 Let us look at the case where if there is a change in the input prices, how it affects the least cost input combination. So, change in input price if you look at it affects the optimal combination of inputs at different magnitude depending on the nature of the input price change. So, if all input price change in the same proportion, so it bound to happen the optimal combination of inputs has to change, if there is a change in the input price either in the different magnitude or in the same magnitude depends on the nature of the input price change. So, if all input price changes in the same proportion the relative prices of input that is the slope of the budget constant or they may remain unaffected. So, if all input prices changes in the same proportion the relative prices of inputs also if you look at the moves in the same proportion and those are remain unaffected, but when the input prices changes at a different rate in the same direction or the opposite direction or the price of only one input changes while the price of other input remain constant the relative price of input will change. Input price when it change in the different rate in the same direction, different rate in the opposite direction, input price of one changes or the remaining constant the relative price of input will change. This change in the relative input out price change in both in the input combination and the level of output as a result of substitution effect of change in the relative prices of input. So, this change in the relative input prices changes both the input combination and the level of output. So, whenever there is a change in the input prices it affects the input combination and the level of output as a result of substitution effect of change in the relative price. So, a change in the relative price of inputs either in the same or the opposite would imply that some input have become cheaper in relation to the others. So, cost minimization firm attempts to substitute relatively cheaper inputs for more expenses one refers to the substitution effect of relative input price change. Because whenever the price of one input changes, it becomes cheaper with respect to the other inputs. And what the cost minimizing producer or cost minimization firms they do it over here, they generally try to replace the expensive input with respect to the cheaper inputs and this is generally known as the substitution effect of relative input price change. So, this numericals we will just look at later that how the quantity of labour and capital changes, before that we will look at the graphical representation that when there is a change in the input prices, how it affects the least cost input, how it affects the level of output or how it affects the level of input combination. So, initially this is the isocost, K L is the isocost, Q is the output, Q 1 is the isoquant and this is the level of output. Now, price of labour input decreases with the help of that now the quantity of the or if you look at the isocost will change from K L to K L dash. So, once it changes from K L to K L dash, then in this case now what is the new input combination? The new input combination is if you look at this becomes L 3 that is L 1 and this is K 2, this is K 1. So, with the change in the input price now the firm will use more of labour and less of capital that is the reason the combination now earlier it is K 1, L 1, now it is K 3, K 2 and L 3. Now, to get to keep it in the same level what the producer will do the may be at least at the same level of output still want to change the input combination. Now, what they will do? They will try to draw a parallel line which also tangent at a point which is at this level. So, suppose this is your point E, this is your point E 1, this is your point E 2. So, at this case if you look at still it uses a higher level of labour as compared to the previous level, but it is use a also a higher level of capital. So, the movement from this E 2, E 1 is the price effect that is in the form of L 1 to L 3. The movement from may be E 1 to E 2 is budget effect, because the producer is trying to keep the income level, the real income level of the producer at the same level that is the reason we got a compensated budget line which is may be K dash and L dash which compensate or which may be the reduce the real income of the producer in term of change in the input prices and that leads to a different combination that is E 2. So, E 1 to E 2, E 1 the movement from E 2, E 1 is the price effect which is in term of the labour input because use we can say L 1 and L 3 movement from E 1 to E 2 is that is L 2 to L 3 is because of the income effect, because the real income is changing and movement from may be E 2, E 2 is the substitution effect because of the change in the real income. So, if you look at the price effect is the combination of the substitution effect and the income effect. So, if you remember your price effect substitution effect and income effect in case of the consumer theory this is nothing but the counter part of the counter part of the counter part of that in the production theory which talks about the change in the input prices. If there is a change in the input prices generally the producer try to substitute that with a cheaper input as compared to expensive input that is the reason they go on using more of that input. So, in this case also the