Find solution using graphical method
MIN z = 20x1 + 10x2
subject to
x1 + 2x2 <= 40
3x1 + x2 >= 30
4x1 + 3x2 >= 60
and x1,x2 >= 0
Solution:
Problem is
MIN `z_x` | `=` | `` | `20` | `x_1` | ` + ` | `10` | `x_2` |
|
subject to |
`` | `` | `x_1` | ` + ` | `2` | `x_2` | ≤ | `40` | `` | `3` | `x_1` | ` + ` | `` | `x_2` | ≥ | `30` | `` | `4` | `x_1` | ` + ` | `3` | `x_2` | ≥ | `60` |
|
and `x_1,x_2 >= 0; ` |
Hint to draw constraints
1. To draw constraint `color{red}{x_1+2x_2<=40 ->(1)}`
Treat it as `color{red}{x_1+2x_2=40}`
When `x_1=0` then `x_2=?`
`=>(0)+2x_2=40`
`=>2x_2=40`
`=>x_2=(40)/(2)=20`
When `x_2=0` then `x_1=?`
`=>x_1+2(0)=40`
`=>x_1=40`
2. To draw constraint `color{green}{3x_1+x_2>=30 ->(2)}`
Treat it as `color{green}{3x_1+x_2=30}`
When `x_1=0` then `x_2=?`
`=>3(0)+x_2=30`
`=>x_2=30`
When `x_2=0` then `x_1=?`
`=>3x_1+(0)=30`
`=>3x_1=30`
`=>x_1=(30)/(3)=10`
3. To draw constraint `color{blue}{4x_1+3x_2>=60 ->(3)}`
Treat it as `color{blue}{4x_1+3x_2=60}`
When `x_1=0` then `x_2=?`
`=>4(0)+3x_2=60`
`=>3x_2=60`
`=>x_2=(60)/(3)=20`
When `x_2=0` then `x_1=?`
`=>4x_1+3(0)=60`
`=>4x_1=60`
`=>x_1=(60)/(4)=15`
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=20x_1 + 10x_2` |
`color{red}{A(4,18)}` | `color{red}{1->x_1+2x_2<=40}` `color{green}{2->3x_1+x_2>=30}` | `20(4)+10(18)=260` |
`color{green}{B(6,12)}` | `color{green}{2->3x_1+x_2>=30}` `color{blue}{3->4x_1+3x_2>=60}` | `20(6)+10(12)=240` |
`color{blue}{C(15,0)}` | `color{blue}{3->4x_1+3x_2>=60}` `color{black}{5->x_2>=0}` | `20(15)+10(0)=300` |
`color{brown}{D(40,0)}` | `color{red}{1->x_1+2x_2<=40}` `color{black}{5->x_2>=0}` | `20(40)+10(0)=800` |
The miniimum value of the objective function `z=240` occurs at the extreme point `(6,12)`.
Hence, the optimal solution to the given LP problem is : `x_1=6, x_2=12` and min `z=240`.
This material is intended as a summary. Use your textbook for detail explanation.
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