Managerial Economics Prof. Trupti Mishra S.J.M. School of Management Indian Institute of Technology, Bombay
Lecture – 10
Now, next we will move into the optimization technique. Now what is optimization technique, if you know till now we have talked about. The relationship between economic variables, and we have understood that how to find out the relationship between two variables; that is through the slope or through the calculus method. Next we will come to the optimization technique, and optimization technique is what. We optimize in such a manner that, we are achieving the desire result or the desire relationship between the economic variable.
So, basically it is a technique of managerial decision making, maximizing or minimizing function, generally this optimization technique is used either for maximizing or for minimizing the function, and it is a technique of finding the value of independent variable which maximizes or minimizes the value of the dependent variable. So, basically we need to maximize the value of independent variable or minimize the value of dependent variable in order to, understand that how particularly those two variables are related. So, generally what are the cases where these optimization techniques are being used; like sometimes some firm may be interested in finding the level of output that maximizes their total revenue. So, it is basically finding the maximum level of output, which maximize their total revenue or the level of output maximize their total revenue. Some firms facing a constant price may want to find the level of output that would minimize the average cost.
So, may be a case, where the firm facing a constant price, or may they are finding a level of output which will minimize their average cost. And if you look most of the firm they are always interested to find out, what should be the level of output which maximizes their profit. So, if you look at this is the basic objective, basic aim of any firm, in order to understand the level of output which maximizes their profit. So, we will see how this optimization technique or what are the thumb rules, or what are the different approach or different methods to use this optimization technique in order to, for solving this managerial decision problem. So, we will take a function here, basically what we maximize; either we maximize the total revenue, or we minimize the cost, because the basic objective is to maximize the profit, and maximization of profit can take place; either by maximization of the total revenue or the minimization of the total cost, because profit is one, it is just the difference between the total revenue and total cost.
Now, what is total revenue, total revenue if you know, then total revenue this is p Q, p is the price and Q is the quantity demanded. Suppose we take that p is equal to 500 minus 5 Q. Now what is total revenue, total revenue is 500 minus 5 Q multiplied by Q, so that comes to 500 Q minus 5 Q square. So, total revenue is p Q, if the value of p is 500 and total revenue is 500, 5 Q multiplied by Q, that comes to 500 Q 5 Q square. Now, what is the role of optimization technique here or what is the role of, or how we can use this optimization technique over here, in order to maximize this total revenue. So, here the optimization problem is, maximization of total revenue; total revenue is p Q. Now, when this total revenue is maximum, total revenue is maximum, when marginal revenue is equal to 0.
So, the optimization problem here is to maximize the total revenue, and when total revenue is maximum, total revenue is maximum when marginal revenue is equal to 0. Now, let us find out marginal revenue. So, from this total revenue function, if you take the first order derivative, then we get the marginal revenue function. So, first order derivative the total revenue function with respect to Q; that will give us the marginal revenue function. So, if you take this, then this is 500 Q minus 5 Q square. We have to take the del of this with respect to q, so that comes to 500 minus 10 Q. Now, what is the thumb rule, the thumb rule is when marginal revenue is equal to 0, total revenue is maximum. So, 500 minus 10 Q has to be 0, if the total revenue has to be maximum. Now, let us see what is the value of Q, when we set marginal revenue is equal to 0, and why we are setting marginal revenue is equal to 0, because the basic economic principle says that, if the total revenue is maximum, then marginal revenue is equal to 0.
So, what is our marginal revenue, marginal revenue is 500 minus 10 Q, which has to be equal to 0, so this is our marginal revenue. Now, if you solve this then it comes to Q is equal to 50. So, what is this Q, when the level of output is equal to 50 units, the total revenue is maximized. There is maximization of total revenue when Q is equal to 50. Now we will find out what are the values of the total revenue. So, our total revenue is500 Q minus 5 Q square. So, if you’re putting the value of Q as 50; that is 550 square by 5 again 50 to the square. So, this comes to 25000 minus 12500, so it comes to 12500. So, the value of the total revenue is 12500. This is the maximum total revenue for the firm. And to achieve this, the Q has to be at least 50 units in order to maximize the total revenue.
