 Welcome to the session. I am Kanika and I am going to help you to solve the following question. The question says, a guide for a sick person must contain at least 4,000 units of the tarmets, 50 units of minerals and 400 units of calories. Two foods A and B are available at a cost of Rs 4 and Rs 3 per unit. Respectively, if 1 unit of A contains 200 units of vitamin, 1 unit of mineral and 40 units of calories, 1 unit of food B contains 100 units of vitamin, 2 units of minerals and 40 units of calories, find what combination of foods should be used to have the least cost. We will solve this question by using corner point method. This method is based on the fundamental extreme point theorem. The theorem states that the optimal solution to a linear programming problem if it exists occurs at an extreme point of the feasible region. So, in corner point method, we have to first find the feasible region of the linear programming problem, then we have to find the coordinates of each vertex of the feasible region and at last we have to evaluate the value of the objective function at each corner point which we have obtained in previous step. If the problem is of maximization, then the solution corresponding to the largest value of z is the optimal solution of the LPP and the value of z is the optimum value. The knowledge of this method is the key idea in this question. Now, begin with the solution. In this question, we want to find the combination of foods A and B which should be used to have the least cost. So, let number of units, food A be x, number of units of food B be y. Now, we will make a table of the information given in the question. So, let's make a table of the information given in the question. We are given that one unit of food A contains 200 units of the tamin and one unit of food B contains 100 units of the tamin. Food A contains 1 unit of mineral water, 1 unit of mineral water and 1 unit of food B contains 100 units of the tamin. Food A contains one unit of mineral and two units of mineral are contained in food B. Food A contains 40 units of calories and food B contains 40 units of calories. Cost of food A is rupees 4 per unit and cost of food B is rupees 3 per unit. Now read the first statement of the question. It says a diet for a sick person must contain at least 4,000 units of vitamins, 50 units of minerals and 1,400 units of calories. So this means minimum requirement of the vitamins is 4,000 units of minerals is 50 units, calories is 1,400 units. Now according to this table I will express X and Y as linear in equation. Now look at the first row. One unit of food A contains 200 units of vitamin and one unit of food B contains 100 units of vitamins. So X units of food A contains 200 X units of vitamin and Y units of food B contains 100 Y units of vitamin. Thus X units of food A and Y units of food B contains 200 X plus 100 Y units of vitamins. Now the vitamins contained in the diet should not be less than 4,000. So 200 X plus 100 Y is greater than equal to 4,000. Now this implies 100 into 2X plus Y is greater than equal to 4,000. This implies 2X plus Y is greater than equal to 40. So this is our first in equation. Now look at the second row. Now according to this row X units of food A and Y units of food B contain X plus 2Y units of minerals. Also the minerals should not be less than 50 units therefore X plus 2Y is greater than equal to 50. This is our second in equation. According to third row X units of food A and Y units of food B contain 40 X plus 40 Y units of calories. And the diet should contain at least 1,400 calories therefore 40 X plus 40 Y is greater than equal to 1,400. This implies 40 into X plus Y is greater than equal to 1,400. This implies X plus Y is greater than equal to 35. X and Y cannot be negative therefore X is greater than equal to 0 and Y is also greater than equal to 0. So we have formed three in equations according to the given constraints. Now we will find the objective function cost of 1 unit of food A is rupees 4 and cost of 1 unit of food B is rupees 3. So cost of X units of food A, food B rupees cost function Z equal to 4X plus 3Y and we have to minimize this cost. The mathematical formulation the linear programming problem minimize equals to 4X plus 3Y subject to the constraints greater than equal to 40 greater than equal to 50 Y greater than equal to 35 greater than equal to 0 and Y is also greater than equal to 0. Equation corresponding to in equation to X plus Y greater than equal to 40 is 2X plus Y equals to 40. Equation corresponding to this is X plus 2Y equals to 50. Equation corresponding to this in equation is X plus Y equals to 35. Points of intersection with X axis and Y axis of this line are 0, 40, 20, 0. Points of intersection with X axis and Y axis of this line are 0, 25, 50, 0. Points of intersection with X axis and Y axis of this line are 0, 35, 35, 0. 0, 0 does not satisfy all these in equations therefore all the points above these lines satisfy this in equations. So, keeping this in mind let us now draw the graph. Let us now draw the lines first line is 2X plus Y equals to 40. This line passes through point 0, 40. This is point 0, 40 and 20, 0. Join these two points. This is the line 2X plus Y equals to 40. Now we will draw the second line. Points of second line are 0, 25, T, 0. Join these two points. So this is the line X plus 2Y equals to 50. Now we will draw the third line. Points are 0, 35 and 35, 0. Let us now join these two points. This is the line X plus Y equals to 35. Now we will find the feasible region. The shaded region is the feasible region. So here feasible region is unbounded above. The lower corners are 0, 40. This point we can obtain by solving equation 2X plus Y equals to 40 and X plus Y equals to 35 simultaneously. On solving these equations, we get this point as 5, 30. This point is 20, 50. We have obtained this point by solving equation X plus 2Y equals to 50 and X plus Y equals to 35 simultaneously and 50, 0. At the corner point method, we know that minimum value of Z will occur at one of these points. So now we will calculate value of Z at each of these points. The first point is 0, 40. Z is equal to 4X plus 3Y. On substituting X as 0 and Y as 40, we get 120. Second point is 5, 30. Z is equal to 4X plus 3Y. I am putting X as 5 and Z as 30. We get 110. Third point is 20, 50. I am putting X as 20 and Y as 50. We get 125. Fourth point is 50, 0. I am putting X as 50 and Y as 0. We get 200. So minimum value of Z is 110 at the point 5, 30. So minimum cost is 110 rupees when units of food A is 5 and units of food B to be used is 30. This is our required answer. So this completes the session. Bye and take care.