same thing has happened the producer is once the price of labour has gone down the producer has try to optimize it and use more of the labour as compared to the capital and that leads to the combination of the change in the input combination or also the change in the level of output. Now, next we will see the input combination or may be the input combination how it changes or how to find out numerically when the production function is given price of inputs is given like W and R and if the production function is given and to produce a specific level of output how to optimize the cost of production and what should be the minimum cost of production. For that we will just take a example of a we will take a numerical example like Q is equal to 100 K that is to the power 0.5 and L to the power 0.5 this production function is in the form of a Cub Douglas production function where W is 30 and R is 40. So, W is the price of labour that is 30 rupees R is the price of capital that is 40 rupees and what we need to do here we need to find the quantity of labour and capital that the firm should use in order to minimize the cost of producing of 144 units of output. So, if we look at this the minimization of case like the second case where the unit of output is given we need to minimize the cost and what is the minimum cost we need to find out that then how we will go for this we will use the we will take the help of the Lagrangian multiplier to solve this. So, Q is equal to 100 K 0.5 and L 0.5 and here if you look at then we have W is equal to 30 and R is equal to 40. So, first we will try to find out the composite function. Now, composite function is rupees 30 L plus rupees 40 K plus lambda dash that is Q 0 minus 100 L to the power 0.5 K to the power 0.5. Now, what we need to do we need to find out the first order condition the first order condition we need to find del z by del L that is 30 minus lambda dash 50 L minus 0.5 K to the power 0.5 K 0.5 has to be equal to 0 or 50 lambda dash L minus 0.5 K 0.5 has to be equal to 30. Suppose this is our equation 1. Now, we will need to look at the first order partial derivative or the first order derivative or the partial derivative with respect to the other input. In the first case we have checked it for L. Now, we will check it for K. So, here we need to find out del partial derivative with respect to K that gives us 40 minus lambda dash 50 L 0.5 K minus 0.5 which has to be equal to 0 or 50 lambda dash L 0 to the power 0.5 K to the power minus 0.5 which is equal to 40 let us call it equation 2. Now, to find out the partial derivative with respect to lambda that is Q 0 minus 100 L 0.5 K 0.5 which is equal to 0. So, you can call it 100 L to the power 0.5 K to the power 0.5 which is equal to Q 0 that is equation 3. Now, if you divide equation 1 by equation 2. Suppose equation 1 by equation 2 then this is 30 by 40 equal to lambda dash that is 50 L minus 0.5 K 0.5 then lambda dash 50 L 0.5 K minus 0.5 simplifying this again. So, if you simplify this then this is 3 by 4 equal to K 0.5 K 0.5 L 0.5 and L 0.5 that comes out to be 0.5. So, this is comes to 3 by 4 by K by L. So, K is equal to we can say 3 by 4 L or 0.75 L. So, in order to take the first order partial derivative once we get the partial derivative out of the first order derivative with respect to L then the first order derivative with respect to K and then the first order derivative with respect to lambda and then we solve for the value K in term of L. Now, what we have got from all this calculation that is K is equal to 3 by 4 L. Now, we will see how we can substitute the value of K into value of this into the equation and find out the value of K and L because ultimately what we need to find out we need to find out ultimately to produce 144 units of output what is the minimum cost because the minimum cost the firm has to incur or what should be the minimum cost or on which ISO cost they have to learn. So, if you substitute the value of K into the given production function for 144 units of output that is 144 that is 100 L 0.5 and 0.75 L because K is equal to 0.75 L. So, that comes to 100 L 0.75 which comes to 1, 4, 4, 100 L and 0.75 then it comes to 14486.6 that comes to 16.67. So, 16.67 is the value of L. Now, we to find out K is equal to 3 by 4 L. So, that comes to 12.5. So, 12.5 is the capital and 16.67 is the labour. So, capital and labour we need we got the value of capital and labour. Next we need to find out what is the cost when the capital is labour is 16.67 and capital is 12.5 because ultimately again let me remind it ultimately we need to find out what is the minimum cost of producing this given level of output. So, C is equal to W L plus R K. So, W is 30, L is 16.67 plus R is 40 and K is 12.5. So, that comes to rupees 100 point 1000.50. So, that comes to 100. So, in order to produce 1, 4, 4 output this is the minimum cost. So, this is the given level of output and this is the minimum cost. So, whether it is a minimization case or maximization case how generally we solve it numerically we solve it numerically with the help of this Lagrangian multiplier method where we take into the where we take the constant in the form of a Lagrangian multiplier. We formulate a composite function then we take the first order partial derivative with respect to 0. Simplifying this that gives us the value of capital and labour we put the value of capital