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`
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 Yaxis
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 Xaxis
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=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`.
This material is intended as a summary. Use your textbook for detail explanation.
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