Elasticity of Demand

Section 5.1 Elasticity of Demand

We begin by analyzing a real example from the air travel industry, and have a detailed look at how the cost of air plane tickets impact the revenue of tickets sold. A simplistic view may lead one to believe that a decrease in the cost for an airplane ticket would cause the revenue to increase and vice versa. In economics, this particular relationship between unit price and revenue is referred to as elastic demand as we will learn later. Particularly in Canada, start-up airlines can collapse more readily under this condition. The Department of Finance in Canada studied the aforementioned relationship and published the research results in Air Travel Demand Elasticities: Concepts, Issues and Measurement: 1 by differentiating between six types of air travel that are associated pairwise: business and leisure, long-haul and short-haul, and international long-haul and North American long-haul air travel. The results of the study corroborate that the demand for business air travel is less elastic than that for leisure air travel. This finding does not come as a surprise, since even a costly booked vacation can be more readily moved to different dates than business travels. The other two results of the study are that the demand for long-haul flights is less elastic than that for short-haul flights, and similarly, the demand for international flights is less elastic than that for North American flights. This make sense, because the further the destination, the less likely it is that an alternative mode of transport can be found as a substitute for an expensive flight.

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We now derive the mathematical model that helps us to analyze the relationship between unit price and revenue, and determines the elasticity of demand of a particular economic situation when the demand function is given.

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In order to aid our analysis, it will be more convenient to write the demand function

f

$f$ in the form

q = f (p) .

$q=f (p)\text{.}$ In other words, we will think of the quantity demanded

q

$q$ of a certain product as a function of its unit price

p .

$p\text{.}$ As is shown in Figure 5.1, the function

f

$f$ is usually a decreasing function of

p,

$p\text{,}$ because the quantity demanded of a product typically decreases as the associated unit price increases.

Note: During problem solving, it is often easier to use the inverse function of

$f\text{,}$ namely

$p = g(q)$ than

$f$ itself.

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Figure 5.1. The demand function

$q=f(p)$ and the effects on this demand from an increase in price by

$h$ dollars.

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We now take a similar approach as in our analysis of the derivative in Chapter 4. Figure 5.1 shows an increase of

$h$ dollars in the unit price

$p$ for some product to a unit price of

$p+h$ dollars. Therefore, the associated quantity demanded changes from

$f(p)$ units to

$f(p+h)$ units with an overall decrease of

$f(p+h)-f(p)$ units. We can now calculate the percentage change in the unit price to be

$\begin{equation*} \dfrac{\text{ Change in unit price } }{\text{ Price } p} \times 100 = \dfrac{h}{p}(100)\text{,} \end{equation*}$

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and the corresponding percentage change in the quantity demanded to be

$\begin{equation*} \dfrac{\text{ Change in quantity demanded } }{\text{ Quantity demanded at price } p} \times 100 = \dfrac{f(p+h)-f(p)}{f(p)}(100)\text{.} \end{equation*}$

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By calculating the ratio of the percentage change in the quantity demanded to the percentage change in price, we can determine the effect the latter has on the former:

$\begin{equation*} \begin{split} \dfrac{\text{ Percentage change in the quantity demanded } }{\text{ Percentage change in the unit price } } \amp = \dfrac{100\dfrac{f(p+h)-f(p)}{f(p)}}{100\dfrac{h}{p}} \\[1ex] \amp = \dfrac{\dfrac{f(p+h)-f(p)}{h}}{\dfrac{f(p)}{p}} \end{split} \end{equation*}$

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We now recognize the difference quotient in this fraction. So, if

$f$ is differentiable at

$p\text{,}$ we can deduce for small

$h$ that

$\begin{equation*} \dfrac{f(p+h)-f(p)}{h} \approx f'(p)\text{.} \end{equation*}$

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In other words,

$\begin{equation*} \dfrac{\text{ Percentage change in the quantity demanded } }{\text{ Percentage change in the unit price } }=\dfrac{f'(p)}{\dfrac{f(p)}{p}} = \dfrac{pf'(p)}{f(p)}\text{,} \end{equation*}$

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when

$h$ is small.