Now, how we can check this, that this is the maximum amount of the total revenue, when the value of Q is equal to 50, we know the total revenue is 12500, but how to check that this 12500, is the maximum total revenue for the firm, which is facing a demand function; like p is equal to, if you remember this is 500 minus 5 Q, how to check this. We will take two different value of Q in order to check this. We will take Q is equal to 51 and we will take Q is equal to 49. So, if you take Q is equal to 51 and putting the value in the T R; total revenue equation, that is 500 Q minus 5 Q square, we get a value of total revenue which is 12495. Suppose we assume that Q is the level of output, if it is not 50, if you produce below this also still we can maximize the total revenue. So, let us assume Q is equal to 49, putting the value of Q as 49 in the total revenue function; we get a value that is 12495. So, we have 2 values; one if 51 and second one is 49. So, one is on a higher side when the level of output increases, whether it has any change in the total revenue, and second when the level of output decreases whether it has any change in the total revenue, and here we found that whether the Q increasing or whether the Q decreasing, the total revenue is decreasing.
If you look at total revenue is decreasing, because this is if what are the total revenue when Q is equal to 50. So, it can be concluded that 50 is that level of output, where the total revenue is maximum any level of output, either more than 50 or less than 50 is showing a decreasing total revenue. So, we can conclude that 50, when the level of output is 50, the total revenue is always maximum; particularly when the demand function is this, and when the total revenue function is this. Now, we will take in the case of the cost minimization, because the first case is revenue maximization, through revenue maximization the firm can increase the profit, and the second one when we can minimize the cost, again the difference between the revenue and cost is more and that leads to increase in the profit, which is in line with the basic objective of a firm, that is maximization of the profit.
So, let us take a case of the cost minimization, in which situation generally there is minimization of cost, particularly when the firm is planning to setup a new production unit. They want to know, what is the minimum average cost through which they can setup a new production unit? When they are planning to expand their skill of production, they are looking for the minimum average cost through which they can expand the scale of production, or planning to raise the price of the product how it is effect the demand. So, these are the case, where the technique of optimizing output is require, by minimizing the average cost. So, here what is the optimization problem, the optimization problem is the cost minimization. Let us take a total cost function, what is total cost function here. Suppose this is 400 plus 60 Q plus 4 Q square. The cost minimization is not with respect to the total cost, rather with respect to the average cost. How to find average cost from here, total cost divided by Q will give us the average cost.
So, what is average cost, this is 400 divided by Q, plus 60 plus 4 Q, this is our average cost. So, we need to minimize the cost, in order to find out the difference to be more between the total revenue or the total cost. So, the first case what we are doing, we are trying to maximize the total revenue in order to maximize the profit. Now what we will try to do, we will try to minimize the cost, so that the difference between the revenue and cost is higher which leads to a higher profit. So, in this case the minimization is not related to total cost, rather the minimization is related to the average cost. Now what is average cost over here, average cost is the total cost divided by the unit of output; that is T C divided by Q, which is 400 by Q plus 60 plus 4 q.
Now, what is the rule of minimization, the rule of minimization is, the derivative must be equal to 0, if you remember in the previous case in order to maximize the T R, the rule was that marginal revenue has to be equal to 0. So, in this case minimization case we always take a thumb rule for this, that this first order derivative with respect to the average cost it has to be equal to 0. So, we need to find out, the derivative of average cost with respect to Q, and that must be equal to 0. Now we will find out what is the derivative of average cost with respect to q. So, that comes to minus 400 by Q square plus 4 which is equal to 0. So, this is again 400 Q square, which is equal to minus 4; that leads to Q square, if you simplify again, then Q square is equal to minus 400 by minus 4, which is equal to 100. So, if Q square is equal to 100, so we need to find out the level of output here.
