APPLICATION OF DIFFERENTIAL CALCULUS IN ECONOMICS

Although introductory economics courses, such as those most college students must complete in the course of their studies, involve little math, an in-depth study of economics requires a rigorous understanding of mathematics, including calculus. Calculus provides the language of economics and the means by which economists solve problems. Calculus is especially significant in illustrating what a leading economist calls a key principle of economics.

IdentificationAs an advanced branch of mathematics, calculus focuses heavily on functions and derivatives. Functions examine the relationship between two or more variables, or entities that take on different values. Mathematicians and economists often use letters, such as X and Y, to symbolize particular variables. If the value of Y changes as the value of X changes, then the two variables have a functional relationship. Derivatives, meanwhile, consider the rate of change in one variable relative to the change in another. Functions and derivatives relate to relevant concepts in economics.

FunctionEconomic research often uses calculus to examine functional relationships. An example includes the relationship between the dependent variable income and various predictors, or independent variables, such as education and experience. If average income rises as years of education and work experience increase, then a positive relationship exists between the variables, namely that income is a function of education and experience. Differential calculus, the process of obtaining derivatives, enables economists to measure the average change in income relative to a single year's increase in education and/or experience.

EffectsDerivatives in calculus, or the change in one variable relative to the change in another, are identical to the economic concepts of marginalism, which examines the change in an outcome that results from a single-unit increase in another variable. Marginal changes relate to an important principle in economics: the notion that people tend to think at the margin, according to Harvard economist Greg Mankiw, author of "Principles of Economics," a popular textbook in college economics courses. Mankiw writes that economists use the term "marginal changes" to describe small, incremental changes, such as incremental changes in work hours or factory output.

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BenefitsCalculus, by determining marginal revenues and costs, can help business managers maximize their profits and measure the rate of increase in profit that results from each increase in production. As long as marginal revenue exceeds marginal cost, the firm increases its profits.

Significance

The amount of interest to be paid on a loan, whether for a home, motor vehicle or capital equipment for a business, is an important consideration for households and firms. Calculus provides a means for determining the amount of interest paid over the life of a loan.

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USING CALCULUS TO LEARN ECONOMICS

It is a premise of this book that a calculus approach to the study of microeconomics pays large dividends. There are several related reasons why this is so.

First, the use of mathematics helps develop problem-solving skills. It imposes a rigor that mandates keeping track of costs and benefits, and it provides a framework for determining which variables and parameters are important. We apply calculus and other mathematics to a wide variety of problems. Our view is that you can’t learn economics without solving problems.

Second, the calculus approach helps in learning to think clearly. You will be forced to translate weak verbal arguments into precise, consistent statements. Making a statement formal clarifies what is being said. When you are called upon to verify what you think is obvious, you may discover that it is not obvious or perhaps not even true.

Third, the use of calculus unifies the material by focusing on the common economic structure of problems. When we strip the specific details away, many problems look surprisingly alike and have common solutions. More generally, this is an argument for the power of abstraction. Calculus makes it easier, not harder, to learn economics. There are, in fact, only a small number of basic ideas in

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intermediate microeconomics; much of what we do is working out the details for particular situations.

Fourth, calculus helps in becoming literate in the language of modern economics. For better or worse, mathematics is the language of economics. If you want to read the economic literature or get an advanced degree in economics, you will have to deal with the mathematics. The best business schools demand mathematical maturity of their students; and if you want to undertake a serious study of finance, for example, you will find that calculus is only the beginning.

Fifth, this approach provides the opportunity to apply mathematics to social science problems. Whereas many of the examples in a first course in calculus come from physics, there are also examples from economics that illustrate most of the same points. This shows the power of calculus and aids in its understanding.

Sixth, this approach allows us to take the subject to a more advanced point and thus gives a better appreciation of the power of economic theory. Economics is ultimately a policy science, and theory plays an essential role. Hal Varian [8] has some enlightening answers to the question, “What use is economic theory?” Among other things, he suggests that theory is useful as a substitute for data that we need but do not have; and he explains how theory can generate useful insights in explaining economic phenomena, even when the theory is only approximately true.

The Economics of a Derivative.