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Note: In Section 5.7 it will be shown that for a decreasing function

$q=f(p)$ on a certain interval

$I\text{,}$ its derivative

$f'(p)\lt 0$ for all

$p \in I\text{.}$ But this means that the value of the quantity

$pf'(p)/f(p)$ is negative. Since it is preferred to work with positive values, economists define the elasticity of demand

$E(p)$ to be the negative of the quantity

$pf'(p)/f(p)\text{.}$

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Definition 5.1. Elasticity of Demand.

Suppose that the demand function $q=f(p)$ is differentiable. Then the elasticity of demand, $E\text{,}$ at price $p$ is defined by

$\begin{equation*} E(p) = -\dfrac{pf'(p)}{f(p)} \end{equation*}$

Note: Some textbooks define elasticity of demand as

$\begin{equation*} E(p,q) = \frac{-\frac{p}{q}}{\frac{dp}{dq}} = -\frac{p}{q} \frac{dq}{dp}\text{.} \end{equation*}$

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Example 5.2. Elasticity of Demand.

The unit price $p$ in dollars and the quantity demanded $q$ of a certain product are related by the equation

$\begin{equation*} p = -0.02q + 400 \ \ \ 0\leq q \leq 20,000 \end{equation*}$

Determine the elasticity of demand $E(p)\text{.}$
Calculate $E(100)\text{.}$ What can you determine from your result?
Calculate $E(300)\text{.}$ What can you determine from your result?

Solution

Writing $q$ in terms of $p\text{,}$ we have

\begin{equation*} q=f(p)=-50p+20,000 \end{equation*}

and so $f'(p) = -50\text{.}$ The elasticity of demand is thus

\begin{equation*} E(p) = -\dfrac{pf'(p)}{f(p)} = \dfrac{50p}{-50p+20,000} = \dfrac{p}{400-p} \end{equation*}
\begin{equation*} E(100) = \dfrac{100}{400-100} = \frac{1}{3} \cdot \end{equation*}

Therefore, when the unit price $p$ is $100 per unit, a small increase in $p$ will lead to a decrease of approximately 0.33% in the quantity demanded $q\text{.}$
\begin{equation*} E(300) = \dfrac{300}{400-300} = 3\text{.} \end{equation*}

Here, we see that a small increase in $p$ from $300 per unit will lead to a decrease of approximately 3% in the quantity demanded $q\text{.}$

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The following economic terminology is useful when describing demand in terms of elasticity.

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Definition 5.3. Elastic, Unitary and Inelastic Demand.

The demand is elastic if $E(p) > 1\text{.}$ That is to say, the demand is elastic if the percentage change in demand is greater than the percentage change in price.
The demand is unitary if $E(p) = 1\text{.}$ That is to say, the demand is unitary if the percentage change in demand and price are relatively equal.
The demand is inelastic if $E(p) \lt 1\text{.}$ That is to say, the demand is inelastic if the percentage change in demand is less than the percentage change in price.

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In Example 5.2, we determined that the demand for the given product is elastic when

$p=300$ and inelastic when

$p=100\text{.}$ These calculations illustrate that a small percentage change in the unit price will result in a greater percentage change in the quantity demanded, i.e. when the demand is elastic; and a small percentage change in the unit price will cause a smaller percentage change in the quantity demanded, i.e. when the demand is inelastic; and lastly, a small percentage change in the unit price will result in the same percentage change in the quantity demanded, i.e. when the demand is unitary.

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Subsection 5.1.1 Elasticity and Revenue

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In the previous section, we developed the notion of elasticity of demand by analyzing the relationship between quantity demanded and unit price in terms of percentage change. Of course this change influences revenue, and so we now have a closer look at the effects of elasticity on revenue. Again we assume that

$q=f(p)$ relates the quantity

$q$ demanded of a certain product to its unit price

$p$ in dollars. When

$q$ units of the product are sold at the price

$p\text{,}$ then the revenue is given by

$\begin{equation*} R(p)=pq=pf(p)\text{.} \end{equation*}$

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We now calculate the marginal revenue with respect to

$p$ and obtain

$\begin{equation*} \begin{split} R'(p)\amp =f(p)+pf'(p)\\ \amp =f(p)\left[1+\dfrac{pf'(p)}{f(p)}\right] \\ \amp =f(p)\left[1-E(p)\right]. \end{split} \end{equation*}$

