Find solution using graphical method
MIN z = 600x1 + 400x2
subject to
3x1 + 3x2 >= 40
3x1 + x2 >= 40
2x1 + 5x2 >= 44
and x1,x2 >= 0
Solution:
Problem is
| MIN `z_x` | `=` | `` | `600` | `x_1` | ` + ` | `400` | `x_2` |
|
| subject to |
| `` | `3` | `x_1` | ` + ` | `3` | `x_2` | ≥ | `40` | | `` | `3` | `x_1` | ` + ` | `` | `x_2` | ≥ | `40` | | `` | `2` | `x_1` | ` + ` | `5` | `x_2` | ≥ | `44` |
|
| and `x_1,x_2 >= 0; ` |
Hint to draw constraints
1. To draw constraint `color{red}{3x_1+3x_2>=40 ->(1)}`
Treat it as `color{red}{3x_1+3x_2=40}`
When `x_1=0` then `x_2=?`
`=>3(0)+3x_2=40`
`=>3x_2=40`
`=>x_2=(40)/(3)=13.33`
When `x_2=0` then `x_1=?`
`=>3x_1+3(0)=40`
`=>3x_1=40`
`=>x_1=(40)/(3)=13.33`
2. To draw constraint `color{green}{3x_1+x_2>=40 ->(2)}`
Treat it as `color{green}{3x_1+x_2=40}`
When `x_1=0` then `x_2=?`
`=>3(0)+x_2=40`
`=>x_2=40`
When `x_2=0` then `x_1=?`
`=>3x_1+(0)=40`
`=>3x_1=40`
`=>x_1=(40)/(3)=13.33`
3. To draw constraint `color{blue}{2x_1+5x_2>=44 ->(3)}`
Treat it as `color{blue}{2x_1+5x_2=44}`
When `x_1=0` then `x_2=?`
`=>2(0)+5x_2=44`
`=>5x_2=44`
`=>x_2=(44)/(5)=8.8`
When `x_2=0` then `x_1=?`
`=>2x_1+5(0)=44`
`=>2x_1=44`
`=>x_1=(44)/(2)=22`
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=600x_1 + 400x_2` |
| `color{red}{A(0,40)}` | `color{green}{2->3x_1+x_2>=40}`
`color{black}{4->x_1>=0}` | `600(0)+400(40)=16000` |
| `color{green}{B(12,4)}` | `color{green}{2->3x_1+x_2>=40}`
`color{blue}{3->2x_1+5x_2>=44}` | `600(12)+400(4)=8800` |
| `color{blue}{C(22,0)}` | `color{blue}{3->2x_1+5x_2>=44}`
`color{black}{5->x_2>=0}` | `600(22)+400(0)=13200` |
The miniimum value of the objective function `z=8800` occurs at the extreme point `(12,4)`.
Hence, the optimal solution to the given LP problem is : `x_1=12, x_2=4` and min `z=8800`.
This material is intended as a summary. Use your textbook for detail explanation.
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