Transcript lecture17&18

2.7 Applications of 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 = 0
x=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 be
P(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.