00:01
In this problem, it is given that there are two processing plants, p1 and p2, which run 7 days a week, and the producer has contracted to provide 120 kg of vhyl, 80kg of medium, and 240k of low -quality beef each week.
00:18
The cost per day of operating the plants, the high -quality, medium -quality and low -quality beef produced by the plants per day is given, and we have to find the number of days the plants should be operated to fulfill the contract most economically.
00:35
So here we are going to use the lpp to solve this problem.
00:40
Let us start with a solution.
00:43
Here we have to minimize the cost of operating the plants and then fulfilling the contract.
00:50
So let us suppose that p1 is operated x days and p2 is operated y days.
01:01
Then the cost of operating the plants would be suppose it is equal to z and z is equal to 4 ,000 multiplied by x plus 3 ,200 multiplied by y because the cost of operating per day of p1 is 4 ,000 and cost of operating per day of p2 is 3 ,200 so we have to minimize the cost that is min z is equal to 4 ,000 x plus 3 ,200 y.
01:38
So this is the problem.
01:41
Now we find the constraints.
01:43
The amount of high quality beef produced would be since 60 kg is produced by p1, so 60 multiplied by x plus 20 kg is produced by p2, so 20 multiplied by y.
02:11
And the requirement is 180 kg so this must be greater than or equal to 18020 kg.
02:18
The next constraint is for medium quality beef and it would be similar manner we will write 20x plus 20y.
02:29
This must be greater than or equal to 80.
02:33
And then we have for low quality beef, the constraint would be that 40x plus 1802y.
02:41
This must be greater than or equal to 140.
02:46
On simplification, these complaints can be written as 3x plus y is greater than or equal to 6.
02:55
Let this be equation 1.
02:58
Then x plus y, it should be greater than or equal to 4.
03:02
This is equation 2...