 Hello and welcome to the session. I am Deepika here. Let's discuss the question which says A formal mix is two brands P and Q of cattle feed Brand P costing rupees 250 per bag contains three units of nutritional element A 2.5 units of element P and two units of element C Brand Q costing rupees 200 per bag contains 1.5 units of nutritional elements A 11.25 units of element B and three units of element C the minimum requirements of nutrients A, B and C are 18 units 45 units and 24 units respectively Determining the number of bags of each brand which should be mixed in order to produce a mixture having a Minimum cost per bag. What is the minimum cost of the mixture per bag? So let's start the solution Let X and Y be the number of bags of brand P and brand Q in the mixture respectively so Obviously X is greater than equal to 0 and Y is greater than equal to 0 Now according to the question of formal mixes two brands P and Q brand P contains three units of nutritional element A and brand Q contains 1.5 units of nutritional element A and The minimum requirement of nutrient A is 18 units So according to the question plus 1.5 Y greater than equal to 18 now this is a constraint on Nutritional element A that is 2X plus Y greater than equal to 12 Again brand P contains 2.5 units of nutritional element B and brand Q contains 11.25 units of nutritional element B and the minimum requirement of the nutrient B is 45 units So again, we have 2.5 X plus 11.25 Y greater than equal to 45 now this is a constraint on nutritional element B that is X plus 4.5 Y greater than equal to 18 or This can be written as 2 X plus 9 Y greater than equal to 36 Now again according to the question brand B contains two units of nutritional element C and brand Q contains three units of Nutritional element C and the minimum requirement of nutrient C is 24 units So now we have 2 X plus 3 Y Greater than equal to 24. So this is a constraint on Nutritional element Now again, we are given the cost of each bag of brand P which is rupees 250 and the cost of each Bag of brand Q which is rupees 200 Now according to the given problem we have total cost in rupees is equal to 250 X plus 205 Let Z is equal to 250 X plus 205 Hence the mathematical formulation the given problem is minimize Z is equal to 250 X plus 200 Y Subject to the constraints Plus Y greater than equal to 12 2 X plus 9 Y greater than equal to 36 2 X Plus 3 Y greater than equal to 24 X greater than equal to 0 and Y greater than equal to 0 Now Z is equal to 250 X plus 200 Y is our objective function Let us take this as number one. Now we have to minimize set Subject to these given constraints. Let us number these constraints as two three four and five 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 2 X plus Y greater than equal to 12 is 2 X plus Y is equal to 12 So we will first draw the line representing the equation 2 X plus Y is equal to 12 clearly the points 0 12 and 6 0 like on the line 2 X plus Y is equal to 12 so the graph of this line can be drawn by plotting points 0 12 and 6 0 and then joining them Let us take a as a point 0 12 and B as a point 6 0 So a be represent the equation of the line 2 X plus Y Is equal to 12 Now this line divides the plane into two half planes We will consider the half plane which will satisfy to So the half plane Which does not contain the origin will satisfy too So we will consider that half plane only Now again the equation of the line corresponding to the inequality 2 X plus 9 Y greater than equal to 36 is 2 X plus 9 Y is equal to 36 Now clearly the points 0 4 and 18 0 lie on the line 2 X plus 9 Y is equal to 36 So we will draw the line representing the equation 2 X plus 9 Y is equal to 36 on the same graph by plotting the points 0 4 and 18 0 and then joining them Now let us take C as a point 0 4 and D as a point 18 0 So CD represents the equation of the line 2 X plus 9 Y is Equal to 36 now CD divides a plane into two half planes We will consider the half plane which will satisfy 3 So we will consider the half plane which does not contain the origin In a similar way We will represent 2 X plus 3 Y greater than equal to 24 graphically by drawing the line 2 X plus 3 Y is equal to 24 Now the points which satisfy this equation are 0 8 and 12 0 So again, we will plot these points on the same graph and then we will join them Now let us take E as a point 0 8 and F as a point 12 0 So E of represents the equation of the line 2 X plus 3 Y is Equal to 24 now again this line divides the plane into two half planes So we will consider the half plane which will satisfy 4 so clearly The half plane which does not contain the origin will satisfy 4 so we will consider that only Again X greater than equal to 0 and 5 greater than equal to 0 implies That the graph lies in the first quadrant only Now the lines a b c g and e f Intercept each other At three points. Let us take this as l This is m and this is m So clearly the coordinates of l r 3 6 the coordinates of m r 9 by 2 3 and the coordinates of n r 9 2 Now here the yellow shaded portion in this graph is the feasible region Satisfying all the given constraints now clearly the feasible region is unpounded with corner points on one side as a l n 90 so here the feasible region is Unbounded with coordinates of the inner points 0 12 3 6 with coordinates 9 2 and D with coordinates 18 0 Now according to the corner point method Minimum value of set will occur at any of these corner points So we will evaluate set at each corner point now at the point a with coordinates 0 12 Z is equal to 250 Into 0 plus 200 into 12 which is equal to 0 plus 2400 and that is equal to 2400 Now at the point L whose coordinates are 3 6 Z is equal to 250 into 3 Plus 200 into 6 which is equal to 750 plus 1200 and that is equal to 1,950 now At the point n whose coordinates are 9 to Z is equal to 250 into 9 plus 200 into 2 which is equal to 2250 plus 400 and that is equal to 2650 now at the point D Whose coordinates are 18 0 Z is equal to 250 into 18 plus 200 into 0 which is equal to 4500 plus 0 and that is equal to 4500 hence minimum value of Z is equal to 1,950 which occurs when x is equal to 3 and y is equal to 6 but here the feasible region Is unbounded So 1,950 may or may not be the minimum value of Z So to decide this issue We will graph the inequality X plus 200 y less than 1,950 To check whether the resulting open half plane has points in common with feasible region or not if it has common points then 1,950 will not be the minimum value of Z. Otherwise 1,950 will be the minimum value of Z now here the green shaded region in this graph excluding the line 250 x plus 200 y is equal to 1,950 or 5x plus 4 y is equal to 39 represents the open half plane 5x plus 4 y less than 39 or Z less than 1,950 Now clearly it has no point common with the feasible region hence Z is not less than 1,950 in the feasible region except at 36 hence 3 bags of brand P and 6 bags of brand Q should be mixed in order to produce a mixture having a minimum cost per pack that is rupees 1,950 So the answer for this question is three bags brand P six bags of brand Q minimum cost of the mixture is rupees 1,950 so this completes our session. I hope the solution is clear to you. Bye and have a nice day