 So, we will continue our discussion on relationship between different economic variables given a quantification or through different methods graph or the mathematical equation. So, if you remember in the last class we have started discussion about the derivative of various function, how to solve the various function then we introduce the optimization technique where we did two type of optimization one is the maximization of revenue or maximization of profit and second one is the maximization of the revenue or maximization of the profit and second one is the minimization of the cost. So, whenever we are doing this optimization technique either it is a maximization or it is a minimization problem we did not consider the case of a constraint and we just optimize the maximization of a profit function or we just optimize the minimization of a cost function. So, today we will discuss the optimization technique with a constraint either in the form of the income or in the form of the cost when it comes to rec cost and when it comes to the revenue either it is maximization or it is a minimization of the cost. So, in case of constraint optimization this is a technique used for achieving a target under constraint situation or condition is called constraint optimization. So, maybe the motivation for optimization is remain same achieving a target either to maximize the profit or to minimize the cost, but here the difference is that there is a constraint along with the objective function and how to do this constraint optimization we generally discussed two type of technology we will talk about the substitution technique and later on we will take the Lagrangian multiplier method. So, taking the substitution technique it can be applied to the problem of profit maximization or it can be for the cost minimization. For a profit maximization one of the variable express in term of the other variables and solve the constraint equation for obtaining the value of one variable. Suppose, there are two variable X and Y. So, the best way for solving it through the substitution technique is represent one variable with the other variable and then you solve for that variable and finally, you substitute the value of one variable in term of what you have solved to the other variable and here the values obtained is substituted in the objective function which is maximized or solve for obtaining the value of the other variables. So, whatever the value is obtained by substituted it will be again substituted back in the objective function which is maximized and solve for obtaining the value for the other variable. So, we will see how we use this substitution technique in case of a profit optimization problem and in case of a cost minimization. And how this is different for cost minimization may be the method again remain same the constraint equation is express in term of any one of these two goods of the variables and the equation is obtained from step one is substituted in the objective function. So, whether it is a cost function, whether it is a profit function, the basic rule for the substitution technique is that we express one variable in term of the other variables, we get the value of one variable and finally, again substitute back to the objective function. So, we will just take an example how generally we do the constant optimization along with the constant with the objective function, whether the objective function is a profit maximization or whether the objective function is a cost minimization. So, we will take a case of profit maximization first and in case of profit maximization we will maximize the profit. So, here profit is equal to 100 x minus 2 x square minus x y plus 180 y minus 4 y square. This is the profit function and the profit maximization, here the optimization problem is the profit maximization. Since, we are saying that this is the case of a constant optimization, there is also a constant attached to this and the constant is in the form of x plus y is equal to 30. So, now what is the optimization problem? The optimization problem is maximization of profit function with respect to the constant that is x plus y which is equal to 30. Now, how we will do this? The first step is we will express x in term of y or we can express y in term of x and after the getting the value of x or y, again we will substitute this value of x and y in the profit function. So, now we will see let us substitute the value of x and y before it converting into the another term or may be the profit maximization problem. So, suppose x plus y is equal to 30, so this can be rewritten as x is equal to 30 minus y. So, in this case x we are representing in term of y or y can be 30 minus x. So, substituting the value of x and y in the profit equation what we can get? Pi is equal to 100, it was 100 x, so x we are representing in term of y, so this is 30 minus y plus 2, 30 minus y square because it was 2 x square minus 30 minus y y because it was x y plus 180 y minus 4 y square. So, if you look at the profit function now all the terms in term of y, there is no x over here in the case of the profit function. Now, again if you will simplify this, then this comes to 1200 plus 170 y minus 5 y square. So, what we did the first step is this where we represent x in term of y. Now, substituting the value of x in the form of y in the profit equation which gives the profit equation which is equal to 1200 plus 170 y plus minus 5 y square. Now, to find out the value of y what we have to do? We have to take the derivative of pi with respect to y and which we need to set equal to 0. So, if you are taking this then this comes as the first order derivative because for any maximization minimization rule in order to get the value always the first order derivative has to be equal to 0. So, del pi by del y is equal to 0 which is like 1200 plus 170 y minus 5 y square which is equal to 0. Now, solving this will give you 170 minus 10 y which is equal to 0 or may be minus 10 y is equal to minus 170 and y is equal to 17. Now, what is our x? x is equal to 30 minus y. So, this is equal to 30 minus 17 which is equal to 13. So, we get a value y which is equal to 17, we get a value of x which is equal to 13. Now, putting the value of y and x in our profit equation we get profit which is equal to 2800. So, here how we maximize the profit with respect to a cost constant or with respect to a value of x and y? The first