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This last equation tells us that elasticity influences revenue. In order to determine what the effects are, we analyze the sign of the marginal revenue. We first note that

$f(p)$ is positive for all values of

$p$ and consider three cases:

Suppose the demand is elastic when the unit price is set at $p$ dollars. Then

$\begin{equation*} E(p)>1 \implies 1-E(p)\lt 0\text{,} \end{equation*}$

and so

$\begin{equation*} R'(p) = f(p)\left[1-E(p)\right] \lt 0\text{,} \end{equation*}$

which means that revenue $R$ is decreasing at $p\text{.}$ In other words, a small increase/decrease in the unit price results in a decrease/increase respectively in the revenue. This is illustrated on the revenue curve of the white region in Figure 5.2.
Suppose the demand is unitary when the unit price is set at $p$ dollars. Then

$\begin{equation*} E(p)=1 \implies 1-E(p)=0\text{,} \end{equation*}$

and so

$\begin{equation*} R'(p) = f(p)\left[1-E(p)\right]=0\text{,} \end{equation*}$

which causes revenue $R$ to be stationary at $p\text{,}$ i.e. neither increasing nor decreasing. This means that a small increase/decrease in the unit price does not affect a change in the revenue. This is visualized on the revenue curve in Figure 5.2 where arrows point to.
Lastly, suppose the demand is inelastic when the unit price is set at $p$ dollars.

$\begin{equation*} E(p)\lt 1 \implies 1-E(p)>0\text{,} \end{equation*}$

and so

$\begin{equation*} R'(p) = f(p)\left[1-E(p)\right] > 0\text{,} \end{equation*}$

which necessitates that revenue $R$ is increasing at $p\text{.}$ This implies that a small increase/decrease in the unit price results in an increase/decrease respectively in the revenue. This is visualized on the revenue curve of the grey region in Figure 5.2.

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Figure 5.2. Inelastic demand corresponds to an increase in revenue (see grey), elastic demand corresponds to a decrease in revenue (see white), and at unitary demand revenue is stationary (see arrows)

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The results of this analysis are summarized below:

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Effects of Unit Price Changes to Revenue.

If the demand is elastic at $p\text{,}$ i.e. $E(p)>1\text{,}$ then a small increase/decrease in the unit price results in a decrease/increase respectively in the revenue.
If the demand is unitary at $p\text{,}$ i.e. $E(p)=1\text{,}$ then a small increase/decrease in the unit price does not affect a change in the revenue.
If the demand is inelastic at $p\text{,}$ i.e. $E(p)\lt 1\text{,}$ then a small increase/decrease in the unit price results in an increase/decrease respectively in the revenue.

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By noticing the following relationships between the unit price and the revenue, you may better be able to remember the effects on a unit price change on the revenue.

When the demand is elastic, then the change in unit price and the change in revenue move in opposite direction.
When the demand is inelastic, then the change in unit price and the change in revenue move in the same direction.

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Example 5.4. Elasticity of Demand.

Refer to Example 5.2.

For $p=100$ and $p=300\text{,}$ calculate whether the demand is elastic, unitary or inelastic.
What can you deduce from your results when $p=100\text{?}$

Solution

From part (b) of Example 5.2, we see that $E(100) = \frac{1}{3} \lt 1\text{.}$ Therefore, the demand is inelastic. From part (c) of Example 5.2, we see that $E(300) = 3 > 1\text{,}$ and so the demand is elastic.
Since the demand is inelastic when $p=100\text{,}$ a slight raise in the unit price will lead to an increase in revenue.

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Example 5.5. Elasticity of Demand.

The demand equation for a certain product is given by

$\begin{equation*} p=-0.02q+300 \ \ \ 0\leq q \leq 15,000 \end{equation*}$

where $p$ denotes the unit price in dollars and $q$ denotes the quantity demanded. The weekly total cost function associated with this product is

$\begin{equation*} C(q)=0.000003q^{3}-0.04q^{2}+200q+70,000 \end{equation*}$

dollars.

