[Music] welcome back to week three in this week we will continue talking about forms we will first do a revision of what we did in the last week and then we will build some new Concepts and then again we will look at some of the problems so that we could check whether we have learned the concepts properly on so in this video we are going to learn A New Concept called marginal rate of technical substitute but first we are going to revise what we did in the last week and remember that Sellers and firm could be representing sellers firms are also individual and we said a black box approach to represent a form and uh we had some sort of technology that took inputs and turns out output that's what we have and then we said that if we go in one dimension that is we have only one input and one output then we can say that this is a way to represent the form it takes in input X and gives that output Y and here we have already included technical efficiency remember two of the efficiencies we talked about technical efficiency and economic efficiency so it already includes technical efficiency how should it look like it should look like let us say why here is X why we we have what we saw here that as X increases y inre that makes sense when we have more inputs have higher output but remember we also talked about something called diminishing marginal return what is diminishing marginal return in this context so we can say marginal productivity is basically f-x as we increase X little bit how much is the change in the production if we use the language of maths we can say the change in production is this is mathematically we can say and d y is equal to we can say f-x and so this is the marginal productivity and we typically assume that f-x is greater than that makes sense and then we have diminishing marginal return what does it mean that f-x is decreasing we can say marginal productivity decreases as amount of input increases what it means that f-x is a decreasing function of X what it means that f-x is less than and that's why we have drawn like this but we don't mean that technology is always like this it is also very much possible that in the beginning it goes like this because remember we talked about dimin finishing marginal productivity doesn't hit right at the beginning it can hit later but as X keep on increasing there would be a point Beyond which marginal productivity will start declining so that's what we had done in the last week exactly the same thing we can have two kind of input let us say labor and capital we have a black box that takes labor capital gives us output Y and here we can talk about marginal productivity of L marginal productivity of K so of course what we have to keep K fixed when we are talking about marginal productivity of Labor and so we assume that again we can say y + Dy - y1 y + Dy is the production when we have kept K fixed at K and changed L from L KN to L + DL so we can say f l + DL a KN minus f l comma so we are keeping Capital fixed at K so here also and this we can again represent Dy we can say this is f sorry this has to be L because L subscript denotes partial derivative with respect to L this is and of course we can also at the same time we can change K and this is so when we are keeping K fixed this is equal to zero so it is equal to this only when we are keeping DL fix this is so here also we assume that f is greater greater than zero and we assume FK is greater here also we have diminishing marginal productivity and what does this diminishing marginal productivity means that as if we keep if we keep K = to K KN and keep on increasing increasing keep on increasing using let us say l the marginal productivity productivity starts declining Beyond a point not right from the beginning it's also possible right from the beginning but there exists definitely a point Beyond which marginal productivity of Le starts declining and that we can represent as f l which mathematically we can write it as what we are saying that this is nothing but and you have done it in math and therefore this is how marginal productivity with respect to labor is changing and so we use assume that f l is less than zero and FK K is so that's what what would be its result on isoquant remember what is an isoquant and we can take a little bit of a space here to talk about isoquant isoquant again gives here is L here is K and let us say we want to produce 10 units and so the combination of all K and L that's without wasting we are not wasting anything we include the notion of technological efficiency technical efficiency the combination of K and L that gives us at least 10 output why I say at least 10 output because it is also very much possible that we use this much of L and this much of K to produce 10 unit how can we do that either we can come to this point or we can come to this point or we can come to this point but when we say that we are not including the notion of technical so when we talk about isoquant the notion of technical efficiency is already included this is you know if we move in this Zone we are not able to produce 10 units of out and so isoquant already include the concept of technical effect but now this is what we did in during the last week now we have to move forward now we want to include the notion of economic efficiency also how can we do that so let us see that this is a point this uses let us say for example five unit of capital and two unit of Labor and this we can say is two unit of capital and five unit of Labor so we can say this isoquant is for k l multiplied is equal to 10 for example or this this particular example is of cob Douglas function and this can be written as k l multi you know that Y is equal to f l comma K which is equal to KL or l k equal and we are drawing here for value now this can be used to produce output Unit 10 this can be used to produce output in fact there are infinite number of combination particularly when you allow for fractional label you know half a half a unit of capital what does it mean so rather than unit using the capital for the whole day you are using only for half a day so you I'm not saying you have to cut the machine into two part and then machine will not be usable but there are mathematical way to say that you are using half of the capital because remember technically we should have include for how long we are using uh labor and capital to produce 10 unit so we can always go into fraction so which particular combination you should be using and we will learn about cost minimization later so I will not get into the details of that as of now but the point is that you want to use the combination because let us say one unit of labor cost you w so the cost of using L unit of Labor is going to be WL and if we are using R unit sorry K unit of capital and let us say rental for capital is R so this much would be the cost so of course to achieve economic efficiency remember for economic efficiency you have