How Do Price and Demand Affect Revenue?

Theory

Revenue and price function

The demand $x$ for a product depends on its price, among other things. Demand is often a function of the price

x = d (p),

where $d (p)$ tends to decrease when $p$ goes up. This correlation can also be written as

p = p (x),

where the price $p$ is a function of the demand $x$ .

Theory

Maximizing revenue through affecting demand

hen you sell

x

units and the price

p (x)

is a function of this demand

x

, revenue from these sales is given by

I (x) = number \cdot price = x \cdot p (x) .

The demand that maximizes the revenue is the demand that yields the highest revenue.

Example 1

A business has determined its price function to be $p (x) = - 5 x + 50$ . What’s the revenue function of this business?

You know that revenue is calculated by the formula $I (x) = x \cdot p (x)$ . If you insert what you know directly into this formula the revenue function is

\begin{array}{l} I (x) & = p (x) \cdot x = (- 5 x + 50) \cdot x \\ = - 5 x^{2} + 50 x . \end{array}

I (x) = p (x) \cdot x = (- 5 x + 50) \cdot x = - 5 x^{2} + 50 x .

Note! Some exercises will ask you to find the level of demand that gives the business the highest revenue. In that case you have to differentiate the revenue function, set it equal to

0

, and solve for

x

.

Example 2

Find the demand that yields the greatest revenue for the business in Example 1. Use that to find the greatest revenue, and find the price of the product when the revenue is at its highest.

When an exercise talks about greatest or least, it’s screaming at you to differentiate! So you differentiate the revenue function

I (x) = - 5 x^{2} + 50 x

and set it equal to zero. In other words, greatest in the exercise is referring to the maximum.

The process goes like this:

\begin{array}{l} I^{'} (x) & = - 10 x + 50 = 0 \\ ⇓ \\ 10 x & = 50 \\ ⇓ \\ x & = 5 \end{array}

This tells you that the demand that yields the greatest revenue is $x = 5$ . You find the greatest revenue by inserting this value for $x$ into the revenue function

I (x) = - 5 x^{2} + 50 x .

That makes the greatest revenue of the business

\begin{array}{l} I (5) & = - 5 {(5)}^{2} + 50 (5) = - 125 + 250 \\ = 125 . \end{array}

I (5) = - 5 {(5)}^{2} + 50 (5) = - 125 + 250 = 125 .

You find the price of the product when the revenue is at its maximum by inserting the same value for $x$ into the price function

p (x) = - 5 x + 50

from Example 1:

p (5) = - 5 (5) + 50 = - 25 + 50 = 25

The price of the product at this point is $25$ .

Theory

Demand and Price

When the demand (the number of units the market wants) depends on the price, you get the expression

d (p)

, which is the demand as a function of the price. This will be a value for

x

, and you get

x = d (p) .

If revenue depends on demand, you can insert $d (p)$ for $x$ in the revenue function $I (x) = x \cdot p (x)$ , making the revenue function look like this:

\begin{matrix} I (x) = I (d (p)) \\ = \\ d (p) \cdot p (x) = d (p) \cdot p, \end{matrix}

I (x) = I (d (p)) = d (p) \cdot p (x) = d (p) \cdot p,

because

p (x) = p

(as you learned above).

Example 3

Find the demand as a function of the price for a product with a revenue function of

I (x) = 600 x - 3 x^{2} .

You know that the formula for revenue is $I (x) = p \cdot x$ , which means that you have to play with the expression to transform it into an expression multiplied by $x$ . You can do that through factorization in this way:

\begin{array}{l} I (x) & = 600 x - 3 x^{2} = x (600 - 3 x) \\ = (600 - 3 x) x \end{array}

I (x) = 600 x - 3 x^{2} = x (600 - 3 x) = (600 - 3 x) x

From this expression you can see that

p = 600 - 3 x

, where

p

is the price and

x

is the number of units sold. In economics the number of units sold depends on the demand, meaning that you can solve for

x

:

\begin{array}{l} p & = 600 - 3 x \\ 3 x & = 600 - p & | : 3 \\ x & = 200 - \frac{p}{3} \end{array}

You can write $x = d (p)$ as $x$ is the demand. That gives you

d (p) = 200 - \frac{p}{3},

just as the exercise asked for.