Determine the revenue function $R$ and the profit function $P\text{.}$
Determine the marginal cost function $C'\text{,}$ the marginal revenue function $R'\text{,}$ and the marginal profit function $P'\text{.}$
Determine the marginal average cost function $\overline{C}'\text{.}$
Calculate $C'(3000)\text{,}$ $R'(3000)$ and $P'(3000)\text{.}$ What can you deduce from your results?
Determine whether the demand is elastic, unitary, or inelastic when $p=100$ and when $p=200\text{.}$

Solution

$\begin{aligned}R(q) \amp = pq \amp \\ \amp =q(-0.02q^{2} + 300q) \amp \\ \amp = -0.02q^{2}+300q \amp \end{aligned}$ $\begin{aligned}P(q) \amp = R(q) - C(q) \amp \\ \amp = -0.02q^{2}+300q -(0.000003q^{3}-0.04q^{2}+200q+70,000) \amp \\ \amp =-0.000003q^{3}+0.02q^{2}+100q-70,000 \ \ \ (0 \leq q \leq 15,000) \amp \end{aligned}$
$C'(q)=0.000009q^{2}-0.08q+200$ $R'(q)=-0.04q + 300$ $P'(q)=-0.000009q^{2}+0.04q+100\text{.}$
The average cost function is

\begin{equation*} \begin{split} \overline{C}(q) \amp = \dfrac{C(q)}{q}\\ \amp = \dfrac{0.000003q^{3}-0.04q^{2}+200q+70,000}{q}\\ \amp = 0.000003q^{2}-0.04q+200 + \frac{70,000}{q} \end{split} \end{equation*}

Therefore, the marginal average cost function is

\begin{equation*} \overline{C}'(q) = 0.000006q-0.04 - \frac{70,000}{q^{2}}\text{.} \end{equation*}
Using the above results, we find

\begin{equation*} C'(3000) = 0.000009(3000)^{2} - 0.08(3000) + 200 = 41 \end{equation*}

That is, when the level of production is already $3000$ units, the actual cost of producing one additional unit is approximately $$41\text{.}$

\begin{equation*} R'(3000) = -0.04(3000)+300 = 180 \end{equation*}

That is, the actual revenue to be realized from selling the $3001$st unit is approximately $$180\text{.}$

\begin{equation*} P'(3000) = -0.000009(3000)^{2}+0.04(3000)+100 = 139 \end{equation*}

That is, the actual profit realized from selling the $3001$st unit is approximately $$139\text{.}$
We use $E(p,q) = -\dfrac{p}{q} \dfrac{dq}{dp}\text{.}$ We first find $\frac{dq}{dp}$ using implicit differentiation:

\begin{equation*} \begin{split} \frac{d}{dp} (p) \amp = \frac{d}{dp} \left(-0.02q + 300\right) \\ 1 \amp = -0.02 \frac{dq}{dp} \\ \frac{dq}{dp} \amp = -50 \end{split} \end{equation*}

Therefore,

\begin{equation*} E(p,q) = \left(-\frac{p}{q}\right) \left(-50\right) = \frac{50 p}{q}\text{.} \end{equation*}

When $p=100\text{,}$ we must have

\begin{equation*} 100 = -0.02q + 300 \implies q = 10,000\text{.} \end{equation*}

Therefore,

\begin{equation*} E(100) = \frac{50 (100)}{10,000} = \frac{1}{2} \lt 1\text{.} \end{equation*}

Similarly, when $p=200\text{,}$ we must have

\begin{equation*} 200 = -0.02q+300 \implies q = 5,000\text{.} \end{equation*}

Therefore,

\begin{equation*} E(200) = \frac{50(200)}{5,000} = 2 > 1\text{.} \end{equation*}

We conclude that the demand is inelastic when $p=100$ and elastic when $p=200\text{.}$

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Exercises for Section 5.1.

Exercise 5.1.1.