and labour in the production function equation. We get the exact value of capital and labour use this in the cost function and that gives us the minimum cost of producing the given level of output. Now, let us see if you remember in the last class we talk about the short run production function that is in term of the law of diminishing return. Next we will see that numerically how we get the value of the three stages the different stages of production and how we find out the value of average product marginal product and at what level generally the diminishing marginal return set in and what level labour is the average product labour is the highest. So, the firm produces the output according to the production function that is Q is equal to 100 k L minus L Q. So, the production function is 100 k L minus L to the power Q capital is fixed at 10 because this is a short run production function. Now, what we need to do we need to find out the value of average product we need to find out the value of marginal product and then maybe we can find out what is the different stages or what is the different level of output or level of labour where the firm or where the producer is achieving the different level of output. So, first we will find out the marginal product of labour. Now, marginal product of labour is del Q by del L. So, that comes to 20 k L minus 3 L square. So, that comes to 200 L minus 3 L square. Now, so this is the marginal product of labour. Now, we will find out the average product of labour. Average product of labour is Q by L. So, that comes to 10 k L minus L square which is equal to since k is equal to 10. So, this is 100 L minus L square. So, average product for average product is 100 L minus L square. Marginal product for labour is 200 L minus 3 L square. Now, second one is we need to find out when MPL is maximum. MPL is maximum where the first order partial derivative with respect to L is equal to 0. So, that comes to 200 minus 6 L which is equal to 0. L is equal to 33.33. Now, what is the significance of this level of labour? The significance of this level of labour is that at this level since marginal product of labour is maximum beyond this generally the law of diminishing returns at in. If you remember your three stages of production function like how the total product curve. So, initially it is convex then it is convex and then it is decreasing. So, this corresponding to this or marginal product of labour is maximum because after this the total product of labour is increasing at a decreasing rate and beyond this MPL is maximum and beyond this if you look at then the total product is increasing at the decreasing rate and marginal product is decreasing. So, we can say corresponding to this point MPL is maximum and this is the point where the law of diminishing return set in. And the second point we look at is when MPL is equal to 0. MPL is equal to 0 when this 200 L minus 3 L square is equal to 0. So, in this case maybe L is equal to if you find out this comes to 66.66 and where this value of L comes the value of L comes at this point because corresponding to this MPL is 0 and TPL is maximum. So, if you remember from till this point this is your point beyond which there is the law of diminishing return set in. Then when a average product of labour is highest average productive labour is highest when DAPL with respect to DL is 0 that comes to L is equal to 50. So, this comes somehow here the average product of labour is maximum when it intersecting the marginal product of labour. So, corresponding to this L is equal to 50. So, now we can say that from 0 to 50 the first stage of production from 50 to 66 unit of labour second stage of production and beyond 66 unit of labour we have third stage of production. And how we have identified this first stage, second stage and third stage? First stage ends till the point the marginal product is equal to the average product and average product is maximum at that point where marginal product is equal to the average product. Where the second stage ends? Second stage ends when the total product level is maximum or the marginal product is 0. So, that achieve at the unit of labour unit at 66 beyond 66 we have the third stage of production. So, depends upon the labour unit we can find out at which level generally the stages of production are decided whether at the first stage, whether at the second stage and third stage. So, first stage is up to a point where the marginal product of labour is equal to the average product of labour. Second stage is the point where total product of labour is maximum marginal product of labour is 0 and third point is beyond this. So, once we find out the value of labour where marginal product of labour is equal to 0 that is the definition of the second stage. Once we find out the maximum value of APL that gives us the may be the beginning of the second stage and when we find the marginal product of labour is maximum that gives us the point beyond which the law of diminishing return setting. So, in the next session we will talk about the cost of production different types of cost, how the cost function is formulated and what is the logic of different shape of the cost function in the short term and in the long run. So, these are the session references generally these are the references that is being used for the preparation of this particular session.