 Hello and welcome to the session, I am Deepika here. Let's discuss the question which says, A dietitian has to develop a special diet using two foods P and Q. Each packet containing 30 gram of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires at least 240 units of calcium, at least 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A? The diet requires at least 240 units of calcium, at least 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet? So let's start the solution. So here we will first formulate the linear programming problem according to the given conditions and then we will solve it graphically. Let X and Y be the number of packets food P and Q respectively. Now according to the question, a dietitian has to develop a special diet using two foods P and Q. Now each packet containing 30 grams of food P contains 12 units of calcium while each packet of the same quantity of food Q contains 3 units of calcium and the diet requires at least 240 units of calcium. So we have 12X plus 3Y greater than equal to 240. Now this is a constraint on calcium. This can be written as 4X plus Y greater than equal to 80. Now again we have each packet of food P contains 4 units of iron and each packet of the same quantity of food Q contains 20 units of iron and the diet requires at least 460 units of iron. So again we have 4X plus 20Y is greater than equal to 460. Now this is a constraint on iron or this can be written as X plus 5Y greater than equal to 115. Also each packet of food P contains 6 units of cholesterol and each packet of food Q contains 4 units of cholesterol and the diet requires at most 300 units of cholesterol. Now this is our first constraint. This is our second constraint. Also we have 6X plus 4Y less than equal to 300. Now this is a constraint on cholesterol or this can be written as 3X plus 2Y less than equal to 150. Now obviously X is greater than equal to zero and Y is greater than equal to zero. These are non-negative constraints. Now according to the given question we have to maximize the amount of vitamin A in the diet. Now each packet of food P contains 6 units of vitamin A and each packet of food Q contains 3 units of vitamin A. So total amount in units of vitamin A is 6X plus 3Y let Z is equal to 6X plus 3Y hence the mathematical formulation of the problem is maximize Z is equal to 6X plus 3Y Now this is our objective function. Let us give this as number 1 subject to the constraints 4X plus Y greater than equal to 80, X plus 5Y greater than equal to 115, 3X plus 2Y less than equal to 150, X greater than equal to zero and Y greater than equal to zero. Now let us number these constraints as 2, 3, 4 and 5. Now we will draw the graph and find the feasible region subject to these given constraints. Now we will first draw the line representing the equation 4X plus Y is equal to 80 corresponding to the inequality 4X plus 5 greater than equal to 80. Now clearly the points 080 and 20 0 lie on the line 4X plus 5 is equal to 80. Therefore the graph of this line can be drawn by plotting points 080 and 20 0 and then joining them. Now let us take A as a point 080 and B as a point 20 0. So A B represents the line 4X plus Y is equal to 80. Now this line divides the plane into two half planes but we will consider the half plane which satisfies two. So we will consider the half plane which does not contain the origin. Again the equation of the line corresponding to the inequality X plus 5Y greater than equal to 115 is X plus 5Y is equal to 115. Again we will draw the line representing the equation X plus 5Y is equal to 115 on the same graph. Now clearly the points 023 and 115 0 satisfies the equation X plus 5Y is equal to 115. So we will plot these points and then join them. Now let us take C as a point 023 and D as a point 115 0. Again C D represents the line X plus 5Y is equal to 115. Now C D divides the plane into two half planes. So we will consider the half plane which satisfies three. So the half plane which does not contain the origin satisfies three. Again the equation of the line corresponding to the inequality 3X plus 2Y less than equal to 150 is 3X plus 2Y is equal to 150. Now clearly the points 075 and 50 0 lie on the line 3X plus 2Y is equal to 150. So we will plot these points on the same graph and then we will join them. Now let us take E as a point 075 and F as a point 50 0. So EF represents the equation of the line 3X plus 2Y is equal to 150. Again EF divides the plane into two half planes. We will consider the half plane which will satisfy 4 that is which will satisfy the inequality 3X plus 2Y less than equal to 150. So the half plane which contains the origin is the graph of 4. Also X greater than equal to 0 and Y greater than equal to 0 implies that the graph flies in the first quadrant only. Now let us take the point of intersection of the line X plus 5Y is equal to 115 and 4X plus Y is equal to 80 as a point M clearly the coordinates of MR 15 20. Again let us take the point of intersection of the line CD and EF as the point N. Now clearly the coordinates of point NR 40 15. Now let us take the point of intersection of AB and EF as the point N. So the coordinates of point L are 272. Hence the shaded region in the graph is the feasible region satisfying the all given constraints. Now here the feasible region is the triangle LM with coordinate of vertices as 272 15 20 and 40 15 respectively. So we will evaluate Z at each corner point. So we have the coordinates of the corner points 272 15 20 40 15 respectively. Now we will evaluate Z at each corner point. So at the point 272 Z is equal to 6 into 2 plus 3 into 72 which is equal to 12 plus 216 and that is again equal to 228. Now at the point 15 20 Z is equal to 6 into 15 plus 3 into 20 which is equal to 90 plus 60 and that is equal to 150. Again at the point 40 15 Z is equal to 6 into 40 plus 3 into 15 and that is equal to 240 plus 45 which is equal to 285. Hence maximum value of Z is equal to 285 which occurs when X is equal to 40 and Y is equal to 15. Hence the amount of vitamin A under the constraints given in the problem will be maximum if 40 packets of food P and 20 packets of food Q are used in the special diet. The maximum amount of vitamin A will be 285 units. Hence the answer for this question is 40 packets of food P and 15 packets of food Q maximum amount of vitamin A is equal to 285 units. So this completes our solution. I hope the solution is clear to you. Bye and have a nice day.