For each demand equation, compute the elasticity of demand and determine whether or not the demand is elastic, unitary, or inelastic at the indicated price, $p\text{.}$

$q=-\frac{1}{2}p+10\text{,}$ $p=10\text{.}$
Answer
1, unitary.
Solution

\begin{equation*} \begin{aligned}E(p) \amp = -\dfrac{pf'(p)}{f(p)}\\ \amp = - \dfrac{p(-\frac{1}{2})}{-\frac{1}{2}p+10}\\ \amp = \dfrac{p}{20-p} \end{aligned} \end{equation*}

Therefore, $E(10)= \dfrac{10}{20-10}=1\text{,}$ and so the demand in unitary at $p=10\text{.}$
$q =-\frac{3}{2}p + 9\text{,}$ $p=1\text{.}$
Answer
$\frac{1}{5}\text{,}$ inelastic.
Solution

We use $E(p,q) = - \dfrac{p}{q} \cdot \diff{q}{p}\text{.}$ We first find $\diff{q}{p}\text{:}$

\begin{equation*} \begin{split} \diff{q}{p} \amp = \diff{}{p} \left(-\frac{3}{2} p + 9 \right)\\ \amp = -\frac{3}{2} \end{split} \end{equation*}

Therefore,

\begin{equation*} E(p) = \left(-\dfrac{p}{-\frac{3}{2}p + 9}\right)\left( -\frac{3}{2}\right) = \frac{p}{6-p}\text{.} \end{equation*}

When the unit price is set at $p=1\text{,}$ we have

\begin{equation*} E(1) = \frac{1}{5} \lt 1\text{,} \end{equation*}

and so the demand is inelastic.
$q+\frac{1}{3}p - 24 = 0\text{,}$ $p=3\text{.}$
Answer
$\frac{1}{5}\text{,}$ inelastic.
$0.4q+p=20\text{,}$ $p=12\text{.}$
Answer
$\frac{3}{2}\text{,}$ elastic.
$p=16-2q^{2}\text{,}$ $p=4\text{.}$
Answer
$\frac{1}{6}\text{,}$ inelastic.
Solution

We use $E(p,q) = - \dfrac{p}{q} \cdot \diff{q}{p}\text{.}$ We first find $\diff{q}{p}$ by implicit differentiation.

\begin{equation*} \begin{split} \diff{}{p} (p) \amp = \diff{}{p} \left(16-2q^2\right) \\ 1 \amp = -4 q \diff{q}{p} \\ \diff{q}{p} \amp = -\frac{1}{4q} \end{split} \end{equation*}

And so $E(p,q) = \dfrac{p}{4q^2}\text{.}$ When $p=4\text{,}$ we must have that

\begin{equation*} \begin{split} 16-2q^2 \amp = 4 \\ q^2 \amp = 6 \\ q \amp = \pm \sqrt{6} = \sqrt{6}, \end{split} \end{equation*}

where we have rejected the negative solution. Therefore,

\begin{equation*} E(4,\sqrt{6}) = \frac{4}{4(6)} = \frac{1}{6} \lt 1\text{.} \end{equation*}

Hence, when $p=4\text{,}$ the demand is inelastic.
$2p = 144-q^{2}\text{,}$ $p=48\text{.}$
Answer
$1\text{,}$ unitary.

Exercise 5.1.2.

It is determined that the demand equation for a certain product is

\begin{equation*} q = \frac{1}{5}(225-p^{2}) \ \ \ \ 0 \leq p \leq 15 \end{equation*}

where $q$ is the quantity demanded in units of hundreds and $p$ is the unit price in dollars.

For $p=8$ and $p=10\text{,}$ determine whether the demand elastic or inelastic. Answer
Inelastic when $p=8\text{.}$ Elastic when $p=10\text{.}$
Solution
We first compute $E(p)\text{:}$
\begin{equation*} \begin{split} E(p) \amp = -\frac{pf'(p)}{f(p)}\\ \amp = -\frac{p (-\frac{2}{5}p)}{\frac{1}{5}(225-p^2)}\\ \amp = \frac{2p^2}{225-p^2} \end{split} \end{equation*}

$E(8) = \dfrac{2\cdot 8^2}{225-8^2}=\dfrac{128}{161} \lt 1\text{,}$ and $E(10)=\dfrac{2\cdot 10^2}{225-10^2}=\dfrac{8}{5} > 1\text{.}$ Thus, the demand is inelastic when $p=8$ and elastic when $p=10\text{.}$
Determine the value of $p$ for which the demand is unitary. Answer
$p=8.66\text{.}$
Solution

