Find solution using graphical method
MAX z = 15x1 + 10x2
subject to
4x1 + 6x2 <= 360
3x1 <= 180
5x2 <= 200
and x1,x2 >= 0

Solution:
Problem is

MAX `z` `=` `` `15` `x_1` ` + ` `10` `x_2`

subject to

`` `4` `x_1` ` + ` `6` `x_2` ≤ `360` `` `3` `x_1` ≤ `180` `` `5` `x_2` ≤ `200`

and `x_1,x_2 >= 0; `

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Hint to draw constraints

1. To draw constraint `color{red}{4x_1+6x_2<=360} ->(1)`

Treat it as `color{red}{4x_1+6x_2=360}`

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

`=>4(0)+6x_2=360`

`=>6x_2=360`

`=>x_2=(360)/(6)=60`

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

`=>4x_1+6(0)=360`

`=>4x_1=360`

`=>x_1=(360)/(4)=90`

`x_1`	0	90
`x_2`	60	0

Put `x_1=0,x_2=0` (origin) in `color{red}{4x_1+6x_2<=360}`, then `0+0<=360`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{red}{4x_1+6x_2<=360}`

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2. To draw constraint `color{green}{3x_1<=180} ->(2)`

Treat it as `color{green}{3x_1=180}`

`=>x_1=(180)/(3)=60`

Here line is parallel to Y-axis

`x_1`	60	60
`x_2`	0	1

Put `x_1=0,x_2=0` (origin) in `color{green}{3x_1<=180}`, then `0<=180`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{green}{3x_1<=180}`

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3. To draw constraint `color{blue}{5x_2<=200} ->(3)`

Treat it as `color{blue}{5x_2=200}`

`=>x_2=(200)/(5)=40`

Here line is parallel to X-axis

`x_1`	0	1
`x_2`	40	40

Put `x_1=0,x_2=0` (origin) in `color{blue}{5x_2<=200}`, then `0<=200`, which is true,

The half plane containing the origin is the region of the solution set of the inequation `color{blue}{5x_2<=200}`

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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=15x_1 + 10x_2`
`color{red}{O(0,0)}`	`color{black}{4->x_1>=0}` `color{black}{5->x_2>=0}`	`15(0)+10(0)=0`
`color{green}{A(60,0)}`	`color{green}{2->3x_1<=180}` `color{black}{5->x_2>=0}`	`15(60)+10(0)=900`
`color{blue}{B(60,20)}`	`color{red}{1->4x_1+6x_2<=360}` `color{green}{2->3x_1<=180}`	`15(60)+10(20)=1100`
`color{brown}{C(30,40)}`	`color{red}{1->4x_1+6x_2<=360}` `color{blue}{3->5x_2<=200}`	`15(30)+10(40)=850`
`color{darkorange}{D(0,40)}`	`color{blue}{3->5x_2<=200}` `color{black}{4->x_1>=0}`	`15(0)+10(40)=400`

The maximum value of the objective function `z=1100` occurs at the extreme point `(60,20)`.

Hence, the optimal solution to the given LP problem is : `x_1=60, x_2=20` and max `z=1100`.

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