 Hello, and welcome to the session I am Deepika here. Let's discuss a question which says, Reshma wishes to mix 2 types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin P. Food P costs Rs 60 per kg and food Q costs Rs 80 per kg. Food P contains 3 units per kg of vitamin A and 5 units per kg of vitamin P while food Q contains 4 units per kg of vitamin A and 2 units per kg of vitamin P determine the minimum cost of the mixture. So, let's start the solution. Let the mixture contains x kg of food P, y kg of food Q. So, obviously we have x greater than equal to 0 and y greater than equal to 0. Now according to the given question, Reshma wishes to mix 2 types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and food P contains 3 units per kg of vitamin A while food Q contains 4 units per kg of vitamin A. So, according to the given question, we have 3x plus 4y is greater than equal to 8. Now, this is a constraint on vitamin A. Again, Reshma wishes that the mixture should contain at least 11 units of vitamin B. Now, food P contains 5 units per kg of vitamin B while food Q contains 2 units per kg of vitamin B. So, we have 5x plus 2y greater than equal to 11. Now, this is a constraint on vitamin B and we are given food P costs rupees 60 per kg and food Q costs rupees 80 per kg. So, we have the total cost in rupees is equal to 60x plus 80y because we have assumed that let the mixture contains x kg of food P and y kg of food Q. Let z is equal to 60x plus 80y. So, the mathematical formulation of the given problem is minimize z is equal to 60x plus 80y subject to the constraints plus 4y greater than equal to 8, 5x plus 2y greater than equal to 11, x greater than equal to 0 and y greater than equal to 0. Now, z is equal to 60x plus 80y is our objective function. We have to minimize z. Let us give this as number 1. Now, this is a constraint on vitamin A. Let us give this as number 2. This is a constraint on vitamin B. Let us give this as number 3 and these are non-negative constraints. Let us give this as number 4. Now, we will draw the graph and find the feasible region subject to these given constraints. Now, the equation corresponding to the inequality 3x plus 4y greater than equal to 8 is 3x plus 4y is equal to 8. So, we will first draw the line representing the equation 3x plus 4y is equal to 8. Now, clearly the points 0, 2 and 8 over 3, 0 lie on the line 3x plus 4y is equal to 8. Therefore, the graph of this equation can be drawn by plotting points 0, 2 and 8 over 3, 0 and then joining them. Now, let us take A as a point, 0, 2 and B as a point, 8 over 3, 0. So, AB represents the equation 3x plus 4y is equal to A line AB divides the plane into 2 half planes. So, we will consider the half plane which will satisfy to clearly the origin does not satisfy this inequality. So, the half plane which does not contain the origin is the graph of 2. Again, the equation of the line corresponding to the inequality 5x plus 2y greater than equal to 11 is 5x plus 2y is equal to 11. So, we will draw the line representing the equation 5x plus 2y is equal to 11 on the same graph. Now, clearly the points 0, 11 over 2 and 11 over 5, 0 satisfy the equation 5x plus 2y is equal to 11. So, we will plot these points on the same graph and then we will join them. Now, let us take C as a point, 0, 11 over 2 and D as a point, 11 over 5, 0. So, the line CD represents the equation of the line 5x plus 2y is equal to 11. Again, line CD divides the plane into 2 half planes. So, we will consider the half plane which will satisfy 3. Now, clearly the origin does not satisfy this inequality. So, the half plane which does not contain the origin is the graph of 3. 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, here the yellow shaded portion in this graph is the feasible region satisfying all the given constraints. Now, the lines AB and CD intersect at a point. Let us take this point as a point E. Now, we can find the coordinates of this point either by inspection or by solving the two equations of the lines intersecting at this point. So, here we observe that the coordinates of ER 2, 1 by 2. So, here the feasible region is unbounded. The coordinates of the corner points B are 0, 11 by 2, 2, 1 by 2 and 8 over 3, 0. According to the corner point method, minimum value of Z will occur at any of these points. Now, we have Z is equal to 60x plus 80y. So, we will evaluate Z at each corner point. Now, at the point 0, 11 by 2, Z is equal to 60 into 0 plus 80 into 11 over 2 and this is equal to 0 plus 40 and this is again equal to 440. Again, 2 1 over 2, Z is equal to 60 into 2 plus 80 into 1 over 2. So, this is equal to 120 plus 40 and this is again equal to 160. Again, at the point 8 over 3, 0, Z is equal to 60 into 8 over 3 plus 80 into 0. So, this is equal to 160 plus 0 which is equal to 160. Now, here we observe that the minimum value of Z is 160 since the feasible region is unbounded. Therefore, 160 may or may not be the minimum value of Z. So, to decide this issue, we will graph the inequality 60x plus 80y less than 160 that is since the feasible region is unbounded. So, we will graph the inequality Tx plus 80y less than 160 to check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then 160 will not be the minimum value of Z. Otherwise, 160 will be the minimum value of Z. Again, the equation of the line corresponding to this inequality is 60x plus 80y is equal to 160. Now, clearly the points 0, 2 and 2, 1 by 2 satisfy the equation 60x plus 80y is equal to 160. So, here the green shaded region in this graph represents the inequality 60x plus 85 less than 160. Clearly, it has no point common with the feasible region. So, the minimum value of Z is 160. Hence, the minimum value of Z is equal to 160 occurs 2 points that is at the point 1 by 2, 8 over 3, 0. Now, we know that if 2 corner points produce the same maximum or minimum value of the objective function, then every point on the line segment joining these points will also give the same maximum or minimum value. Hence, the minimum value of Z occur at all points of the line segment joining the points 2, 1 by 2 and 8 over 3, 0. Hence, the answer for this question is minimum cost equal to rupees 160 at all points lying on segment joining 8 over 3, 0 and 2, 1 by 2. So, this completes our session. I hope the solution is clear to you. Bye and have a nice day.