[Music] welcome back in the last video we learned about mrts marginal rate of technical substitute in this video we are going to talk about cost Min and we will use different approaches to do cost Min so first let us try to understand what are we trying to the idea is very simple let us say your production technology is that you are using n inputs yes I have written here one input and one output I will come to that but let us say you have n input and the cost of I input is wi per unit so if you're using a particular combination how much would be the total cost the total cost is going to the summation of w IX I and I has to be equal to from 1 to n this would be the cost but of course it depends on the different combination that you would as we have already seen that different combination would cost you print the idea is to produce the output while incurring the minimum cost so of course here if you this minimization is very easy pick all XI is equal to Zer but if you pick all XI is equal to Z your output is going to be zero cost is going to be zero output is also going to be so this is not a normal minimization problem you have to figure out if different combination of input such that this particular is entity is minimized and at the same time we can write such that you are able to produce at least Y amount of output so y has to be less than so we can see in two Dimension here we have X1 we have X we can write the same problem minimize with respect to X1 and X2 you will have here W1 X1 W2 X2 and you have to pick a particular combination so that you are able to produce at least Y so all these combination can also produce y by wasting some of the in so of course here we write less than or equal to but when we are minimizing we can say that there is no point wasting input because wasting inut input is wasteful and you are not minimizing the cost so basically this is what you in two Dimension so we have done n Dimension and two Dimension and we are going to talk about all these but let us begin when you have one input and one output technology your production function looks like this and here of course we can give output as well as input this is the key now if you want to produce let us say why not amount of output so to produce why not amount of output you have to use at least X not amount of input more you can take and waste some so of course the best thing to do is technical efficient thing to do is to use not amount of out but here you do not have any bigle room here you do not have any choice because if you use X not amount of input your cost is going to be WX because you using X not amount of input and each unit of input X cost you w and we can in fact we can figure out from here Y is equal to FX KN let us say for example it is Ln X then we can say x has to be equal to e to the power y or in other word for any general function if you have y is equal to FX then X is = to F inverse y so this is going to be your cost total cost to produce is going to be W * by F inverse Y in fact you do not have any choice here at you know in a in a way you have choice that you can use all different combination all this X's Which is higher than x not but that's wasteful you are ultimately at this point and you do not get to choose anything and so we will see the situation would change when we have two inputs and only one output we have this is the problem y isal to X1 X2 and if we take this although we are not going to solve this problem that we will do slightly later we will take much simpler you can use this combination this combination this combination all will give you y or Y not can say Y and you have to pick the combination of X1 and X2 that gives the same output at lowest cost possible so here we are using linear technology and we can do let us say that we have linear technology is this what does it mean that if you want to produce one unit of output you can use either half unit of input one why half unit because it's getting multiplied by two productivity is higher one unit of input gives you two unit of output and as you want only one unit of output you can use one unit of or you can use one unit of input two because increase it by one while keeping it at zero or any one level and is it output is going to increase by one unit or or any combination combination of input one and two such that 2 X1 + X2 is = that is also you can use combination so let us see which one you should be using as you can use one half unit of input one to produce one unit of output what does it mean how much will it cost if you are using using input one it costs W1 by 2 unit or W1 by two cost to produce produce if you use input two then it cost W2 to produce one unit and it doesn't matter if in fact you are producing y not using only input one then it will produce it will cost W1 by 2 multiplied by y not and if you are using only second input then it it will cost W2 i y not of output using only input 2 so which one you should use very clearly if W1 by 2 Y is greater than W2 y y not in fact will get cancelled or we can write here that if W1 is greater than 2 W2 then it is good idea to use only input 2 and if W1 by 2 is less than W2 or we can say that W1 is less than 2 W2 then it is good idea to use only input one and what if there is a third situation also third let me change the color third situation what if W1 by 2 is equal to W2 then it doesn't matter any combination of X1 and X2 such that 2 X1 KN + X2 KN is equal to Y will work equally well so what we are saying the cost is either going to be this one or it is going to be this one why what if the combination is this then clearly that it's equal the costs are equal so it would not matter but I want you to do it mathematically also and check for yourself so we can say the cost to produce or one should say minimum cost to produce why not amount of output output output is going to be let us write here either minimum of color so clear minimum of W1 by 2 y comma W2 Y and let me write here here in this particular case minimum of W1 by 2 y or W2 y or we can write it like this minimum of W1 by 2 comma W2 multipli by and this is the way you will get the minimized let us look at another technology that is leonti technology remember what we had in the leonti technology Y is equal to minimum of let us take X1 comma X and we had figured out that you know if X1 is less than X2 then minimum will be equal to X1 and if X2 is less than X1 minimum is going to be X2 so if we want to produce y not the idea here is very simple that we need X1 is = y and X2 is = y we can check what happens if we take X1 isal to Y -1 or y - K to be General and X2 as y we will get minimum of Y minus K comma y isal to Yus K we are not able to produce y not amount so that we have to keep in mind we have to use X1 isal to Y X2 isal to Y so this problem is very simple in other word if we use graph let us see this is the graph if we have here and this is and this is equal to Y you want to produce at this point and at this point this is X1 isal X2 and at this point we can say X1 is = X2 is equal to Y this X1 axis this is X2 so here the problem is very simple there is only one combination