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3. Graphical method example ( Enter your problem )
  1. Algorithm & Example-1
  2. Example-2
  3. Multiple optimal solution example
  4. Unbounded solution example
  5. Infeasible solution example
Other related methods
  1. Simplex (BigM) method
  2. Two-Phase method
  3. Graphical method
  4. Primal to dual conversion
  5. Dual simplex method
  6. Integer simplex method
  7. Branch and Bound method
  8. 0-1 Integer programming problem

2. Example-2

Find solution using graphical method
MAX z = 15x1 + 10x2
subject to
4x1 + 6x2 <= 360
3x1 <= 180
5x2 <= 200
and x1,x2 >= 0

Problem is
MAX `z_x``=````15``x_1`` + ``10``x_2`
subject to
```4``x_1`` + ``6``x_2``360`
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=?`




When `x_2=0` then `x_1=?`





2. To draw constraint `color{green}{3x_1<=180 ->(2)}`

Treat it as `color{green}{3x_1=180}`


Here line is parallel to Y-axis

3. To draw constraint `color{blue}{5x_2<=200 ->(3)}`

Treat it as `color{blue}{5x_2=200}`


Here line is parallel to X-axis

The value of the objective function at each of these extreme points is as follows:
Extreme Point
Lines through Extreme PointObjective function value
`z=15x_1 + 10x_2`

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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