00:01
To maximize the objective function p as x1 plus of x2, subject to the given equation that is 2 of x1 plus x2 is less than or equal to 4, x1 plus 2 of x2 is less than or equal to 3, x1 and x2 is greater than or equal to 0.
00:22
We can solve the linear program this way.
00:26
First, we will plot the lines corresponding to the inequality.
00:30
So, just imagine.
00:31
We will write it as a 2x1 plus x2 equals to 4 x1 plus 2x2 equals to 3 just suppose this is our line 1 and this is our line okay so then we will evaluate this okay so how we will solve this equation we will put both the equal so because this one and this one we will solve this system equation to find the first power point now the x1 x1 is where x2 will be 0 this point will be from there 2 ,0 and the x2 x2 means where x1 will be 0 this point is 0 and 1 .5 from these two equations now let's find the value of p.
01:42
So for the first corner point 2x1 plus x2 we will solve this system.
01:49
This one and this one.
01:51
First we need to find out first corner point we will solve this equation 2x1 plus of x2 equals to 4 and the another one is x1 plus of 2x2 equals to 3.
02:04
So how we will solve it? we will solve it 2x1 plus x2 equals to 4 is the answer.
02:10
Multiply this 2 it will become 2x1 plus 4x2 equals to 6 minus minus minus this 2 district 8 times 3 of x2 equals to minus of 2 so x2 will be 2 by 3 put this value out of any of these we will put we are getting x1 plus 2 into 2 by 3 divided by 3.
02:40
This one will be 3 minus 4 by 3...