Find solution using graphical method
MIN z = 200x1 + 400x2
subject to
x1 + x2 >= 20
x1 + 3x2 >= 40
x1 + 2x2 <= 35
and x1,x2 >= 0
Solution:
Problem is
| MIN `z_x` | `=` | `` | `200` | `x_1` | ` + ` | `400` | `x_2` |
|
| subject to |
| `` | `` | `x_1` | ` + ` | `` | `x_2` | ≥ | `20` | | `` | `` | `x_1` | ` + ` | `3` | `x_2` | ≥ | `40` | | `` | `` | `x_1` | ` + ` | `2` | `x_2` | ≤ | `35` |
|
| and `x_1,x_2 >= 0; ` |
Hint to draw constraints
1. To draw constraint `color{red}{x_1+x_2>=20 ->(1)}`
Treat it as `color{red}{x_1+x_2=20}`
When `x_1=0` then `x_2=?`
`=>(0)+x_2=20`
`=>x_2=20`
When `x_2=0` then `x_1=?`
`=>x_1+(0)=20`
`=>x_1=20`
2. To draw constraint `color{green}{x_1+3x_2>=40 ->(2)}`
Treat it as `color{green}{x_1+3x_2=40}`
When `x_1=0` then `x_2=?`
`=>(0)+3x_2=40`
`=>3x_2=40`
`=>x_2=(40)/(3)=13.33`
When `x_2=0` then `x_1=?`
`=>x_1+3(0)=40`
`=>x_1=40`
3. To draw constraint `color{blue}{x_1+2x_2<=35 ->(3)}`
Treat it as `color{blue}{x_1+2x_2=35}`
When `x_1=0` then `x_2=?`
`=>(0)+2x_2=35`
`=>2x_2=35`
`=>x_2=(35)/(2)=17.5`
When `x_2=0` then `x_1=?`
`=>x_1+2(0)=35`
`=>x_1=35`
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=200x_1 + 400x_2` |
| `color{red}{A(5,15)}` | `color{red}{1->x_1+x_2>=20}`
`color{blue}{3->x_1+2x_2<=35}` | `200(5)+400(15)=7000` |
| `color{green}{B(25,5)}` | `color{green}{2->x_1+3x_2>=40}`
`color{blue}{3->x_1+2x_2<=35}` | `200(25)+400(5)=7000` |
| `color{blue}{C(10,10)}` | `color{red}{1->x_1+x_2>=20}`
`color{green}{2->x_1+3x_2>=40}` | `200(10)+400(10)=6000` |
The miniimum value of the objective function `z=6000` occurs at the extreme point `(10,10)`.
Hence, the optimal solution to the given LP problem is : `x_1=10, x_2=10` and min `z=6000`.
This material is intended as a summary. Use your textbook for detail explanation.
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