 Hello and welcome to the session. In this session, we discuss the following question which says, a chemical industry produces two compounds C and D. The following table gives the units of ingredients P and Q per kg of compounds C and D as well as minimum requirements of P and Q and costs per kg of C and D. So, this table shows the units of ingredients P and Q presently compounds C and D. It also gives the cost per kg of the compounds C and D and also the minimum requirement of the ingredients in the compounds. And we have to find the quantities of C and D which would minimize the cost. Let's now proceed with the solution. Now as we are supposed to find the quantities of C and D, so for this we suppose let X kg of C and Y kg of D be produced. Now in this table we have as the minimum requirement of the ingredient P in the compounds C and D is 80. So, this means if we have X kg of C is being produced and ingredient P per kg in compound C is 1 unit. So, we have 1 into X that is X plus. Now as Y kg of B compound is being produced and ingredient P per kg in the compound D is 2 units. So, it would be 2Y. So, here we have X plus 2Y is the total ingredient P present in both the compounds C and D and this should be greater than equal to 80 as the minimum requirement of the ingredient P is 80. Now next we consider the ingredients Q. Now ingredient Q per kg present in compounds C is 3 units and in compound D is 1 unit and the minimum requirement of the ingredient Q in both the compounds is 70 units. So, here we have 3X plus Y should be greater than equal to 70. Now cost per kg of compound C is 4 dollars and of compound D is 6 dollars and as X kg of compound C is produced. So, which is the total cost of the X kg of the compound C produced. Now as 6 dollars is the cost of 1 kg of compound D and Y kg of compound D is produced. So, 6Y is the total cost of Y kg of the compound D produced and let this be equal to Z and this is the cost function. Now we have to find the quantities of C and D such that it would minimise the cost. So, we have to minimise Z equal to 4X plus 6Y to the constraints 2Y greater than equal to 80, 3X plus Y greater than equal to 70, X greater than equal to 0 and Y greater than equal to 0. These are the non-negative constraints. So, this is the problem that we are supposed to solve as this is the case of minimising the cost. So, we will use the isocost method for this. For this first of all we will find out the feasible region. So, the graphs of these constraints, before we start drawing a graph of these in equations we have to draw the graphs of the equations X plus 2Y equal to 80, 3X plus Y equal to 70, X equal to 0 and Y equal to 0. This is the line X plus 2Y equal to 80 and this is the line 3X plus Y equal to 70. Now X equal to 0 is the Y axis and Y equal to 0 is the X axis. Now next we will graph these constraints that is these in equations for this. We first consider the in equation X plus 2Y greater than equal to 80 and let's see if the 0.00 satisfies this in equation or not. On putting X equal to 0 and Y equal to 0 we get 0 is greater than equal to 80 which is not true and therefore we say that the 0.00 does not satisfy the in equation X plus 2Y greater than equal to 80 and hence the origin O does not lie in the region of X plus 2Y greater than equal to 80. So, the line X plus 2Y equal to 80 and the region to the right hand side of this line which does not contain the origin represents the region X plus 2Y greater than equal to 80 and this should be above the X axis and to the right of the Y axis because we have the non-negative constraints X greater than equal to 0 and Y greater than equal to 0. Next we consider the in equation 3X plus Y greater than equal to 70. Now putting X equal to 0 and Y equal to 0 in this we have 0 is greater than equal to 70 which is again not true therefore the point 0.00 does not satisfy the in equation 3X plus Y greater than equal to 70 and hence we can say that the origin O does not lie in the region 3X plus Y greater than equal to 70. So, the line 3X plus Y equal to 70 and the region to its right hand side but above the X axis and to the right of the Y axis would be the region that represents the in equation 3X plus Y greater than equal to 70. Also X greater than equal to 0 is above the X axis and Y greater than equal to 0 is to the right of the Y axis and so the region common to all these constraints is this shaded region that is the region A B C this shaded region is the feasible region. In the next step we will assign some constant value to the objective function Z. So, let the objective function Z equal to 4X plus 6Y be given a constant value say 100 then we get a line 4X plus 6Y equal to 100 or you can say 2X plus 3Y is equal to 15. Let this be equation 1 now we will draw a line p1 q1 to represent this equation 1. So, we have drawn this line p1 q1 representing this equation 1. In the same way we assign another constant value to the objective function Z. So, let objective function Z equal to 4X plus 6Y be given a constant value say 300. So, we get a line 4X plus 6Y is equal to 300 that is 2X plus 3Y is equal to 150. Now, let this be equation 2 and we will draw a line p2 q2 to represent this equation 2. This is the line p2 q2 which represents the equation 2 and this line p2 q2 is parallel to the line p1 q1. Now, as this constant value 100 is less than this constant value 300. So, we will move this line p2 q2 from p2 q2 to p1 q1 parallel to itself that is we will move the line p2 q2 in the decreasing direction towards the origin. So, that we obtain a line which is nearest to the origin and has at least one point common to the feasible region. Thus, we obtain a line p3 q3 which is nearest to the origin and passes to the point B of the feasible region. Now, as you can see this point B is the point of intersection of the lines 3X plus Y equal to 70 and X plus 2Y equal to 80. So, equal to 80 plus Y equal to 70 we get the value of X as and Y as 34. The point B has coordinates 12, 34 and so now we can say that we have a line p3 q3 which is to the origin the point B with coordinates 12, 34 feasible region. So, we can say that the optimal solution R X equal to 12 and Y equal to 34 the optimal value of your objective function Z equal to 4X plus 6Y is given as 4 into 12 plus 6 into 34 which is equal to 48 plus written 4 which is equal to 252. So, now the optimal value of the objective function Z is equal to 252 or we can say that Z is minimum at the point B with coordinates 12, 34 and hence we can say that for a minimum cost 12 kgs of 34 kgs of taken. So, this is our final answer. This coming is the session that we have understood the solution of this question.