The methods of differentiation find great application in estimating various quantities of interest. Furthermore, economics has differentiation tools like marginal cost and marginal revenue as its basic necessities. From calculating the change in demand for a product to the change in its cost price to estimating the rate of change in revenue with an increase in selling price; everything in practice can be efficiently found out by taking the derivative of the dependent variable of interest with respect to the independent variable. Let us learn more!

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## The Cost Function

The total cost of production of *x *number of products, represented by *C(x)*, can be written as –

C(x) = F(x) + V(x)

where,

*F(x)*= The fixed cost, independent of the number of products being manufactured. For eg – The salary paid by the company to its workers, the property rents and the maintenance costs etc. could be independent of the number of items they produce in a given month.*V(x)*= The variable cost, which depends on the number of products being manufactured. For eg – The cost of packaging and transportation of the manufactured goods, which is obviously dependent on the number of goods produced.

The Average Cost per product (AC) = \(\frac{C(x)}{x}\)

Average Fixed Cost per product (AFC) = \(\frac{F(x)}{x}\)

Average Variable Cost per product (AVC) = \(\frac{V(x)}{x}\)

## The Marginal Cost

The term marginal* *comes into play when we need to ascertain the increase in any dependent variable with a unit change of the independent variable. Thus, we define the marginal costs as –

If C(x) is the total cost of producing x units, then the change in the total cost if one additional unit needs to be produced, at an output level of x units, is given by –

Marginal Cost = \(\frac{dC(x)}{dx}\)

Similarly, we may calculate the marginal average costs as – \(\frac{d[AC(x)]}{dx}\), \(\frac{d[AFC(x)]}{dx}\) and so on.

## The Revenue and the Profit Functions

Denoted by *R(x)*, the revenue function represents the amount of money earned (the total turnover) by a company, by selling *x* number of products. If the selling price of every unit is equal to S*P,* the revenue function would be –

R(x) = SP × x

Now if the cost function and the revenue function for x products take on the value C(x) and R(x) respectively, then we can write the profit function *P(x)* as –

P(x) = R(x) – C(x)

## The Marginal Revenue and Marginal Profit

Similar to the definition of Marginal Cost, we can define the Marginal Revenue and the Marginal Profit as follows –

- The rate of change of revenue per unit change in the output (number of products) is the Marginal Revenue, given by \(\frac{dR(x)}{dx}\)
- The rate of change of profit per unit change in the output (number of products) is the Marginal Profit, given by \(\frac{dP(x)}{dx}\)

## The Consumption Functions and the Marginal Propensities

The Consumption function C(Y) relates the total consumption (or the total expenditure incurred) C to the total income Y. We give the difference in the total income and the total consumption as the Saving function S(Y).

Thus, we can define the following marginal propensities –

- Marginal Propensity to Consume (MPC): It is the rate of change of the total consumption per unit change in the income – \(\frac{dC(Y)}{dY}\)
- Marginal Propensity to Save (MPS): It is the rate of change of the saving per unit change in the income – \(\frac{dS(Y)}{dY}\)

This concludes our discussion of the important financial terms that we’ll be dealing with in the general market. Now go through the solved examples below to get a feel for the topic –

## Solved Examples for You

#### Question 1

A firm is selling 100 units at a price of Rs 250. However, to sell 110 units, they need to cut the price down to Rs 240. What is the level of marginal revenue at this higher level of sales?

**Solution: **This question simply requires a good understanding of the economic terms we have encountered so far. We can simply solve it as –

We can write the total revenue function for 100 units as – R(100) = 100 × 250 = Rs 25000

Similarly, for 110 units – R(110) = 110 × 240 = Rs 26400

The marginal revenue is then simply: The difference between the total revenue at 110 units and the total revenue at 110 units, divided by 10 (the number of additional units), to get the extra cost per unit. Thus,

Marginal Revenue = \(\frac{26400 – 25000}{10}\)

= Rs 140

#### Question 2

The cost function for the manufacture of x number of goods by a company is C(x) = \(x^3 – 9x^2 + 24x \). Find the level of output at which the marginal cost is minimum. Further, if the selling price of a unit is \(2x^3 + 9x^2\), find the average profit.

**Solution:**We calculate the marginal cost as –

\(\frac{dC(x)}{dx} = \frac{d}{dx}(x^3 – 9x^2 + 24x)\)

\( = 3x^2 – 18x + 24\)

In order to be a minimum at x = x_{0}(say), its derivative must vanish at x_{0}. Thus –

\(\frac{d}{dx}_{x = x_0}(3x^2 – 18x + 24)\)

\(x_0 = +2,+4\)

By the seond derivative test, we can conclude that at x = +4, the function assumes a minima. Thus, for an output = 4 finished goods, the marginal cost would be minimum. Now, since we are have the selling price already, we can calculate the average profit as –

\(\frac{P(x)}{x} = \frac{SP(x) – CP(x)}{x}\)

⇒ \(= \frac{(2x^3 + 9x^2) – (x^3 – 9x^2 + 24x)}{x}\)

⇒ \(= \frac{x^3 + 24x}{x}\)

\(= x^2 + 24\)

We can see clearly that the average profit depends on the number of goods manufactured (i.e. x), which should definitely be so in any practical example.

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