So, if Q square is equal to 100, then Q is equal to 10. So, when the unit of output or when the level of output is 10, this is the optimum level of output where the cost is minimum. So, at this level of output, the firm minimizes the cost, and if there guiding principle is on the basis of minimization of cost, the firm should follow a level of output that is equal to 10 units in order to minimize the cost. So, what we checked over here, optimization technique is used, either to maximize the revenue or to minimize the cost. So, we took a optimization problem which maximize the total revenue and there the thumb rule was to, maximizing the total revenue when marginal revenue is equal to 0.
And we took the second one second optimization technique which was the minimization of the cost, and here the optimization of problem is to minimize the average cost of production in order to maximize the profit. And here the thumb rule to minimize the cost, was to minimize the level of output, and for that is first order derivative of average cost with respect to Q has to be 0. Following that we got the level of output, and we say that this is the level of output, what the firm should follow in order to minimize the cost. If you look at all the business firm, they have a common objective. The common objective for all business firm is to maximize the profit. So, if you look at indirectly in the last two cases, last two optimization problem also we are trying to do. So, we are trying to in one case, we are trying to maximize the revenue, so that profit can be more, because their difference between the total revenue and total cost would be more, and the second case we minimize the cost.
So, that again the difference between the total revenue and total cost can be more which will maximize the profit. Now, we will take a problem where we will maximize the profit, rather than maximizing the total revenue or minimizing the cost. Let us see how we can do this by taking a profit function. And the basic need for this is if you look at the goal or the objective of the firm is to always to maximize the profit. So, now take a profit function and what is profit function; that is pi is equal to total revenue minus total cost. There are two conditions to maximize the profit; one is the necessary or the first order condition, which says that marginal revenue should be equal to the marginal cost. This is the first condition for the profit maximization, and second condition for the profit maximization, is the sufficient condition or the second order condition which says that; the second order derivative that is del square T R and del Q square should be less than del square T C and del Q square.
So, essentially it means, the slope of the marginal revenue function has to be less than the slope of the marginal cost function. So, for profit maximization, there are two conditions; one is necessary and the first order condition; that is marginal revenue equal to marginal cost. Second one is the sufficient condition or the second order condition where it says that; the second order derivative of the total revenue function should be less than the second order derivative of the total cost function, or on in other word the slope of the marginal revenue should be less than the slope of the marginal cost. So, let’s take a profit function in order to understand, that how the profit is maximized, and how the first order and second order condition gets fulfilled when the profit gets maximized.
So, we will take a function that is total revenue, which is equal to 600 Q minus 3 Q square. So, what is marginal revenue, marginal revenue is first order derivative of this. So, this comes to 600 minus 6 Q. Then we will take a total cost function, total cost function is 1000 plus 100 Q plus 2 Q square, what is marginal cost function. The first order derivative of the total cost function. So, that comes to 100 plus 4 Q. Now, what is the first order or the necessary condition? The marginal revenue should be equal to marginal cost. This is the first order condition or necessity condition what is our marginal revenue that is 600 minus 6 Q is equal to, what is our marginal cost, 100 plus 4 q. So, if you simplify this is 6 Q minus 4 Q is minus 600 plus 100. So, minus 10 Q is equal to minus 500 and Q is equal to 50. So, the outcome of the first order condition is, we found out the level of output; that is Q is equal to 50. Now what is the second order condition, second order condition is that. The second order derivative of the total revenue function has to be less than the second order derivative of the total cost function.
Now, let us see whether, particularly in this functional form whether we are fulfilling the second order condition or not. So, second order condition is del square T R, del Q square equal to del M R with respect to q. So, this is minus 6 del square T C del Q square. So, this is del M C with respect to del Q which is equal to 4 So, if you look at, this is less than this, and if the sum of both of this is also less than 0. So, this is del square T R del Q square minus del square T C, del Q square there is also less than 0. So, we know that, the second order condition gets fulfilled. So, we know that, profit is maximum when the necessary conditions get fulfilled; that is Q is equal to 50. Again we can do a random checking the way we did it for the other optimization problem; that you take any level of output which is more than 50 or less than 50. In order to understand, whether this is the level of output, which actually maximize the profit or not.
