Multiple optimal solution
The optimal value of the objective function occurs at more than 1 extreme points, then the problem has multiple optimal solution.

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Example
Find solution using graphical method
MAX z = 10x1 + 6x2
subject to
5x1 + 3x2 <= 30
x1 + 2x2 <= 18
and x1,x2 >= 0

Solution:
Problem is

MAX `z` `=` `` `10` `x_1` ` + ` `6` `x_2`

subject to

`` `5` `x_1` ` + ` `3` `x_2` ≤ `30` `` `` `x_1` ` + ` `2` `x_2` ≤ `18`

and `x_1,x_2 >= 0; `

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Hint to draw constraints

1. To draw constraint `color{red}{5x_1+3x_2<=30} ->(1)`

Treat it as `color{red}{5x_1+3x_2=30}`

When `x_1=0` then `x_2=?`

`=>5(0)+3x_2=30`

`=>3x_2=30`

`=>x_2=(30)/(3)=10`

When `x_2=0` then `x_1=?`

`=>5x_1+3(0)=30`

`=>5x_1=30`

`=>x_1=(30)/(5)=6`

`x_1`	0	6
`x_2`	10	0

Put `x_1=0,x_2=0` (origin) in `color{red}{5x_1+3x_2<=30}`, then `0+0<=30`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{red}{5x_1+3x_2<=30}`

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2. To draw constraint `color{green}{x_1+2x_2<=18} ->(2)`

Treat it as `color{green}{x_1+2x_2=18}`

When `x_1=0` then `x_2=?`

`=>(0)+2x_2=18`

`=>2x_2=18`

`=>x_2=(18)/(2)=9`

When `x_2=0` then `x_1=?`

`=>x_1+2(0)=18`

`=>x_1=18`

`x_1`	0	18
`x_2`	9	0

Put `x_1=0,x_2=0` (origin) in `color{green}{x_1+2x_2<=18}`, then `0+0<=18`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{green}{x_1+2x_2<=18}`

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The value of the objective function at each of these extreme points is as follows:

Extreme Point Coordinates (`x_1`,`x_2`)	Lines through Extreme Point	Objective function value `z=10x_1 + 6x_2`
`color{red}{O(0,0)}`	`color{black}{3->x_1>=0}` `color{black}{4->x_2>=0}`	`10(0)+6(0)=0`
`color{green}{A(6,0)}`	`color{red}{1->5x_1+3x_2<=30}` `color{black}{4->x_2>=0}`	`10(6)+6(0)=60`
`color{blue}{B(0.86,8.57)}`	`color{red}{1->5x_1+3x_2<=30}` `color{green}{2->x_1+2x_2<=18}`	`10(0.86)+6(8.57)=60`
`color{brown}{C(0,9)}`	`color{green}{2->x_1+2x_2<=18}` `color{black}{3->x_1>=0}`	`10(0)+6(9)=54`

The maximum value of the objective function `z=60` occurs at 2 extreme points.

Hence, problem has multiple optimal solutions and max `z=60`.

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