In my textbook, it's stated that:

When $\epsilon < -1$, demand is elastic and raising price will result in smaller income, while lowering price will result in bigger income.

When $\epsilon = -1$, demand is neither elastic nor inelastic and change in price won't result in change in income.

When $\epsilon > -1$, demand is inelastic and raising price will result in bigger income, while lowering price will result in smaller income.

$\epsilon = \%\Delta Q / \%\Delta P$.

This is the exercise I found confusing:

Old price: 5

New price: 6

Old quantity: 25

New quantity: 20

Calculate elasticity

This is my solution:

$\% \Delta P = \frac{\text{new price } - \text{ old price}} {\text{old price}} = \frac{6 - 5} 5 = 0.2$

$\%\Delta Q = \frac{\text{new quantity } - \text{ old quantity}} { \text{old quantity}} = \frac{20 - 25} {25} = -0.2$

$\epsilon = \%\Delta Q / \%\Delta P = -0.2 / 0.2 = -1$

$$$$ This is why I am confused:

$\text{Old income} = \text{old price} \times \text{old quantity} = 5 \times 25 = 125$

$\text{New income} = \text{new price} \times \text{new quantity} = 6 \times 20 = 120$

Old income does not equal new income even though elasticity is -1!

What am I doing wrong? Am I misunderstanding the textbook?

$$$$ Edit: the answer provided is $\epsilon = 1.22$ but I have no idea where it comes from.


2 Answers 2


We discourage numeric questions as they are unlikely to be useful for future visitors but this is a very good example of why using non-marginal quantities can be misleading.

The exact definition of the price elasticity of demand is $$ \epsilon(p) = \frac{ \text{d} D(p)}{ \text{d} p} \cdot \frac{p}{D(p)}. $$ (In your notation $D(p) = Q$.)
By straightforward calculation you can show \begin{eqnarray*} \max_p & & p \cdot D(p) \\ \\ p \cdot \frac{ \text{d} D(p)}{ \text{d} p} + D(p) & = & 0 \\ \\ p \cdot \frac{ \text{d} D(p)}{ \text{d} p} & = & - D(p) \\ \\ \frac{p}{D(p)} \cdot \frac{ \text{d} D(p)}{ \text{d} p} & = & - 1 \\ \\ \epsilon(p) & = & - 1. \end{eqnarray*} Whether a marginal price change increases or decreases revenue when $\epsilon(p) \neq - 1$ can be seen from a similar calculation.

Note that the reasoning used above only stated that revenue reached its local maximum. It is possible that several local maxima exist and the global maxima is elsewhere.

So $\epsilon(p) = - 1$ implies that the revenue cannot be increased by small (marginal) changes in $p$. So far I have used the concept of point elasticity. But your textbook uses arc elasticity, which measures non-marginal changes: $$ \frac{\Delta Q}{Q} / \frac{\Delta p}{p} $$ Some reasons to do this:

  1. Arc elasticity can be measured empirically, point elasticity can only be approximated.
  2. In case of small changes the two should be reasonably close.
  3. Point elasticity requires some knowledge of calculus.

There are however also some problems with this:

  1. No guarantuees are made by the $\epsilon(p) = - 1$ condition for non-marginal price changes.
  2. What base you should use for arc elasticity is not clear at all. You used $$ \frac{Q_{new} - Q_{old}}{Q_{old}} \cdot \frac{p_{old}}{p_{new} - p_{old}}. $$ Why is the old quantity and price the reference for the rate of change? Why is it not the new quantity and price $$ \frac{Q_{new} - Q_{old}}{Q_{new}} \cdot \frac{p_{new}}{p_{new} - p_{old}} $$ or the average of the two $$ \frac{Q_{new} - Q_{old}}{\frac{Q_{new} + Q_{old}}{2}} \cdot \frac{\frac{p_{new} + p_{old}}{2}}{p_{new} - p_{old}}? $$

These bases give you different values for arc elasticity. The second base gives you $1.2\dot{2}$ so it seems to be the one your textbook expects. I guess it is best if you see the difference between point elasticity that is used by the rule and arc elasticity. Other than this you have to ask your professor what definiton of elasticity she/he expects you to use.

  • $\begingroup$ Thanks. I will consult my economics teacher who is also a mathematics teacher (thus also knows how to calculate derivatives and antiderivatives of functions) at the first chance I get. For know, I will use the arithmetic mean as the reference for the rate of change as it seems it is the one the textbook is using, too. Now I know that elasticity is simply a function and I hope to find out how to derive it very soon. $\endgroup$ Commented Feb 15, 2016 at 13:45

The point elasticity of a function with respect to its argument, is useful mainly to characterize some specific point of interest, for example the profit maximizing point of a monopolist (where we learn that the point elasticity has to be greater than unity in absolute terms, see this post).

But when discrete changes are considered, the point elasticity is at best an approximation, and we are into trouble anyway, since the "intuitive" approach to consider the elasticity as being "percentage change over percentage change"

$$\varepsilon_{q,p} = \frac {\%\Delta q(p)}{\%\Delta p} = \frac {(q_1-q_0)/q_0}{(p_1-p_0)/p_0} \tag{1}$$

(and which is what the OP attempted), has the problem of not being symmetric : if we consider moving from $q_1$ to $q_0$ (instead of moving from $q_0$ to $q_1$), we will obtain a different value using $(1)$.

This is why the arc elasticity was devised which divides each interval of change by its mid-point, rather than its boundary:

$$\text {Arc} \;\varepsilon_{q,p}=\frac {\frac{(q_1-q_0)}{(q_1+q_0)/2}}{\frac{(p_1-p_0)}{(p_1+p_0/2}}$$

See this post also, on Arc elasticity and its properties.

Applying the Arc elasticity on the specific numerical problem, gives indeed $1.22$ (in absolute terms), which is the answer provided.

Note that this measure also calculates correctly the change in Total Revenue. Using the Difference operator $\Delta$ we have that

$$\Delta (pq) = \Delta p \cdot q + \Delta q \cdot p$$

$$\implies \frac {\Delta (pq)}{pq} = \frac {\Delta p}{p} + \frac {\Delta q}{q}$$

$$\implies \frac {\Delta (pq)}{pq} = \frac {\Delta p}{p} \cdot \Big [1+\frac {\Delta q/q}{\Delta p/p}\Big]$$

Using the mid-point calculations associated with the arc-elasticity we obtain for the specific numerical example,

$$\frac {\Delta (pq)}{pq} = \frac {1}{5.5}\cdot [1-1.22] = -0.04$$

And indeed,

$$\frac {120-125}{125} = \frac {5}{125} = -0.04$$


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