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3. Graphical method example ( Enter your problem )
  1. Algorithm & Example-1
  2. Example-2
  3. Multiple optimal solution example
  4. Unbounded solution example
  5. Infeasible solution example
Other related methods
  1. Simplex (BigM) method
  2. Two-Phase method
  3. Graphical method
  4. Primal to dual conversion
  5. Dual simplex method
  6. Integer simplex method
  7. Branch and Bound method
  8. 0-1 Integer programming problem

2. Example-2

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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_x``=````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; `




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`090
`x_2`600




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`6060
`x_2`01




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`01
`x_2`4040









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 PointObjective 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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