 Hello and welcome to the session. In this session we discussed the following question which says a diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs rupees 4 per unit and F2 costs rupees 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these two foods and also needs the minimal nutritional requirements. Let's move on to the solution now. First of all we assume let the diet contain x units of food F1 and y units of food F2. Now clearly we have x greater than equal to 0 and y greater than equal to 0. We make this following table from the given data. So according to the given problem we have two foods F1 and F2 and we had assumed that the food F1 contains x units and food F2 contains y units. Now here we have vitamin A and minerals. In the question it's given to us that one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. So in this table we write vitamin A in food F1 is 3 units and minerals in food F1 is 4 units. Then in food F2 according to the question we have that one unit of food F2 contains 6 units of vitamin A and 3 units of minerals. So vitamin A in food F2 is 6 units and 3 units of minerals in food F2. In the question we have that the diet is to contain at least 80 units of vitamin A and 100 units of minerals. So in this table in the column of requirement we'll write 80 for vitamin A and for minerals it would be 100. Then in the question it's also given that the food F1 costs rupees 7 per unit and the food F2 costs rupees 6 per unit. So here we would write 4 and here we would write 6. Then in the question we have the cost of food F1 per unit is rupees 4 and the cost of food F2 per unit is rupees 6. Now according to the question we have that the diet must contain at least 80 units of vitamin A and 100 units of minerals. So these are the constraints 3x plus 6y should be greater than equal to 80 3y should be greater than equal to 100. These are the two constraints. Now the total cost say z of the diet containing x units of food F1 and y units of food F2 is given by z is equal to now it's given in the question that rupees 4 per unit is the cost of food F1 and rupees 6 per unit is the cost of food F2. Now since we are purchasing x units of food F1 and y units of food F2 so we would have z would be equal to 4x plus 6y. Hence we say the mathematical formulation of the problem minimize z is equal to 4x plus 6y let this be equation 1. Subject to constraints plus 6y greater than equal to 80 let this be equation 2x plus 3y greater than equal to 100 let this be equation 3 and x and y greater than equal to 0 let this be equation 4. Now we will graph the inequalities 2, 3 and 4 that is these three inequalities first we consider the equation 3x plus 6y is equal to 80 this is the table of values for the equation 3x plus 6y equal to 80 so for x equal to 0 we have y equal to 40 upon 3 and for x equal to 80 upon 3 we have y is equal to 0 now we draw the graph for this equation this is the graph of the equation 3x plus 6y is equal to 80 since the 0.00 that is the origin does not satisfy the in equation 3x plus 6y greater than equal to 80 so the region above this line would represent the inequality 3x plus 6y greater than equal to 80 so this region is the inequality 3x plus 6y greater than equal to 80 next consider the equation 4x plus 3y equal to 100 in this equation when we put x equal to 0 we get y as 100 upon 3 and for x equal to 25 we get y as 0 we plot these points to get the graph of this equation this line joining the point C and D represents the graph of the equation 4x plus 3y is equal to 100 now as the point origin that is 00 does not satisfy the in equation 4x plus 3y greater than equal to 100 so the region above this line not containing the point 00 would represent the inequality 4x plus 3y greater than equal to 100 so this shaded region represents the inequality 4x plus 3y greater than equal to 100 now x greater than equal to 0 is represented by the y axis and the region to its right hand side that is this region then y greater than equal to 0 is represented by the x axis and the region above this so this region in red is the inequality 4x plus 3y greater than equal to 100 and this region in yellow that is above this line including this red region is the inequality 3x plus 6y greater than equal to 80 so this region with lines is the feasible region including the points C, D and B now we also need to find the point of intersection of these two equations so we have equation 3x plus 6y equal to 80 now we multiply this equation by 2 so we have 8x plus 6y is equal to 200 now we subtract these two equations so as to get minus 5x equal to minus 120 and from here we get x is equal to 120 upon 5 now 524 times is 120 therefore we get x is equal to 24 now to get the value for y we substitute the value for x in any of the above two equations so now we have 3 into x that is 24 plus 6 into y is equal to 80 thus further we have 72 plus 6y is equal to 80 this gives us 6y is equal to 80 minus 72 that is 8 so now y is equal to 8 upon 6 now 2 3 times is 6 and 2 4 times is 8 therefore y is equal to 4 upon 3 thus x is equal to 24 and y is equal to 4 upon 3 so the point of intersection of the two lines 3x plus 6y equal to 80 and 4x plus 3y equal to 100 is this point say point E with coordinates 24 comma 4 upon 3 thus the corner points of this feasible region are C E and the point B so the corner points of the feasible region are point C with coordinates 0 comma 100 upon 3 E with coordinates 24 comma 4 upon 3 and point B with coordinates 80 upon 3 and 0 now we have z is equal to 4x plus 6y now we will evaluate z at these corner points now at the corner point C with coordinates 0 comma 100 upon 3 let's find the value of z which would be equal to 4 into 0 plus 6 into 100 upon 3 now 3 2 times is 6 and this is equal to 200 now next at the corner point E with coordinates 24 and 4 upon 3 we have the value of z equal to 4 into 24 plus 6 into 4 upon 3 now 3 2 times is 6 and so this is equal to 104 then at the corner point B with coordinates 80 upon 3 and 0 we have the value of z equal to 4 into 80 upon 3 plus 6 into 0 and this is equal to 106.6 now out of these 3 values of z we have 104 is minimum therefore we say the minimum cost is rupees 104 when we have x is equal to 24 that is the x coordinate of the point E that is x is 24 units and y is 4 upon 3 units so this is the required answer that the minimum cost is rupees 104 when x is 24 units and y is 4 upon 3 units so this completes the session hope you have understood the solution of this question