In economics a derivative typically represents a marginal concept. If benefits (B) and costs (C) depend on the level of an activity (x), Then the derivative of B with respect to x representsmarginal benefit and the derivative of C with respect to x represents marginal cost. If revenue depends on the quantity of a good sold, then the derivative of revenue with respect to quantity represents marginal revenue. If production depends on labor input, then the derivative of production with respect to labor input represents the marginal product of labor. If resource cost depends on labor input, then the derivative of resource cost with respect to labor input represents the marginal resource cost of labor. We will see many such examples throughout the book.

We examine one of these concepts, marginal cost (MC), more carefully. In a first course in economics, such an expression is usually defined as MC = _C/_x and represents the cost of producing one more (or the last) unit. In the formula it is understood that x takes on only the discrete values 1, 2, 3, etc. This is precisely

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the point of view that must be taken, at least in principle, when a good is indivisible. Supertankers and dams, for example, come only in whole units.

Seeing the formula for marginal cost in the previous paragraph, students sometime wonder why they must divide by the change in x. The reason is to obtain the extra cost per unit of x. If $10 of extra cost is incurred in producing two more bushels of output, then the marginal cost is $5 per bushel.

We will assume that the variable x is continuous, i.e., that it can take on any real value (or at least any nonnegative value) and not just 1, 2, 3, etc. This is appropriate when we want to measure, in pounds, the amount of cheese produced in a day. Although commercial scales may not be capable of measuring cheese to the one-millionth of a pound, we are not misled by presuming that we could do so. Even when we deal with goods such as automobiles that are less homogeneous than cheese, we can still make sense of requiring a production of 61.5 automobiles per day if we interpret that to mean 123 automobiles over a two day production run. The advantage of working with a continuous variable is that it greatly simplifies the mathematics.

For a continuous variable x the appropriate definition of marginal cost is as the derivative of the cost function: MC = dC/dx or MC = C0(x). An advantage of using the C0(x) notation is that it emphasizes that marginal cost is itself a function of x.

In a continuous setting we often speak loosely of marginal cost representing the extra cost incurred by increasing x by one unit or the cost saved by decreasing it by one unit. We should remember that, strictly speaking, the kinds of increments we are dealing with are infinitesimal onper bushel, dollars per pound, dollars per automobile, etc.

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APPLICATION OF DIFFERENTIAL CALCULUS IN ECONOMICS

We have learnt in calculus that when 'y' is a function of 'x', the derivative of y w.r.to x i.e. dy/dx measures the instantaneous rate of change of y with respect to x. In Economics and commerce we come across many such variables where one variable is a function of the other. For example, the quantity demanded can be said to be a function of price. Supply and price or cost and quantity demanded are some other such variables. Calculus helps us in finding the rate at which one such quantity changes with respect to the other. Marginal analysis in Economics and Commerce is the most direct application of differential calculus. In this context, differential calculus also helps in solving problems of finding maximum profit or minimum cost etc., while integral calculus is used to find he cost function when the marginal cost is given and to find total revenue when marginal revenue is given..BASIC FUNCTIONSBefore studying the application of calculus, let us first define some functions which are used in business and economics.

Cost FunctionThe total cost C of producing and marketing x units of a product depends upon the number of units (x). So the function relating C and x is called Cost-function and is written as C = C (x).

The total cost of producing x units of the product consists of two parts(i) Fixed Cost(ii) Variable Cost i.e. C (x) = F + V (x)

Fixed Cost : The fixed cost consists of all types of costs which do not change with the level of production. For example, the rent of the premises, the insurance, taxes, etc.

Variable Cost : The variable cost is the sum of all costs that are dependent on the level of production. For example, the cost of material, labour cost, cost of packaging, etc.

Demand FunctionAn equation that relates price per unit and quantity demanded at that price is called a demand function. If 'p' is the price per unit of a certain product and x is the number of units demanded, then we can write the demand function as x = f(p) or p = g (x) i.e., price (p) expressed as a function of x.

Revenue functionIf x is the number of units of certain product sold at a rate of Rs. 'p' per unit, then the amount derived from the sale of x units of a product is the total revenue.