step is always to represent one variable in term of the other variable. So, in this case what we did? We represented x in term of y and after represent the value of one variable in term of the other variables then we put the value in the objective function. So, if you remember in the previous slide what I am showing that we represent the profit function only in term of variable y. Then after getting the profit function we took the first order derivative equal to 0 in order to get the value of y and through that we got the value of y which is equal to 17 and from there we got the value of x which is equal to 13. Now, putting the value of x and y in the original profit function we get a profit which is equal to 2800. So, by substitution technique following the two step we got the profit, we got the value of x and we got the value of y. Now, we will see through substitution technique how we can do a cost minimization problem. So, here what is the optimization problem? The optimization problem is to minimization of the total cost. Now what is total cost? Let us take total cost is equal to 2 x square minus x y plus 3 y square. Now the form here what is the constant? The form has to get a 36 units of x and y as the combined order. Now what is the optimum combination? Optimum combination is to what is the what should be the minimum cost to produce this 36 unit of x and y. So, in this case what should be the constant? The constant is again if you look at x plus y is equal to 36. So, optimization problem is to minimize the total cost with respect to or may be subject to x plus y is equal to 36. Now following the substitution technique what is the first step? The first stage we have to represent one variable in term of the other variable. So, x is equal to 36 minus y because if you remember the first step substitution technique is always representing one variable in term of the other variable. So, here x is equal to 36 minus y. Now putting the value of x in the cost equation 2 x square. So, this is 2 36 minus y square x and y. So, this is 36 minus y y plus 3 y square. So, this is 3 y square. So, if you again simplify this this comes to 2592 minus 180 y plus 6 y square. So, after putting the value of x in the cost function in term of y we get a total cost function which is equal to 2592 180 y plus 6 y square. Now, in order to get the value of y and in order to get the optimum combination or the optimum cost what we have to do? We have to take the first step derivative of total cost function with respect to y and we have to set it equal to 0 in order to get the value of y. So, now we have to take a derivative the first order derivative of 2592 minus 180 y plus 6 y square and this has to be equal to 0. So, if you do this then we get the value 180 plus 12 y which is equal to 0 which you further simplify then it is minus 12 y is equal to minus 180 and y is equal to 15. And if it is y is equal to 15 then x is equal to 36 minus y which is equal to 21. So, y is equal to 15 x is equal to 21. Now, this is the optimum combination the firm should produce 15 unit of y and 21 unit of x and this is the optimum combination for the firm. Now, what is the next best task for us? The next best task for us is to whether producing this combination the firm is incurring the minimum cost of production or not what should be the minimum cost to produce this combination. So, for that what we need to do? We need to put the value of y we need to put the value of x in the cost equation and we need to find out the minimum cost. So, what was your cost equation? The cost equation is 2 x square minus x y plus 3 y square. So, putting the value of x is equal to 21 and y is equal to 15 this comes to 882 minus 315 plus 675 which is equal to 1 to 4. So, this is the minimum cost what the firm incurs in order to produce 15 unit of y and 21 unit of x. So, what is the optimization problem here? The optimization problem here is to minimize the cost with a constraint that at any cost the firm has to produce 36 unit of both the goods that is x and y. So, this is the optimum combination for the firm and this is the minimum cost to produce the optimum combination of the firm. Next we will see the second method for this constant optimization and that is the Lagrangian multiplier method. So, apart from substitution technique the most popular or may be the most commonly used technique to do a constant optimization is always a Lagrangian multiplier method. So, what is Lagrangian multiplier method? It is again one up kind of method to solve the constant optimization and it involves combining of both the objective function and the constant equation and solving by using the partial derivative methods. Basically, it takes the partial derivative with respect to both the variables and then he gets the value of x and y and you know by getting the value of x and y it maximizes or the profit or minimizes the cost. So, we will see how it works for the Lagrangian methods. Let us take a profit maximization case. Suppose, the profit equation is 100 x minus 2 x square minus x y plus 1, 1, 80 y minus 4 y square. Again subject to a constant optimization x plus y is equal to 30. The same profit equation what we took for the substitution technique and the same constant what we took for by using the substitution technique method. So, x plus y is 30 that is constant and profit is what we take for the substitution method. Now, how it is different from the other method? In case of other method we are substituting the value of x and x for y or y for x. Here we will not do that, but rather we will use partial derivative method to solve this profit maximization problem. In this case what we do? So, x plus y is equal to 30. So, we will find another variable that here that is x plus y minus 30 is equal to 0 and lambda x plus y minus 30. So, now, we will reframe the objective function using the adding a Lagrangian multiplier over here. And what is the Lagrangian multiplier here? That is lambda x plus y minus 30, this is the another term what we are getting here. So, what is our new profit function? New profit function is 100 x minus 2 x square minus x y plus 1, 80 y minus 4 y square minus 4 y square. This is our original profit function. Along with that we add a Lagrangian multiplier that is lambda x plus y minus 30. So, if you look at now the constant also we have added in the objective function. So, this is our Lagrangian