Find dual from primal conversion
MIN z = x1 + 2x2
subject to
2x1 + 4x2 <= 160
x1 - x2 = 30
x1 >= 10
and x1,x2 >= 0

Solution:
Primal problem is

MIN `z_x` `=` `` `` `x_1` ` + ` `2` `x_2`

subject to

`` `2` `x_1` ` + ` `4` `x_2` ≤ `160` `` `` `x_1` ` - ` `` `x_2` = `30` `` `` `x_1` ≥ `10`

and `x_1,x_2 >= 0; `

Since objective function is minimizing, all `<=` constraints (1) can be converted to `>=` type by multipling both sides by -1

MIN `z_x` `=` `` `` `x_1` ` + ` `2` `x_2`

subject to

` - ` `2` `x_1` ` - ` `4` `x_2` ≥ `-160` `` `` `x_1` ` - ` `` `x_2` = `30` `` `` `x_1` ≥ `10`

and `x_1,x_2 >= 0; `

In primal, There are `2` variables and `3` constraints, so in dual there must be `2` constraints and `3` variables

In primal, The coefficient of objective function `c_1=1,c_2=2` becomes right hand side constants in dual

In primal, The right hand side constants `b_1=-160,b_2=30,b_3=10` becomes coefficient of objective function in dual

In primal, objective function is minimizing, so in dual objective function must be maximizing

Let `y1,y2,y3` be the dual variables

Since `2^(nd)` constraint in the primal is equality, the corresponding dual variable `y_2` will be unrestricted in sign.

Now, Dual problem is

MAX `z_y` `=` ` - ` `160` `y_1` ` + ` `30` `y_2` ` + ` `10` `y_3`

subject to

` - ` `2` `y_1` ` + ` `` `y_2` ` + ` `` `y_3` ≤ `1` ` - ` `4` `y_1` ` - ` `` `y_2` ≤ `2`

and `y_1,y_3 >= 0; ``y_2` unrestricted in sign

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Click here to get solution of this Dual problem by Simplex method

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