00:01
Hello everyone here in this question we have the information of max p is maximum speed is 8 x1 plus 15 x2 15 x2 so the subject to this is subject to 5 x1 plus x2 less than or equals to 10 so x1 plus 5 x2 is less than or equals to 50 so x1 will be greater than or equals to 0 which is x2 is also x2 is also greater than or equals to 0 is also greater than or equals to 0 now the introduction of two slack variables variables is is 5x1 plus 5x1 plus x2 x2 plus 4 equals to 10 which is x1 plus another one is x1 plus 5x2 plus v equals to 15.
01:50
So now we can deformulate the problem as deformulate the problem as 5x1 plus x2 plus x2 plus x2 plus 4 equals to 10 and x1 plus 5 x2 plus v equals to 15 so this will come as minus 8 x minus 15 x 2 plus minus 8 x1 minus 15 x2 plus p which is equal to 0 so here x1 comma x2 comma 4 comma v is greater than 0 sorry it is u this not 4 it is u equals to so this is greater than 0 greater than or equals to 0 so by using the flex method simplex method we will get the data as matrix x1 x2 uvp and here uvp matrix close 51101 1515010 and minus 8 minus 5 0 0 0 0 0 which is multiplied with 10, 15, 0.
04:19
So this is 10 divided by 1 and 15 divided by 5, which is equals to 3 and this is equal to 10.
04:30
So this is the minimum value.
04:35
The minimum value.
04:39
And here 1 divided by 5, r1 goes to r2 will be x1, x2, uvp.
04:58
Here u v p which will be 5 1 divided by 5 1 1 1 0 1 0 0 0 0 0 0 0 0 so this will be so p will be equals minus minus 15 0 0 0 0 1 divided by 5 0 so this will be so p will be equals minus 8 minus 15 0 0 0 0 1.
05:46
So here it will be 10, 3 and 0.
05:53
So here minus r2 plus r1 goes to r1, where 15r2 plus r3 goes to r3 will be come as 15 goes to r2.
06:20
It will become as like this x1, x1, x2, x2, u, v.
06:26
So here uvp will be 1 .05 divided by 24 minus 1 divided by 24 0 .1 divided by 5, 1 divided by 5, 0...