 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question 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 food 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. Firmulate this as linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the mineral nutritional requirements. Let us now start with the solution. Now first of all we will formulate the given conditions as linear programming problem. First of all let us assume that number of units of food F1 in the mixture is equal to x. So we can write, let number of units of food F1 in the mixture be x. Also let us assume that number of units of food F2 in the mixture be y. So here we can write and number of units of food F2 in the mixture be y. Now we have given that food F1 costs rupees 4 per unit and F2 costs rupees 6 per unit. Also we know that one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. And one unit of food F2 contains 6 units of vitamin A and 3 units of minerals. We also know that a diet is to contain at least 80 units of vitamin A and 100 units of minerals. How using the given data we can make this table? We know one unit of food F1 contains 3 units of vitamin A and one unit of food F2 contains 6 units of vitamin A. Similarly one unit of food F1 contains 4 units of minerals and one unit of food F2 contains 3 units of minerals. Also food F1 costs rupees 4 per unit and food F2 costs rupees 6 per unit. And minimum requirement of vitamin A is 80 units and minimum requirement of minerals is 100 units. Now we know diet must contain at least 80 units of vitamin A and 100 units of minerals. So we have the constraints 3x plus 6y is greater than equal to 80 and 4x plus 3y is greater than equal to 100. Clearly we can see units of vitamin A in x units of food F1 and y units of food F2 will be 3x plus 6y. And the minimum requirement of vitamin A is 80 units. So we get the first constraint 3x plus 6y is greater than equal to 80. Now again units of minerals in x units of food F1 and y units of food F2 will be 4x plus 3y. And minimum requirement of minerals is equal to 100 units. So we get the second constraint 4x plus 3y is greater than equal to 100. Also units of food F1 and F2 in the mixture will be greater than equal to 0. So xy is greater than equal to 0. Now total cost z of purchasing x units of food F1 and y units of food F2 is z is equal to 4x. Plus 6y clearly we can see cost of purchasing x units of food F1 and y units of food F2 is equal to 4x plus 6y. Now let us assume that this total cost is equal to z. So we can write z is equal to 4x plus 6y. Now the mathematical formulation of the given problem is minimize z is equal to 4x plus 6y subject to the constraints 3x plus 6y is greater than equal to 80. 4x plus 3y is greater than equal to 100. x is greater than equal to 0 and y is greater than equal to 0. Now let us name this expression as 1 and these inequalities as 2, 3, 4 and 5. Now we will graph these inequalities 2 to 5. Now for drawing the graph we shall first of all draw the line representing the equation 3x plus 6y is equal to 80 corresponding to this inequality on the graph. Now we find that points 0 40 upon 3 and 80 upon 3 0 lie on the line 3x plus 6y is equal to 80. Therefore the graph of the line can be drawn by plotting these 2 points and then joining them. Now clearly we can see this line is dividing the plane into 2 half planes. Now we shall consider the half plane that satisfies the condition 3x plus 6y is greater than 80 that is the half plane that does not contains origin. Now we will draw line 4x plus 3y is equal to 100 corresponding to this inequality. Now we find that points 0 100 upon 3 and 25 0 lie on the line 4x plus 3y is equal to 100. So we will plot these 2 points on the same graph and obtain the line by joining these 2 points. Now this line divides the plane into 2 half planes and we will consider the half plane satisfying 4x plus 3y is greater than 100. That is the plane which does not contains 0 0 or we can say plane that does not contains origin. Also x is greater than equal to 0 and y is greater than equal to 0 implies that the graph lies in the first quadrant only. Now clearly we can see these 2 lines intersect at point 24 4 upon 3 and this graph is lying in first quadrant only. This blue shaded portion in the graph is the feasible region satisfying all the given constraints. Clearly we can see this feasible region is unbounded with corner points on the one side as 80 upon 3 0 24 4 upon 3 and 0 100 upon 3. According to the corner point method minimum value of z will occur at any of these 3 points. Now we will calculate the value of the cost z is equal to 4x plus 6y at these corner points. Now first corner point is 0 100 upon 3. Now value of z at this point is equal to 4 multiplied by 0 plus 6 multiplied by 100 upon 3 which is further equal to 200. Now next corner point is 24 4 upon 3. Value of z at this point is equal to 4 multiplied by 24 plus 6 multiplied by 4 upon 3 which is further equal to 104. Now next corner point is 80 upon 3 0. Value of z at this corner point is equal to 4 multiplied by 80 upon 3 plus 6 multiplied by 0. Now simplifying we get 106.67. Clearly we can see out of these 3 values of z this value is the minimum value. So we get the minimum value of z at the corner point is 104. Now as the feasible region is unbounded we have to check if the region represented by z less than 104 has any point common to the feasible region. Now z is less than 104 implies 4x plus 6y is less than 104. Now dividing both sides by 2 we get 2x plus 3y is less than 52. Now we will draw a line 2x plus 3y is equal to 52 on the same graph. Now points 0 52 upon 3 and 24 4 upon 3 lie on the line 2x plus 3y is equal to 52. So we can obtain this line by joining these two points. This line represents the line 2x plus 3y is equal to 52 or we can say this line represents 4x plus 6y is equal to 104. Now this gray shaded region represents 4x plus 6y less than 104 or we can say it represents z less than 104. Now clearly we can see this gray shaded region has no point common with the feasible region. So we get minimum value of z is equal to 104 which occurs at point 24 4 upon 3. Or we can say the minimum cost is equal to rupees 104. Now we can write the minimum value of z is 104 at the point 24 4 upon 3. Thus required minimum cost is equal to rupees 104. This is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.