Infeasible solution
If there is no feasible area (there is no any point that satisfy all constraints of the problem), then this solution is called infeasible solution.

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Example
Find solution using graphical method
MAX Z = 6X1 - 4X2
subject to
2X1 + 4X2 <= 4
4X1 + 8X2 >= 16
and X1,X2 >= 0

Solution:
Problem is

MAX `Z` `=` `` `6` `X_1` ` - ` `4` `X_2`

subject to

`` `2` `X_1` ` + ` `4` `X_2` ≤ `4` `` `4` `X_1` ` + ` `8` `X_2` ≥ `16`

and `X_1,X_2 >= 0; `

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Hint to draw constraints

1. To draw constraint `color{red}{2X_1+4X_2<=4} ->(1)`

Treat it as `color{red}{2X_1+4X_2=4}`

When `X_1=0` then `X_2=?`

`=>2(0)+4X_2=4`

`=>4X_2=4`

`=>X_2=(4)/(4)=1`

When `X_2=0` then `X_1=?`

`=>2X_1+4(0)=4`

`=>2X_1=4`

`=>X_1=(4)/(2)=2`

`X_1`	0	2
`X_2`	1	0

Put `X_1=0,X_2=0` (origin) in `color{red}{2X_1+4X_2<=4}`, then `0+0<=4`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{red}{2X_1+4X_2<=4}`

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2. To draw constraint `color{green}{4X_1+8X_2>=16} ->(2)`

Treat it as `color{green}{4X_1+8X_2=16}`

When `X_1=0` then `X_2=?`

`=>4(0)+8X_2=16`

`=>8X_2=16`

`=>X_2=(16)/(8)=2`

When `X_2=0` then `X_1=?`

`=>4X_1+8(0)=16`

`=>4X_1=16`

`=>X_1=(16)/(4)=4`

`X_1`	0	4
`X_2`	2	0

Put `X_1=0,X_2=0` (origin) in `color{green}{4X_1+8X_2>=16}`, then `0+0>=16`, which is false,

The half plane not containing the origin is the region of the solution set of the inequation `color{green}{4X_1+8X_2>=16}`

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Problem has an infeasible solution.

Note: If there is no feasible area (there is no any point that satisfy all constraints of the problem), then this solution is called infeasible solution.

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