Linear Programming Problem Using Graphical Methods.

Suppose you are given an Objective function :

                Max Z = 6x+5y

Subject to,

                X+Y ≤  5
                3X+2Y ≤ 12
                X,Y ≥ 0

Solution :-

Step 1: You have to put 0 in place of X to calculate Y. Again, you have to put 0 in place of Y and calculate X, by removing the less-than sign.

 So, Putting X=0 in equation X+Y ≤ 5.
The equation becomes, [0+Y=5], So Y comes out to be Y=5. 
So the solution is (0,5).................................................................................{I}

Now, putting Y=0 in the equation X+Y ≤ 5.
The equation becomes,[X+0=5], So X comes out to be 5.
So the solution is (5,0)..................................................................................{II}

Now, Putting X=0 in equation 3X+2Y ≤ 12.
The equation becomes,[3*0+2Y=12], So Y comes out to be Y=6.
So the solution is (0,6)..................................................................................{III}

Now, Putting Y=0 in equation 3X+2Y ≤ 12.
The equation becomes,[3X+2*0=12], So Y comes out to be X=4.
So the solution is (4,0)..................................................................................{IV}

Now, we have to draw X and Y co-ordinates taking X as a horizontal line while Y as vertical.


Now you have to marks the point (0,5) & (5,0) and draw a line joining these two points.


Now, you have to mark the point (0,6) & (4,0) and draw a line joining these two points.


                                         Now, Shade the common part of the co-ordinate axis.


Clearly, we can see that there are 4 corner points touching the colored part B(4,0) C(0,5), Origin(0,0) and one corner point is, where the two lines cut each other. So, we have to calculate the 4th corner point first.
So, We have to solve the two equations.
                                                  X+Y=5.............(i)
                                                  3X+2Y=12.......(ii)
  Multiplying equation (i) by 3 we get the equation as follows.
                                                  3X+3Y=15
                                                  3X+2Y=12
Now,subtracting equation (ii) from equation (i),we get ,
                                                  Y=3
Putting Y=3 in equation (i) we get, X=2.
So, the 5th corner point comes out to be (2,3).   

Now, check all the corner points One by One in the objective function.

Checking (4,0)

Z=6*4 + 5*0 => Z=24

Checking (0,5)

Z=6*0 + 5*5 => Z=25

Checking (0,0)

Z=6*0 + 5*0 => Z=0

Checking (2,3)

Z=6*2 + 5*3 => Z=27

 Clearly, we see that we are getting the Zmax 27 at point (2,3) .
So,

Zmax = 27     (Answer)



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