function. Lagrangian function comes from the constant and what we add in the objective function in order to maximize the profit. Now, this is the profit function now. Now, we have to find out the value of unknowns over here. What are the unknowns over here? The unknown is x, the unknown is y and the unknown is lambda. So, we need to solve for the value of x, we need to solve for the value of y, we need to solve for the value of lambda. Now, how we will do that? We will take a partial derivative of the profit function with respect to x, we will take a partial derivative with respect to profit with respect to y and we will take a partial derivative with respect to lambda, what is the Lagrangian function or which one is the Lagrangian multiplier. So, we will take the first one that is L may be first derivative of this with respect to x. So, this we will get as 100 minus 4 x minus y plus lambda which is equal to 0 and let us call it the equation 1. Similarly, for the second one we have to take the derivative with respect to y. So, this comes as lambda, sorry this as x plus 180 minus 8 y plus lambda which is equal to 0 and this is equation 2. The third unknown is with respect to lambda. So, this is del L pi with respect to lambda that gives you x plus y minus 30 and this is equation 3. Now, if you make a summation and if you can make it 2 equation then it comes to 100 minus 4 x minus y plus lambda is equal to 0 and if you add the second 2 equation this is 180 minus x minus 8 y plus lambda is equal to 0. So, if you do a subtraction from 1 to 2 over here then you would get minus 80 minus 3 x plus 7 y plus 0 which is equal to 0. So, if you look at what we did over here we basically in order to get the value of x and y we got a got 2 equation may be 2 joint equation in order to get the value of the unknown. So, in the first case this is 100 minus 4 x minus y plus lambda and second case it is 180 minus x minus 8 y plus lambda which is equal to 0. If you subtract the second one from the first one when we get minus 80 minus 3 x plus 7 y plus 0 which is equal to 0. Now, if you look at the equation 3 x plus y minus 30 this is our equation 3 will multiply 3 with this equation. So, this is x plus y minus 30 which is equal to 0. So, if you multiply 3 in equation 3 this comes to 3 x plus 3 y minus 90 equal to 0 and what was your previous equation when we did for this this is minus 3 x plus 7 y and minus 80 and if you take this again this comes to minus 3 x plus 7 y minus 80 which is equal to 0. From this 2 equation if you sum it then it comes to 10 y minus 170 minus 0. So, we get x plus y which is equal to 0 and y is equal to 17. So, we got the first unknown value that is y is equal to 17. Now, we can get the value of x from here because x plus y is equal to 13. So, from that we can get the value x which is equal to 13 and now we can get a value from the. So, this is our first unknown this is our second unknown our third unknown is lambda. So, from the value of x and y we can get the value of lambda and lambda will come to minus 31. So, we have the we know the value of all these 3 unknown once we put in the value in the profit equation we get the profit and we can see whether the profit is maximum or not. Similarly, using this Lagrangian method we can also solve a cost minimization problem where the optimization problem is to minimize the cost. So, let us look at to the cost minimization problem. Now, the cost function over here is 100 x square plus 150 y square and this is subject to x plus y which is equal to 500. Now, Lagrangian method what is the first step? The first step is to get the Lagrangian function from the constant equation then again form a cost function adding the Lagrangian multiplier or the Lagrangian function. So, in this case the Lagrangian function is lambda 500 minus x minus y. So, this is our Lagrangian function taking this what is our Lagrangian cost function that is 100 x square plus 150 y square plus lambda 500 minus x minus y. So, if you look at now how we have again 3 unknown that is x y and lambda in order to find that what we will do we will follow the same maybe formula what we do did for the profit maximization. We will find out del L c with respect to x we will find out del L c with respect to y and we will find out the del L c with respect to lambda. And after getting the equation again we can get the value of x as 1.5 y. So, that comes to y is equal to 200 and x is equal to 300 and from there we can get the value of the cost and we can get the value of the Lagrangian multiplier. So, what is the essential difference between the substitution technique and the Lagrangian technique? We use both the methods to solve the constant optimization problem and what is a constant optimization problem a constant optimization problem is 1, where we maximize the profit function or the minimize the cost function with a constant that is in the form of the other variable. So, in case of substitution technique what we do we substitute the we represent the value of one variable in term of the other variables and then we substitute that value in term of the other variables in the objective function whether it is a profit function or whether it is a cost function. Then we solve for it and we get when we get one value we represent that in term of others and get the final value of the other variable also. And in case of Lagrangian multiplier method we form a Lagrangian function on the basis of a constant add that to our objective function and solve the objective function through the partial derivative method in order to know the value of unknowns and what are the unknowns over here unknowns always in term of the variable 2 variable those are in the objective function here typically in the case of the x and y. So, we started our discussion for this typical topic we started our discussion with the relationship between different variables whether it is linear, collinear, maybe non-linear or curvilinear. Then we discuss different function attached to different kind of function linear, non-linear and the curvilinear. Then we show then I think we discuss the method of how to solve the different functions and then we talked about the optimization problem where we maximize the profit and minimize the cost and today's class we have talked about the optimization with a constant using both the method that is Lagrangian method and the substitution technique.