`R_2`(new)`= R_2`(old) `xx16/5`

`R_2`(old) =	`7/2`	`0`	`0`	`1/8`	`-1`	`5/16`
`R_2`(new)`= R_2`(old) `xx16/5`	`56/5`	`0`	`0`	`2/5`	`-16/5`	`1`

`R_1`(new)`= R_1`(old) - `1/32 R_2`(new)

`R_1`(old) =	`3/4`	`1`	`0`	`-3/16`	`0`	`1/32`
`R_2`(new) =	`56/5`	`0`	`0`	`2/5`	`-16/5`	`1`
`1/32 xx R_2`(new) =	`7/20`	`0`	`0`	`1/80`	`-1/10`	`1/32`
`R_1`(new)`= R_1`(old) - `1/32 R_2`(new)	`2/5`	`1`	`0`	`-1/5`	`1/10`	`0`

`R_3`(new)`= R_3`(old) + `3/32 R_2`(new)

`R_3`(old) =	`7/4`	`0`	`1`	`1/16`	`0`	`-3/32`
`R_2`(new) =	`56/5`	`0`	`0`	`2/5`	`-16/5`	`1`
`3/32 xx R_2`(new) =	`21/20`	`0`	`0`	`3/80`	`-3/10`	`3/32`
`R_3`(new)`= R_3`(old) + `3/32 R_2`(new)	`14/5`	`0`	`1`	`1/10`	`-3/10`	`0`

Iteration-4		`C_j`	`2000`	`1500`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`	MinRatio
`x_1`	`2000`	`2/5`	`1`	`0`	`-1/5`	`1/10`	`0`
`S_3`	`0`	`56/5`	`0`	`0`	`2/5`	`-16/5`	`1`
`x_2`	`1500`	`14/5`	`0`	`1`	`1/10`	`-3/10`	`0`
`Z=5000`		`Z_j`	`2000`	`1500`	`-250`	`-250`	`0`
		`Z_j-C_j`	`0`	`0`	`-250`	`-250`	`0`

Since all `Z_j-C_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=2/5,x_2=14/5`

Min `Z=5000`

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