 Hello and welcome to the session. Let's work out the following problem. It says Kellogg is a new cereal form of mixed of bran and rice that contains At least 88 grams of protein and at least 36 milligrams of iron knowing that gram contains 80 grams of protein and 40 milligrams of iron per kilogram and that rice contains 100 gram of protein and 30 milligrams of iron per kilogram find the minimum cost of producing this new cereal If bran cost rupees five per kilogram and rice cost rupees four per kilogram So let us now move on to the solution Now we are given that the mixture is made up of bran and rice so let the mixture kg of bran y kg of Five x as we are given that bran cost rupees five per kilogram So x kg of bran will cost rupees five x and similarly y kg of rice Will cost rupees four y as rice cost rupees four per kilogram So y kg of rice will cost rupees four y now we have to minimize the cost of producing the new cereal At least we have to minimize the cost therefore the total cost is equal to Five x plus four y we have to minimize z is equal to five x plus four y At this mixture must contain At least 88 grams of protein right and we know that bran contains 80 grams of protein and rice contains 100 grams of protein right and As we are given that the mixture must contain at least 88 grams of protein So first constraint is plus 100 y should be greater than equal to 88 As we have given that bran contains 80 gram of protein and rice contain 100 gram of protein So 80 x plus 100 y is greater than equal to 88 as the mixture contains at least 88 grams of protein This is subject to the constraint At the mixture must contain at least 36 milligrams of iron and we are given that the bran contains 40 milligram Of iron and rice contains 30 milligrams of iron So we have 40 x plus 30 y should be greater than equal to 36 So we have to minimize the cost subject to these two constraints And also x is greater than equal to zero and y is greater than equal to zero Now we'll draw the lines corresponding to the constraint 88 plus 100 y greater than equal to 88 and 40 x Plus 30 y greater than equal to 36. So we have drawn the lines This is the line corresponding to the inequality 80 x plus 100 y Is equal to 88 and this is the line Corresponding to the constraint 40 x plus 30 y Greater than equal to 30 now we have to shade the region corresponding to both the constraints Now the constraint it is 80 x plus 100 y greater than equal to 80. So we will take any point not lying on the line 80 x plus 100 y is equal to 80 and we'll see whether that point satisfies This inequality or not if that point satisfies this inequality will shade the region Which contains that point and if that point does not satisfy The inequality will shade the region which Does not contain that point now the point zero zero does not lie on this line And we see that if x is zero y is zero then the inequality becomes zero plus zero greater than equal to 88 That means zero zero does not satisfy this inequality. So we'll shade the region For this inequality, which does not contain the point zero zero That means this is the region which just does not contain the point zero zero for the inequality 80 x plus 100 y greater than equal to 88 Similarly, we'll put x is equal to zero y is equal to zero Here in the inequality 40 x plus 30 y greater than equal to 36 and we'll see and this also does not satisfy The inequality. So we'll shade the region which does not contain the point zero zero. So this is the region and we see that This is the common region for both the constraints or inequalities Now by the corner point method or the fundamental extreme extreme point theorem The optimal solution to a linear programming problem If it exists occur at an extreme point of the feasible region now, this is the feasible region And the extreme points are zero one point two Point six point four one point one zero So We'll find the cost at these corner points and we'll see Which is the minimum So now we find z equal to five x plus four y the point zero one point two Means x is zero y is one point two. So this is five into zero plus four into one point two So this is 4.8 So the cost is 4.8 Now we find the cost When x is point six and y is point four So z At the point point six point four the z is five into point six plus four into point four this is 3.0 plus 1.6. So this is 4.6 Now we find z at The point one point one and zero So z is equal to five into one point one. That is x is one point one and 4 y y is zero. So this is 5.5 And we see that the cost is minimum When x is point six and y is point four and the minimum cost is 4.6 hence the minimum cost 4.60 when x is 4.6 kg and y is Point four kg So The minimum cost is rupees four point six zero So this completes the question and the session. My friend now take care. Have a good day