Find solution using graphical method
MIN z = 3x1 + 2x2
subject to
5x1 + x2 >= 10
x1 + x2 >= 6
x1 + 4x2 >= 12
and x1,x2 >= 0
Solution:
Problem is
| MIN `z_x` | `=` | `` | `3` | `x_1` | ` + ` | `2` | `x_2` |
|
| subject to |
| `` | `5` | `x_1` | ` + ` | `` | `x_2` | ≥ | `10` | | `` | `` | `x_1` | ` + ` | `` | `x_2` | ≥ | `6` | | `` | `` | `x_1` | ` + ` | `4` | `x_2` | ≥ | `12` |
|
| and `x_1,x_2 >= 0; ` |
Hint to draw constraints
1. To draw constraint `color{red}{5x_1+x_2>=10 ->(1)}`
Treat it as `color{red}{5x_1+x_2=10}`
When `x_1=0` then `x_2=?`
`=>5(0)+x_2=10`
`=>x_2=10`
When `x_2=0` then `x_1=?`
`=>5x_1+(0)=10`
`=>5x_1=10`
`=>x_1=(10)/(5)=2`
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`
3. To draw constraint `color{blue}{x_1+4x_2>=12 ->(3)}`
Treat it as `color{blue}{x_1+4x_2=12}`
When `x_1=0` then `x_2=?`
`=>(0)+4x_2=12`
`=>4x_2=12`
`=>x_2=(12)/(4)=3`
When `x_2=0` then `x_1=?`
`=>x_1+4(0)=12`
`=>x_1=12`
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=3x_1 + 2x_2` |
| `color{red}{A(0,10)}` | `color{red}{1->5x_1+x_2>=10}`
`color{black}{4->x_1>=0}` | `3(0)+2(10)=20` |
| `color{green}{B(1,5)}` | `color{red}{1->5x_1+x_2>=10}`
`color{green}{2->x_1+x_2>=6}` | `3(1)+2(5)=13` |
| `color{blue}{C(4,2)}` | `color{green}{2->x_1+x_2>=6}`
`color{blue}{3->x_1+4x_2>=12}` | `3(4)+2(2)=16` |
| `color{brown}{D(12,0)}` | `color{blue}{3->x_1+4x_2>=12}`
`color{black}{5->x_2>=0}` | `3(12)+2(0)=36` |
The miniimum value of the objective function `z=13` occurs at the extreme point `(1,5)`.
Hence, the optimal solution to the given LP problem is : `x_1=1, x_2=5` and min `z=13`.
This material is intended as a summary. Use your textbook for detail explanation.
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