So, taking Q is equal to 50, we get the total revenue which is equal to 22,500 putting the value of Q, we get the total cost which is equal to 11,000. So, in this case, the profit is 11500. Suppose, you take a value Q is equal to 51 and Q is equal to 49. In the first case, the profit is equal to 11,495 and second case the profit is equal to again 11,495. So, we can conclude here that since the first order condition gets fulfilled, the profit is maximum when Q is equal to 50, because when we increase the level of output from 50 to 51, the profit is less; and when we decrease the level of output from 50 to 49 still the profit is less. So, we can say that, Q is that level of output which maximizes the profit. So, till now we are taking the optimization problem, and we are maximizing the revenue or profit or minimizing the cost without the constraints. So, next class, we will introduce the constraints, and then we will see how to use this optimization technique, in order to solve for the profit maximization or for the cost minimization.
Managerial Economics Prof. Trupti Mishra S.J.M School of Management Indian Institute of Technology, Bombay
Lecture -11
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 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 constrained, and we just optimize the maximization of a profit function or we just optimize the minimization of a cost function.
So, today I will discuss the optimization technique with a constrained, either in the form of the income or in the form of the cost, when it comes to cost, and then it comes to the revenue, either its maximization or it is a minimization of the cost.
So, in case of constrained optimization, this is a technique used for achieving a target under constrained situation or condition, is called constrained optimization. So, may be the motivation for optimization is remains same. It is achieving a target, either to maximize the profit or to minimize the cost, and but here the difference is that there is a constrained along with the objective function, and how to do this constrained optimization? We generally discuss 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 expressed 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 to 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 value is 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 remains same. The constrained equation is expressed 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 this substitution technique, is that we expressed 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 constrained optimization, along with the constrained with the objective function, whether the objective function is a profit maximization or whether the objective function is the 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, here the optimization problem is the profit maximization. Since we are saying that this is the case of a constrained optimization, there is also a constraint attached to this, and the constraint 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 constraint, 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 another term, or may be the profit maximization problem.
So, suppose x plus y is equal to 30. So, this can be written 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 has hundred 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 2x square, minus 30 minus y , because it was x y, plus 180 y minus 4y 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 minus 5y 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, dell pi by dell y is equal to 0, which is like 1200 plus 170 y minus 5 y square, which is equal to 0. Now solving this, 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 constrained, and 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 have 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. By putting with a 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 steps 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 3y square. Now, the firm here what is the constrained. The firm has to get a 36 units of x and y as the combine order, now what is the optimum combination. Optimum combination is to, what is what should be the minimum cost to produce this 36 unit of x and y. So, in this case, what should be the constrained? The constrained 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 order 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 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 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 our 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 1242. 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 constrained, 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 to for this constrained 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 constrained optimization is always a Lagrangian multiplier method. So, what is Lagrangian multiplier method, it is again one of kind of method to solve the constrained optimization, and it involves combining of both the objective function and the constrained equation, and solving by using the partial derivative methods. Basically, it takes the partial derivative with respect to both the variables, and then it gets the value of x and y, and by getting the value of x and y, it maximizes the profit or minimizes the cost. So, we will see how it works for the Lagrangian method.
Let us take a profit maximization case. Suppose, the profit equation is, hundred x minus 2 x square minus x y, plus 180 y minus 4 y square, again subject to x plus y is equal to 30.The same profit equation what we took for the substitution technique, and the same constrained what we take for y using the substitution technique method. So, x plus y is 30; that is constrained, 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; rather we will use a 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 here; that is x plus y minus 30 is equal to 0, and the lambda x plus y minus 30. 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 180 y, 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 constrained also we have added in the objective function. So, this is our Lagrangian function. Lagrangian function comes
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