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Thus, if R represents the total revenue from x units of the product at the rate of Rs. 'p' per unit thenR= p.x is the total revenue. Thus, the Revenue function R (x) = p.x. = x .p (x)

Profit FunctionThe profit is calculated by subtracting the total cost from the total revenue obtained by selling x units of a product. Thus, if P (x) is the profit function, then P(x) = R(x) - C(x)

Break-Even PointBreak even point is that value of x (number of units of the product sold) for which there is no profit or loss.i.e. At Break-Even point P (x ) = 0or R ( x)-C( x )= 0 i.e. R ( x) = C ( x )

SOME EXAMPLES. For a new product, a manufacturer spends Rs. 1, 00,000 on the

infrastructure and the variable cost is estimated as Rs.150 per unit of the product. The sale price per unit was fixed at Rs.200. Find (i) Cost function (ii) Revenue function (iii) Profit function, and (iv) the break even point.

Solution : (i) Let x be the number of units produced and sold,then cost function

C ( x) = Fixed cost + Variable Cost = 1,00,000 + 150 x

(ii) Revenue function = p.x = 200 x

(iii) Profit function P ( x ) = R ( x )-C( x ) =200 x - (100,000+150 x ) =50 x -100,000

(iv) At Break-Even point P ( x ) = 050 x -100,000 = 0 x = 100,000/50 = 2000

Hence x = 2000 is the break even point.i.e. When 2000 units of the product are produced and sold, there will be no profit or loss.

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A Company produced a product with Rs 18000 as fixed costs. The variable cost is estimated to be 30% of the total revenue when it is sold at a rate of Rs. 20 per unit. Find the total revenue, total cost and profit functions.

Solution : (i) Here, price per unit (p) = Rs. 20

Total Revenue R ( x ) = p. x = 20 x where x is the number of units sold.

(ii) Cost functionC ( x) = 18000 + 30/100 R(x)

=18000 + 30/100 x 20 x=18000+6x

(iii) Profit function P ( x )= R ( x ) -C( x )=20 x-(18000+6x)=14 x-18000

A manufacturing company finds that the daily cost of producing x items of a product is given by C(x)=210x+7000.

(i) If each item is sold for Rs. 350, find the minimum number that must be produced and sold daily to ensure no loss.

(ii) If the selling price is increased by Rs. 35 per piece, what would be the break even point.

Solution :(i) Here, R( x ) =350x and C( x )= 210x+7000 P( x ) =350x-210x-7000

=140x -7000

For no loss P( x )= 0 40x - 7000= 0 or x=50

Hence, to ensure no loss, the company must produce and sell at least 50 items daily.

(ii) When selling price is increased by Rs. 35 per unit,R(x)= (350 +35)x =385 x

P(x) =385x-(210x+7000)=175 x - 7000

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At Break even point P(x) =0 175 x - 7000 =0x = 7000/175 = 40

Average and Marginal FunctionsIf two quantities x and y are related as y = f (x), then the average function may be defined as f (x)/ x and the marginal function is the instantaneous rate of change of y with respect to x. i.e.Marginal function is dy/dx or d/dx (f(x))

Average Cost : Let C = C(x) be the total cost of producing and selling x units of a product, then the average cost (AC) is defined as AC = C/xThus, the average cost represents per unit cost.

Marginal Cost : Let C = C(x) be the total cost of producing x units of a product, then themarginal cost (MC), is defined to be the rale of change of C (x) with respect to x. Thus

MC = dC/dx or d/dx(C(x))

Marginal cost is interpreted as the approximate cost of one additional unit of output.

For example, if the cost function is C = 0.2x2 + 5, then the marginal cost is MC= 0.4x

The marginal cost when 5 units are produced is[MC] x=5 = (0.4) (5) = 2

i.e. when production is increased from 5 units to 6, then the cost of additional unit is approximately Rs. 2.

However, the actual cost of producing one more unit after 5 units is C(6) - C(5) = Rs. 2.2

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EXAMPLE The cost function of a firm is given by C =2x2 +x-5.

Find (i) the average cost (ii) the marginal cost, when x = 4

Solution :(i) AC = (2x2 + x - 5)/x

=2x +1 – 5/x

At x = 4 , AC = 2(4) + 1- 5/4= 9 -1.25 = 7.75

(ii) MC = d/dx (C) = 4x + 1 MC at x = 4 = 4(4) + 1 = 16 + 1= 17Example 41.5

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