 method of analysis and the interpretation of the Lagrange multiplier. Lagrange multiplier that was basically devised to solve the solutions in economics and when we have to decide our objective function respective to the constraint and through the procedure of the optimization so if we look at some things in the previous module in which we see what are the necessary conditions for the optimal vector what are the equations for that so when we see that if we have an objective function and a constraint then we have a third equation that comes with the constraint with lambda and now we have three unknown equations first we have an objective function now we have an objective function that value that we have calculated which will satisfy the objective function with respect to the constraint and the second will be this and the third will be the value of lambda which will help us to solve the objective function in an optimal way and now the lambda here is our with constraint as an exogenous parameter it is not from within the system and that is why when we solve the unknown function of parameter B now when we want to solve it then through the envelope theorem and implicit function we directly calculate their values so the optimized value is the value of x that will be called the optimized value of the objective function i.e. there will be many functions of x but the best solution of the value of x is the same as the value of x2 and the same as the value of lambda so the values that satisfy the constraint and help us to sort out the optimal conditions that condition is called the optimized value of the objective function so the value of the objective function which will be at the optimized point or at the optimal point we show it through steric which we have v steric which is the function of x1 steric and x2 steric and similarly it is possible that this is lambda so now if we say this v steric it depends upon the values and those values which we directly through implicit function or envelope theorem we calculate now if we take the Lagrange function as an example in a simple form we can solve it when Lagrange is set equal to 0 and after that we calculate its derivatives with respect to first explanatory variable then second and then with respect to lambda if we want to solve an example of this then we can say that because in microeconomics we have examples of either consumption or production so here if I want to take an example in which we say that the production quantity that is equal to 20x plus 30l this is our production function for which we say that the constraint that we have is that the wage rate or multiply by labour plus capital or similarly and that they both should be equal to our budget line or cost so if we set this as 0 then we will say that wl plus rk minus c is equal to 0 so this can be our function and here if we call it as Lagrange so the Lagrange will always be 20x plus 30l and here we can see that it will be equal to minus lambda into wl plus rk minus c and we will take the first derivative with respect to if I say here so if we want to multiply it by xt so we can keep it as 20k so if we take it with respect to k is the first derivative then that will be equal to 20 and when I will take the derivative with respect to labour small l that will be 30 and when it will be equal to change in lambda and here now I have part of this 20 but now with respect to k when we will come here that multiplied by the lambda that will be equal to minus lambda k and this will be equal to minus lambda w and with respect to lambda this whole function it will come here and equating these three equations equal to 0 and now even if single equation if I will bring the lambda on one side that will be equal to that 20 is equal to lambda k or other form that lambda is equal to 20 by k and here again we can say that 30 minus lambda w will give us equal to lambda w and then again lambda will be equal to 30 by w and this equation will give us w l plus r k minus c is equal to 0 so from here either we can take the value of k that will give us the 20 by lambda or like this we can equate this value in this equation and we can solve so basically the lambda is the Lagrange multiplier that gives us the rate of change in the dependent variable with respect to our that explanatory variable with respect to which we are going to solve the objective function so that Lagrange multiplier is basically the derivative of the optimal value with respect to our that parameter value b and or in other words we can say it is that rate at which the changes in the constraint parameter b occurs and that rate provides us the solution to our all the objective function that we want to solve with respect to our constraint and so according to this Lagrange multiplier theory we can calculate the effect through this Lagrange on the solution of the value value and this are the value solution gives us through two points means optimized values of x1 and optimized values of x2 so this Lagrange multiplier will give us the rate of change and the solution to our economic theoretical problems thank you