Derivatives to Business and Economics

Our textbook tells us that business analysts and economists have increasingly turned to mathematical models to help them describe what is happening, predict the effects of various policy alternatives, and choose reasonable courses of action.

The derivative is one of the mathematical tools employed by economists and business analysts.

Section 2.7 is devoted to illustrating a few of the many applications of the derivative to the problems of business and economics.

Applications in our textbook are centered around the theory of the firm.

We will study the activity of a business or industry and restrict our analysis to a time period during which background conditions (supplies, wage rates, etc.) are fairly constant.

There are three main functions we will work with in this section

C(x) = cost of producing x units of a product

R(x) = revenue generated by selling x units of a product

P(x) = R(x) – C(x) = the profit (or loss) generated by producing and selling x units of the product.

NOTE: These functions are often only defined for nonnegative integers. Why?

Often, these functions give rise to a set of discrete points on a graph.

In studying these functions, economists usually assume that C(x) is defined for all values of x and draw a smooth curve through the points.

If we assume that C(x) has a smooth graph, then we can use the tools of calculus (derivatives) to study it.

Suppose that the cost function for a manufacturer is given by

C(x) = (10-6)x3 - .003x2 +5x +1000 dollars

(a)Describe the behavior of the marginal cost.

(b)Sketch the graph of C(x)

For (a), recall that marginal is the word for derivative in business and economics.

For (b), we can get information about the graph from calculating first and second derivatives.

In sketching the graph of the marginal cost y = C’(x), we can use y’ = C’’(x) to provide us with information about the graph.

We can tell that the marginal cost graph will be a parabola that opens upward. How?

The relative minimum point of the marginal cost graph can be discovered by examining y’ = C’’(x) = 0.

Solving C’’(x)=.000006(x-1000) = 0 for x uncovers a horizontal tangent at x = 1000.

The associated value of y in the marginal cost graph is calculated by

(3)(10-6)(1000)2 - .006(1000) + 5= 3 – 6 + 5 = 2

The graph of the marginal cost function.

Since the graph of the marginal cost function never reaches 0, then there are no relative extreme points in the graph of the cost function.

We also note that the graph of C’(x) is never negative. What does this tell us about the graph of C(x)?

What else can we tell about the graph of C(x) from looking at the graph of C’(x)?

The graph of the cost function.

Revenue Functions

R(x) is the revenue received from the sale of x units of some commodity.

We call R’(x) the marginal revenue.

Economists use marginal revenue to measure the rate of increase in revenue per unit increase in sales.

If x units are sold at a price p per unit, the total revenue R(x) is given by

R(x) = xp

Some economic concepts for revenue

If a firm is small and has many competitors, then its sales have little effect on the market price of its commodity.

The price is then constant as far as the one firm is concerned, and the marginal revenue R’(x) equals the price p.

However, if the firm’s production of a commodity has a significant impact on the availability of the commodity (for example, when the firm has a monopoly), then an interesting problem arises.

Consumers will purchase more of the commodity when the price is low, and fewer units when the price is high.

Let f(x) be the highest price per unit at which all x units can be sold to customers. Since consumers will buy less when the price increases, the graph of f(x) will be decreasing.

The demand equation p = f(x) determines the total revenue function.

If a firm wants to sell x units, the highest price it can set is f(x) dollars per unit.

The revenue function becomes R(x) = xp = xf(x)

The concept of the demand curve applies to an entire industry, as well as a single monopolistic firm. In this case, many producers offer the same commodity for sale.

If x denotes the total output of the industry, then f(x) is the market price per unit of output and xf(x) is the total revenue earned by the industry from the sale of x units.

The demand equation for a certain product is

p = 6 – ½ x dollars

Find the level of production that maximizes revenue.

We want to begin by developing our revenue function R(x). R(x) = xp so

The marginal revenue is given by R’(x)

Note that R(x) is the graph of a parabola that opens downward. How do we know?

We use R’(x) = 0 to identify the relative maximum point.

6 –x = 0x = 6

The corresponding value of y is

The graph of R(x) appears as

Profit Functions

Once we know the cost function C(x) and the revenue function R(x), we can compute the profit function P(x).

P(x) = R(x) – C(x)

Suppose that the demand equation for a monopolist is p = 100 - .01x and the cost function is

C(x) = 50x + 10,000.

Find the value of x that maximizes the profit and determine the corresponding price and total profit for this level of production.

We need to construct the profit function P(x) = R(x) – C(x). C(x) has been given, so we compute R(x) using the demand equation

Then, our profit function is

The graph of P(x) is a parabola that opens downward. How do we know?

The relative extreme point of P(x) will be the point at which profit is the highest. We need to compute the marginal profit function P’(x), and solve P’(x) = 0 for x.

Solving P’(x) = 0, we have

-.02(x-2500) = 0

which yields x = 2500

If 2500 units are produced, the profit would beP(2500) = -.01(2500)2 + 50 (2500) – 10,000

= 52, 500 dollars

The highest price at which the 2500 units can be sold is

P = 100 - .01(2500)= 100 – 25 = 75 dollars

Answer:Produce 2500 units and sell them at $75 per unit. The profit will be $52, 500.

Setting Production Levels

Suppose that a firm has cost function C(x) and revenue function R(x).

In a free-enterprise economy, the firm will set production x in such a way as to maximize the profit function P(x) = R(x) – C(x).

If P(x) has its maximum at x = a, then P’(a) = 0. It follows that since P’(x) = R’(x) – C’(x) that R’(a) – C’(a) = 0 or R’(a) = C’(a)

Thus profit is maximized at the production level for which marginal revenue equals marginal cost.