 Hello and welcome to the session, I am Deepika here. Let's discuss a question which says, are dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin B. The vitamin content of 1 kg food is given below. Food X contains 1 unit of vitamin A, 2 units of vitamin B and 3 units of vitamin C and food Y contains 2 units of vitamin A, 2 units of vitamin B and 1 unit of vitamin C. 1 kg of food X costs rupees 16 and 1 kg of food Y costs rupees 20, find the least cost of the mixture which will produce the required content. So, let's start the solution. Let us assume that the mixture contains X kg of food X and Y kg of food Y. So, we have X greater than equal to 0 and Y greater than equal to 0. Now, according to the question, are dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A and food X contains 1 unit of vitamin A per kg and food Y contains 2 units of vitamin A per kg. So, we have X plus 2 Y greater than equal to 10. So, this is a constraint on vitamin A. Again, the dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 12 units of vitamin B and food X contains 2 units per kg of vitamin B and food Y contains 2 units of vitamin B per kg. So, the constraint on vitamin B is 2 X plus 2 Y greater than equal to 12 that is X plus Y greater than equal to 6. Again, the dietician wishes that the mixture should contains at least 8 units of vitamin C and the vitamin C content of 1 kg of food X is 3 units and vitamin C content of 1 kg of food Y is 1 unit. Now, this is a constraint on vitamin A. This is a constraint on vitamin B and we have the constraint on vitamin C as 3 X plus Y greater than equal to 8. Now, the total cost Z of purchasing X kg of food X and Y kg of food Y is Z is equal to 16 X plus 25. Hence, the mathematical formulation of the problem is minimize Z is equal to 16 X plus 25 subject to the constraints X plus 2 Y greater than equal to 10 X plus Y greater than equal to 6, 3 X plus Y greater than equal to 8, X greater than equal to 0 and Y greater than equal to 0. So, our objective function is Z is equal to 16 X plus 25 and we have to minimize Z. Let us take this as number 1. These are the given constraints. Let us take this as number 2, 3, 4 and 5. Now, we will draw the graph and find the feasible region subject to these given constraints. Now, the equation of the line corresponding to the inequality X plus 2 Y greater than equal to 10 is X plus 2 Y is equal to 10. So, we will first draw the line representing the equation X plus 2 Y is equal to 10. Now, clearly the points 0 5 and 10 0 lie on the line X plus 2 Y is equal to 10. Therefore, the graph of this line can be drawn by plotting points 0 5 and 10 0 and then joining them. Let us take A as the point 0 5 and V as the point 10 0. So, AB is the line which represents the equation X plus 2 Y is equal to 10. Now, AB divides the plane into two half planes. Now, the half plane which does not contain the origin is the graph of 2. Again, the equation of the line corresponding to the inequality X plus Y greater than equal to 6 is X plus Y is equal to 6. Clearly, the points 0 6 and 6 0 satisfy the equation X plus Y is equal to 6. So, the graph of this line can be drawn by plotting the points 0 6 and 6 0 and then joining them. Now, let us take C as the point 0 6 and D as the point 6 0. So, C D represents a line X plus Y is equal to 6. Again, the line C D divides the plane into two half planes. We will consider the half plane which satisfies 3. So, the half plane which does not contain the origin is the graph of 3. Now, again the equation of the line corresponding to the inequality 3 X plus 5 greater than equal to 8 is 3 X plus 5 is equal to 8. So, we can draw this line by plotting the point which satisfies the equation 3 X plus Y is equal to 8. Now, clearly the points 0 8 and 8 over 3 0 satisfies the equation 3 X plus 5 is equal to 8. Now, let us take E as the point 0 8 and F as the point 8 over 3 0. So, E F represents the line 3 X plus Y is equal to 8. Now, clearly the half plane which does not contain the origin satisfies 4. So, we will consider that half plane only. Again, X greater than equal to 0 and Y greater than equal to 0 implies that the graph flies in the first quadrant only. Now, we know that the point of intersection of two lines can be determined either by observing the graph or by solving the two equations of the lines. So, we observe that the lines A B C D and E F intersect at 3 points. Let us take these as L M and N. So, the coordinates of L R 1 5 and the coordinates of M R 6 by 5 22 by 5 the coordinates of N R 2 4. Now, here the yellow shaded portion in the graph is the feasible region satisfying all the given constraints. Now, here the feasible region is unbounded with coordinates of the corner points 0 8 L with coordinates 1 5 and with coordinates 2 4 and B with coordinates 10 0. Now, we will evaluate set at these corner points. Now, at the point E with coordinates 0 8 Z is equal to 16 into 0 plus 20 into 8 which is equal to 0 plus 160 and that is equal to 160. Now, at the point L with coordinates 1 5 Z is equal to 16 into 1 plus 20 into 5 and this is equal to 16 plus 100 which is equal to 116. Now, at the point N whose coordinates are 2 4 Z is equal to 16 into 2 plus 20 into 4 and this is equal to 32 plus 80 which is equal to 112. Again, at the point B with coordinates 10 0 Z is equal to 16 into 10 plus 20 into 0 which is equal to 160. Hence, minimum value of Z is equal to 112 which occurs when X is equal to 2, Y is equal to 4. But if the region has been bounded, this smallest value of Z is the minimum value of Z but here we see that the feasible region is unbounded. Therefore, 112 may or may not be the minimum value of Z. Since the feasible region is unbounded, we graph the inequality plus 20 Y less than 112 or 4 X plus 5 Y less than 28. Now, this is a line representing the equation 4 X plus 5 Y is equal to 28. Now, here the green shaded region in this graph excluding the line 4 X plus 5 Y is equal to 28 represents the open half plane 4 X plus 5 Y less than 28. Now, we have to check whether this resulting open half plane has points in common with feasible region or not. If it has common points then 112 will not be the minimum value of Z otherwise 112 will be the minimum value of Z. But here it has no point common with the feasible region. Thus, the minimum value Z 112 which occurs when X is equal to 2 and Y is equal to 4. So, the optimal mixing strategy for the dietician would be to mix 2 kg of food and X and 4 kg of food Y and with this strategy the minimum cost of the mixture will be rupees 112. Hence, the answer for this question is least cost of the mixture is rupees 112 and this is obtained by mixing 2 kg of food X and 4 kg of food Y. So, this completes our session. I hope the solution is clear to you. Bye and have a nice day.