graphically we are able to help solve that you have to be here there is no point being on this side of the arm or this side of the arm you would be wasting some of the inputs and so therefore your cost is going to be to produce y not amount of output the minimum cost required is going to be W1 y plus W2 y y y not here because you need X Y not amount of X1 and Y not amount of X2 so what did we get the cost here is W1 + W2 by y so what we say here we start with this technology which looks like leonti technology and here we get linear cost function in the previous case we start with a linear technology we get a cost function which has the same form as leonti production so keep that in mind it's very very important to understand that let us move to the cob Douglas technology and we will take Y is = to X1 to the power Alpha X2 to the power beta and that's what we have to do we will see how does it go so here the idea is that minimize W1 x1+ W2 X2 of course you have to pick different X1 and X2 to do this minimization such that X1 to the power Alpha X2 to the power beta is equal to Y remember that this is already technically so there are several ways and in fact we will do more than one ways to we will use more than one ways to solve this problem but first we do mathematically and see how cumbersome it is so sometime in economics it's good idea to take help from graph and other techniques and we will do that but let us do it pure using pure mathematics so what it means it's a two-dimensional problem and this equation can help us make it one dimens so we can figure out X2 has to be equal to Y / X1 to the power Alpha whole thing to the power 1 by Beta so we have to now minimize with respect to X1 only W1 X1 + y by X1 to the^ Alpha whole thing to the power 1 by Beta because that's what X2 is multiplied by and it is a simple optimization problem in one dimension you can differentiate with respect to X1 what do you get the first order condition is going to be W1 and this because X1 is in denominator we will get this Alpha by Beta y to the^ 1 by Beta we will have W2 and we will also have X1 we have this minus Alpha Beta + 1 that's what we will get and this is going to be equal to and if you solve this this is what you will get as X sorry as X1 so rather than taking time to do all the step I have already WR you this is the optimal amount of X1 required this is the optimal amount of X2 required now the minimum cost is W1 X1 Star Plus W2 X2 star so you can multiply here by W1 you can multiply here by W2 and you can get this check it for yourself but what I'm am trying to say that it is mathematically intense now we are going to use slightly different technique like in the previous case we have have taken some help from the graph we can do that again and let us look at it how to obtain the cost minimization in case of cop Douglas so the here the idea is let me do it for a general case and then we will solve it for uh cob Douglas function here the idea is that we are doing minimize W1 X1 plus W2 X2 such that we can vary X1 and X2 and such that F we can say Y is equal to F of x1a so what we have this we have already learned that in X1 X2 plane this particular expression can be represented by isoquant different ISO and here we can say that we are not using isoquant map we are interested in producing a particular label and to emphasize that we can say this is why not we can do it in previous cases also we can um it helps us understand so this is what is fin now if we pay attention to W1 x1+ W2 X2 it looks like it's a linear combination so we can think of different combination of X1 and X2 that gives us the fixed cost and so we can draw it how would it look like if we draw it it's going to look like let us say it's going to look like this for a particular value of and if we vary K there will be parallel shifts there will be a parallel shifts here and get an of course K is changing here it is K K1 it is K2 K3 and things like that so this is what it is changing and it is the cost is increasing in this direction K3 is greater than K2 K2 is greater than K so of course this is just isocost these are isocost line at each point the cost remain the same but now we are bringing two things together one is what is is technologically feasible visible and what second thing is what does the market say which is the list combination so of course if you are producing here you are able to get y not because in fact if we don't do it efficiently at all these places we are able to produce why not but what's the point of was some that's not efficient so we should be on isoquant this combination can also give you why not but we it will cost you K3 if you move to let me draw another isoquant there could be an isoquant here so at this point also you get to produce y not but the cost is lower so you can keep on moving but there would be a point here where one particular isoquant is tangent to one sorry one particular isocost line is tangent to this isoquant if you move further in this direction you are not able to produce y not your level of production will go below y not that is not allowed so you can produce at this particular point and that will give you the minimum cost and the idea is very simple you can see this is the I shaped of things as long as you have scope to move in this direction you are not doing economically efficient once you reach to a point where an isocost line is barely touching sorry I one isocost line is barely touching the isoquant then only you are doing the cost minimization so what is the requirement because at that is the tangency crer what is happening the slope of isoquant isoquant at that point point is equal to slope of an isocost now the thing is we have so many isocost line but the good thing is that when we have this isocost line the slope of isocost line does doesn't depend on K we can write it like this that W2 X2 is = k- W1 X1 or X2 is = k / W2 minus W1 by W2 and if you have difficulty with this coordinate geometry in live session we can talk about it I hope you don't have any so the slope of line is given by minus W1 light W so the slope of isocost line is minus W1 and what is the slope of isoquant at that point it is equal to Marginal rate of that's very very important optimization criteria that mrts has to be equal to slope of cost let us also try to understand why does it work what does ISO what what happens on the isoquant the mrts says that you are allowed to exchange one input with other input with this particular race mrts is still so let us say if it allows you to exchange five units of X1 for one unit of X2 and it so happens that five units of X1 cost less than one unit of X2 of course you should do that exchange so you should keep on doing you know this these exchanges till mrts becomes equal to slope it means that the market rate of exchange and input substitution rate they both are equal equal and then it works beautifully so I am going to stop here we will in the next video we will continue with cost minimization we will take example of cob Douglas and learn many more things about cost Min so one can say this is part one of two-part video thank you