We wish to find $p$ such that $E(p)=1\text{:}$

\begin{equation*} \begin{split} E(p) \amp = 1\\ 1\amp = \frac{2p^2}{225-p^2}\\ 225-p^2\amp = 2p^2\\ p^2 \amp = \frac{225}{3}\\ p \amp = \pm \sqrt{\frac{225}{3}} \end{split} \end{equation*}

And so the demand is unitary when $p\approx 8.66\text{.}$
If the unit price is lowered slightly from $10, will the revenue increase or decrease? Answer
Increase
Solution
Since the demand is elastic at $p=10\text{,}$ lowering the unit price slightly will result in an increase in revenue.
If the unit price is increased slightly from $8, will the revenue increase or decrease? Answer
Increase
Solution
Since the demand is inelastic at $p=8\text{,}$ raising the unit price slightly will results in an increase in revenue.

Exercise 5.1.3.

It is estimated that the quantity $q$ of fair tickets purchased is related to the ticket price $p$ by the demand equation

\begin{equation*} q = \frac{2}{3}\sqrt{36-p^{2}} \ \ \ 0 \leq p \leq 6\text{.} \end{equation*}

Currently, the price is set at $2 each.

Is the demand elastic or inelastic at this price? Answer
Inelastic
Solution

Let $q=f(p)=\frac{2}{3}\sqrt{36-p^2}\text{.}$ To determine if the price is elastic or inelastic when $p=2\text{,}$ we compute

\begin{equation*} \begin{split} E(p) \amp = -\frac{pf'(p)}{f(p)} \amp = -\frac{p\frac{1}{3}(36-p^2)^{-1/2}(-2p)}{\frac{2}{3}\sqrt{36-p^2}} \amp = \frac{p^2}{36-p^2} \ \ \ \ \ 0 \leq p \lt 6. \end{split} \end{equation*}

So $E(2) = 4/32 = 1/8 \lt 1\text{,}$ and the ticket price is therefore inelastic.
If the ticket price is increased, will the revenue increase or decrease? Answer
Increase
Solution
Since the unit price is inelastic at $p=2\text{,}$ a small increase from $2 will result in an increase in revenue.

Exercise 5.1.4.

The demand function for a certain product is

\begin{equation*} p = \sqrt{9-0.02q} \ \ \ \ 0 \leq q \leq 450 \end{equation*}

where $p$ is the unit price in hundreds of dollars and $q$ is the quantity demanded per week.

Calculate the elasticity of demand. Answer
$E(p) = \dfrac{2p^{2}}{9-p^{2}}\text{.}$
Solution

The demand function is given by

\begin{equation*} p = \sqrt{0-0.02q} \ \ \ \ \ 0 \leq q \leq 450\text{,} \end{equation*}

where $p$ is the unit price in hundreds of dollars and $q$ is the quantity demanded. We first find

\begin{equation*} q = f(p) = -50(p^2-9) \ \ \ \ \ 0 \leq p \leq 3\text{.} \end{equation*}

We are now in a position to calculate the elasticity of demand:

\begin{equation*} \begin{split} E(p) \amp = - \frac{pf'(p)}{f(p)} \amp = -\frac{p(-100p)}{-50(p^2-9)} \amp = \frac{2p^2}{9-p^2} \ \ \ \ \ 0 \leq p \lt 3 \end{split} \end{equation*}
Determine the values of $p$ for which the demand if inelastic, unitary and elastic. Answer
For $p \lt \sqrt{3}\text{,}$ demand is inelastic. For $p=\sqrt{3}\text{,}$ demand is unitary. And for $p > \sqrt{3}\text{,}$ demand is elastic.
Solution

The demand is unitary when

\begin{equation*} \begin{split} E(p) \amp = 1 \frac{2p^2}{9-p^2} \amp = 1 2p^2 \amp = 9-p^2 p^2 \amp = 3 p \amp = +\sqrt{3}, \end{split} \end{equation*}

or approximately $173 (note we reject the negative solution). Additionally, we see that for $\sqrt{3} \lt p \lt 3\text{,}$ $E(p) > 1$ and the price is elastic; for $0 \lt p \lt \sqrt{3}\text{,}$ $E(p) \lt 1$ and the price is inelastic.