Find solution using graphical method
MAX z = 2x1 + x2
subject to
x1 + 2x2 <= 10
x1 + x2 <= 6
x1 - x2 <= 2
x1 - 2x2 <= 1
and x1,x2 >= 0

Solution:
Problem is

MAX `z` `=` `` `2` `x_1` ` + ` `` `x_2`

subject to

`` `` `x_1` ` + ` `2` `x_2` ≤ `10` `` `` `x_1` ` + ` `` `x_2` ≤ `6` `` `` `x_1` ` - ` `` `x_2` ≤ `2` `` `` `x_1` ` - ` `2` `x_2` ≤ `1`

and `x_1,x_2 >= 0; `

---

Hint to draw constraints

1. To draw constraint `color{red}{x_1+2x_2<=10} ->(1)`

Treat it as `color{red}{x_1+2x_2=10}`

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

`=>(0)+2x_2=10`

`=>2x_2=10`

`=>x_2=(10)/(2)=5`

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

`=>x_1+2(0)=10`

`=>x_1=10`

`x_1`	0	10
`x_2`	5	0

Put `x_1=0,x_2=0` (origin) in `color{red}{x_1+2x_2<=10}`, then `0+0<=10`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{red}{x_1+2x_2<=10}`

---

2. To draw constraint `color{green}{x_1+x_2<=6} ->(2)`

Treat it as `color{green}{x_1+x_2=6}`

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

`=>(0)+x_2=6`

`=>x_2=6`

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

`=>x_1+(0)=6`

`=>x_1=6`

`x_1`	0	6
`x_2`	6	0

Put `x_1=0,x_2=0` (origin) in `color{green}{x_1+x_2<=6}`, then `0+0<=6`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{green}{x_1+x_2<=6}`

---

3. To draw constraint `color{blue}{x_1-x_2<=2} ->(3)`

Treat it as `color{blue}{x_1-x_2=2}`

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

`=>(0)-x_2=2`

`=>-x_2=2`

`=>x_2=-2`

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

`=>x_1-(0)=2`

`=>x_1=2`

`x_1`	0	2
`x_2`	-2	0

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

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

---

4. To draw constraint `color{brown}{x_1-2x_2<=1} ->(4)`

Treat it as `color{brown}{x_1-2x_2=1}`

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

`=>(0)-2x_2=1`

`=>-2x_2=1`

`=>x_2=(1)/(-2)=-0.5`

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

`=>x_1-2(0)=1`

`=>x_1=1`

`x_1`	0	1
`x_2`	-0.5	0

Put `x_1=0,x_2=0` (origin) in `color{brown}{x_1-2x_2<=1}`, then `0+0<=1`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{brown}{x_1-2x_2<=1}`

---

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=2x_1 + x_2`
`color{red}{O(0,0)}`	`color{black}{5->x_1>=0}` `color{black}{6->x_2>=0}`	`2(0)+1(0)=0`
`color{green}{A(1,0)}`	`color{brown}{4->x_1-2x_2<=1}` `color{black}{6->x_2>=0}`	`2(1)+1(0)=2`
`color{blue}{B(3,1)}`	`color{blue}{3->x_1-x_2<=2}` `color{brown}{4->x_1-2x_2<=1}`	`2(3)+1(1)=7`
`color{brown}{C(4,2)}`	`color{green}{2->x_1+x_2<=6}` `color{blue}{3->x_1-x_2<=2}`	`2(4)+1(2)=10`
`color{darkorange}{D(2,4)}`	`color{red}{1->x_1+2x_2<=10}` `color{green}{2->x_1+x_2<=6}`	`2(2)+1(4)=8`
`color{darkviolet}{E(0,5)}`	`color{red}{1->x_1+2x_2<=10}` `color{black}{5->x_1>=0}`	`2(0)+1(5)=5`

The maximum value of the objective function `z=10` occurs at the extreme point `(4,2)`.

Hence, the optimal solution to the given LP problem is : `x_1=4, x_2=2` and max `z=10`.

---

---