to minimize the cost the idea is to minimize this particular expression how can you minimize you can of course minimize by using k equal to 0 L is equal to Z and you don't produce anything in that case cost would be zero but that's useless the idea is to minimize cost while producing at least let us say here 10 unit so your FL L comma K has to be greater than or equal to we are going to talk about this cost minimization in detail little later but here I want just to motivate that as a manager of a firm you would like to change the input mix to produce the same output why because sometime it's economically efficient for you that so now we need to figure out that how can we do that so let us look at and we will change you know we can take this let us say that we have we can draw so 8 10 12 these are different isoquant so our production function remains LK but these are drawns for8 then for 10 and then for 12 so let us say you are at k = 4 and L equal to 3 so let us say you are roughly here this is four what happens if you decrease K by 1 unit now you have K = to 3 and L = to 3 so I think it's good idea to keep Market 9 anyway these are qualitative description so of course once you decrease this is where you come because L is fixed now your output is 9 instead of 12 so one thing that we should understand how much is the marginal productivity of capital at this point marginal productivity of capital is three because if you move from three to four output goes up by three units but that we have already seen I just thought it's good idea to revise it now if we want to come back to the same level with this New Capital label we have to increase L and we have to move here and this if K is three because we are moving to isoquant 12 it has to be four so what we are saying that and in this particular at this particular point one unit of capital capital can be exchanged can be exchanged for one unit of labor but that is not the case all the time consider that you are at K is equal to 6 and L is equal to 2 you are at 12 and so if you give up K by one unit you reach here to five your L has to be 12 by 5 and so it means for one unit of capital capital how many units of Labor you need L 12 by 5 - 2 so you just need 2x 5 units of capital me write it here again so that you can see 2x 5 unit so at different place the rate of substitution is different so how does it work let us try to understand let me repeat what we have obtained we saw that at L is = to 3 and K = 4 if we want to reduce K by 1 unit we need to increase L by one unit so capital and labor can be exchanged with one another in ratio of one another thing that we saw that when L is = to 2 and K is equal 6 and if we want to reduce K by 1 unit L has to be increased by 2 by 5 unit we have to be careful about two things that we are talking about exchanging one input or the other so there are two ways one can happen one that we are changing one unit of K with X unit of L or we are changing X sorry we are changing one unit of L or X unit of K so what are we defining one has to be very clear otherwise we will always make a mistake of a ratio so here is the isoquant and let us still District to 12 what we are obtaining is basically this is what we have said that here we are decreasing this by one unit and then this is also one unit so in this case this angle and this angle let us say that this is a triangle that both would be equal but when we use this at let us say here this is 1 and this is 2x 5 so the ratio one ratio is going to be 2x 5 1 or 2x 5 other ratio is going to be 5x 2 so which one we are trying to obtain we have to be clear about so that is one issue Clarity which input is in numerator and thus which other one is in denominator that's one thing second thing that we have to be clear about that we are incorporating change of one unit in a particular input this as we have already been talking about this one is a random number why what is so special about one why not we can change it by the small Delta K unit and then of course if we decrease K by let us say k KN comes down to K KN minus DK and DK indicates it is a very very small number and then of course we have to increase L by something how much is that increase let us say that increase is Delta L and then if we calculate we are trying to obtain this angle and in that case let us say it becomes Delta l so in that case we are obtaining or DL we are obtaining DK by DL so in our earlier discussion we are obtaining this 5x2 ratio then we will get this angle and what would it mean that substituting substituting L for K we are bringing we are increasing L for K that's what we are trying to do so we have to be clear about one can Define in exactly opposite manner also we can one can very clearly write here that substituting Ting k for L and that would be valid but we have to follow a convention so when we say simply that ratio that we are obtaining whether we are talking about this one or this one that Clarity we have to so that's one thing and the second is of course as we were talking about that rather than talking about a difference of one that we have to decrease one particular input by one unit and then see how many more units of other input we have to bring in we can simply figure out the rate at which this exchange has to take place so we can do that mathematically in calculus you must have done we can keep on decreasing Delta K to zero or here we can do Delta L to Z to zero and see what happens okay of course you can can not decrease both at the same time if you decrease both at the same time then your output will decrease so you are decreasing only one of them and as a result you are increasing the other one so that Clarity you have to have and so we can say what we are here we are if we draw this isoquant here is K here is L so basically we are interested in obtaining the slope because that slope gives us the rate of this exchange so we can say here small this is of course approximation this is DK and this is DL so we are interested in obtaining DK by D what we are doing here here we are substituting l for that's what we are and that we have to be very clear so that convention we are going to so once we have let us say once we have this again the same isoquant l k is equal to 12 once we are on this isoquant one thing we have to be certain of that we are not allowed to vary L and K simultaneously if we have decided L let us say l has to be equal to l KN then K has to be equal to 12 by L we cannot simultaneously decide L and K both if we want to be on this isopon so that's another Clarity we have so if we are saying that our production function is this and then we have a particular isoquant in mind so then basically we have only one variable in hand we can pick either l or we can pick K whichever we want to but we are as L is on x axis we can say to maintain output at F not let us say if L is varying this K has to become a function of and then we can go further we can say if we differentiate this expression with respect to L on both side what do we get y not is a fixed number so it doesn't vary with L so this is going to be equal to zero L appears as the first argument and the second argument is also a function of L so to figure out how a change in L changes F first we have to differentiate f with respect to L and this is what we get and then we have to differentiate f with respect to K and then this is one and so we can bring this on the other side this is DF by DL by the way what is this this is partial derivative of f let me write it that is partial derivative of f f with respect to L in our economic term this is nothing but marginal productivity of Labor that's what we have and this is equal to d f by d k which is marginal productivity of capital DK by DL that's what we are interested in in obtaining DK by D L is basically the ratio of two marginal productivity and this DK by DL what is the rate of exchange between K and L that is what we Define as if we use a different color then this is what we Define as marginal rate of technical substitute so the idea here is that we already have technical efficiency we are not playing with technical efficiency we want to do production efficiently but we want to see that to produce the same level of output in technically efficient manner how one input can be exchanged with the other input and that's what is given by marginal rate of technical substitution in short this is called Mr and this is same as here if we had X2 here this is X1 here but this would have been same as derivative of X2 with respect to X1 and how can we do that because once we fix the isoquant X2 becomes a function of X1 particularly when a production function has only two inputs of production let us try to obtain mrts for some of the technology that we have seen so the First Technology we will take is linear technology and what was linear technology if we have just two input the production function is a linear function what we have here is X1 amount of input 1 X2 amount of input 2 A and B are productive productivity parameter and we assume that a is greater than zero and B is greater than further we can obtain both ways using calculus and without Calculus if you increase input one by one unit how much output will increase by a unit of course the assumption is that amount of X2 is not being changed so the marginal productivity of input one we can say mp1 is a and if you use calculus this is a case you use calculus you get get marginal productivity as one it Remains the Same we will see in some of the cases these two techniques may not give the same answer the reason is very simple that in this technique we change the input by one whole unit and in this technique we are changing input marginally at very small amount and seeing the rate of change in output production so that's why they could be different but here they are the same and similarly we can in in calculate that mp2 is B and if we want to see how much is let us say here is here is the production function this is X2 this is X1 a X1 + B X2 is equal to K so clearly if we want to obtain let us say Delta X2 Delta X1 what are we going to get here clearly if we decrease if we decrease if we if we want to obtain Delta X2 by Delta X1 basically this angle what are we going to get so of course let us say here change is one unit it means you are decreasing here now we move in this way we are decreasing X1 by one unit you know if you decrease X1 by one unit output will go down by a unit and to compensate how many more unit of X2 you need you need B by sorry a by B unit you would require because then if you bring a by B unit here marginal productivity is b b will get cancelled and you will get a more you will get a by B and this is what you are going to get so marginal rate of technical substitution is a by B and if you had used the calculus technique mrts is given as minus of mp1 divided by mp2 remember we had MPL and in numerator and mpk in denominator here X1 is on x axis so mp1 and this is again then it will come out minus AB of course here also you can put minus sign as why we are having the minus sign notice that marginal productivity we are assuming is greater than zero so if you are increasing the amount of one input you have to decrease the amount of other input to bring back on the same label of isoon so it doesn't matter which technique you use you will get exactly the same answer in this particular let us take some other technology and that is Bob Douglas techology and what did we see that we had taken you know let us take Y is equal to KL or l k the same thing we use you know and let us say that we are at LK is equal to 12 K is = 4 L isal 3 or K isal 3 l is equal 4 so at this level you see that marginal rate of technical substitution is coming out to be minus one if you want to increase Capital by one unit you have to decrease labor by one unit if you want to increase labor by one unit you have to decrease Capital by one unit so you will get minus1 but let us use calculus technique and see what happens we have have to calculate here MPL that is marginal productivity of Labor so if you use marginal productivity of Labor the formula is this and you are going to leftt with K and if you calculate mpk this is going to be and you are going to left with L and so mrts is minus DK by DL or minus MPL L / mpk and this is coming out to be minus K by L so if you are here it's going to be -4 by3 which is not same as minus what is happening is wrong so the there is nothing wrong this formulation is an approximation what we have here is a curve like this L A and we are calculating this particular angle but when we are using calculus technique then we are calculating tangent and this particular point and they are different this is more accurate this is approxim so this is what I also wanted to emphasize remember the third technology the third technology that we had talked about is Leon Tiff technology and it looked like how it looked like L saved okay I'm not going to obtain marginal rate of technical substitution for you I urge you that on your own you try to obtain marginal rate of technical substitution for leonti technology see what happens you have to be little bit of careful let me give you the hint why you need to be careful if you look at poob Douglas technology the isoquant is smooth if you look at the linear technology isoquant is smooth if you look at the leonti technology it is smooth everywhere here and it is smooth everywhere here but at this point there is a king and that causes a problem so be careful about it you should get mrts should be same every like you will get one value here here one value here and at this point you should not be able to Define it because there is an Abrupt so that is preliminary discussion on mrts I also gave you some example we will come back to these this very important thing and use it in optimization great we will meet in next week