<p>Like anything derived from calculus, the statement is guaranteed to hold true <strong><em>only for infinitesimal change</em></strong>. Unfortunately, since I don't know the entire demand function, I can't say much more than this.</p>

<p>A clear example of why this wouldn't work: starting at the same point \$100 and Q = 1000, we can decrease price to zero and total revenue would go down.</p>

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2019-10-25T05:49:31.643Z
"we can decrease price to zero and total revenue would go down" regardless of the increase in quantity sold.
Giskard
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2019-10-25T03:49:57.927Z
2019-10-25T03:49:57.927Z
Art
<blockquote>
  <p>My understanding is that when demand is elastic, a price reduction will increase total revenue.</p>
</blockquote>

<p>Some readers may want to know that an elastic demand can either be <a href="https://sites.google.com/site/economicsbasics/unitary-elastic-demand" rel="nofollow noreferrer"><em>unitary elastic</em></a> (<span class="math-container">$E_d = −1$</span>) or <a href="https://sites.google.com/site/economicsbasics/relatively-elastic-demand" rel="nofollow noreferrer"><em>relatively elastic</em></a> (<span class="math-container">$−∞ &lt; E_d &lt; −1$</span>). When unitary elastic, the above statement is not true, strictly speaking.</p>

<hr>

<blockquote>
  <p>Can anyone reconcile this with the point that when demand is elastic a price reduction should increase total revenue?</p>
</blockquote>

<p>Actually, to be logically compatible with your "point" (in which the observed change in total revenue is the objective reference of accuracy), your elasticity should be smaller than <span class="math-container">$1$</span> (in absolute value). Which you do not get since you are using <em>an approximation</em> formulae, not good enough in this case. Indeed, as outlined in the other answer, this approximation formulae is only good for infinitesimal changes. </p>

<p>That being explained, note that the <a href="https://en.wikipedia.org/wiki/Elasticity_(economics)#Mathematical_construct" rel="nofollow noreferrer">mathematical construct behind elasticities</a> is based on continuous sets. A better approximation is then</p>

<p><span class="math-container">$E_d = \frac{\ln(1240/1000)}{\ln(80/100)} = \frac{.215}{-.223} = -.964$</span></p>

<p>so <span class="math-container">$−1 &lt; E_d &lt; 0$</span>, which caracterizes a <a href="https://sites.google.com/site/economicsbasics/relatively-inelastic-demand" rel="nofollow noreferrer"><em>relatively inelastic demand</em></a>. In this case, as summarized on <a href="https://en.wikipedia.org/wiki/Price_elasticity_of_demand#Effect_on_total_revenue" rel="nofollow noreferrer">wikipedia</a>, when the price goes down, the total revenu decreases... <strong><em>which is what you see.</em></strong> Put differently, the "reconciliation" you are searching lies in the fact that you are actually <strong><em>not</em></strong> dealing with an elastic demand.</p>

<p><hr>
<strong><em>NB</em></strong>: The better approximation <em>here</em> lies in computing variations as <span class="math-container">$\left[\ln\frac{a}{b}\right]$</span> instead of <span class="math-container">$\left[-1 + \frac{a}{b}\right]$</span>. See <a href="https://economics.stackexchange.com/questions/23790/using-an-elasticity-to-evaluate-changes-in-a-dependent-variable/23792#23792">this answer of mine</a> for explanations.</p>

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2019-10-25T10:57:00.613Z
Any question @Gary ?
keepAlive
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2019-10-25T20:14:31.473Z
The Wikipedia article you yourself link to says "*The demand for a good is said to be elastic (or relatively elastic) when its PED is greater than one*", which is the concept used by the OP.
Giskard
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2019-10-25T20:41:33.980Z
@Giskard Sorry, I do not understand what you say. The PED of the OP is wrong. Correctly computed, it is lower than one *(in absolute value)*, which reconciles this with the change in total revenue she sees.
keepAlive
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2019-10-25T21:30:54.417Z
Seems I have misunderstood what you meant by "you do not precise how elastic your demand is".
Giskard
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2019-10-25T22:11:58.053Z
@Giskard ok, is it the wording better now?
keepAlive
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2019-10-25T22:27:17.693Z
Actually this is how I understood it. But why are you pointing out that the statement is only true for relatively elastic demand, when the OP already says that it is only true for elastic demand and as the Wikipedia article you yourself link to says these are synonyms? (see the quote in the comment above)
Giskard
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2019-10-25T22:44:51.553Z
@Giskard They are hyponyms of "elastic", but not synonyms: an elastic demand can either be unitary or relative. When unitary, the statement is not true, strictly speaking. Do you think that I should simply omit that? To me, it looks like the OP does not have this nuance in mind.
keepAlive
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2019-10-26T05:17:13.213Z
I don't think "elastic" is generally used in this way (i.e. $\geq 1$ rather than $>1$), the Wikipedia page certainly doesn't use it like this. There it is clearly treated as a synonym. Anyway, this is but a nuance.
Giskard
CC BY-SA 4.0
2019-10-25T10:53:05.717Z
2019-10-26T01:47:59.937Z
2019-10-26T01:47:59.937Z
keepAlive
<p>Like @Art said, I think the key thing to realise is that the elasticity is only well defined for small intervals. When you have a big change in price and quantity, the statement of &quot;revenue increases when price decreases if demand is elastic&quot; are no longer accurate / well defined.</p>
<p>We can see the problem mathematically. Let R, P, Q be the old revenue, price, and quantity respectively. Let R' be the new revenue, and <span class="math-container">$\Delta P$</span>, <span class="math-container">$\Delta Q$</span> to be the change in price and quantity, respectively. Let <span class="math-container">$\epsilon_D$</span> be the price elasticity of demand.</p>
<p><span class="math-container">\begin{align}
R &amp; \equiv P Q \\
R' &amp; \equiv (P + \Delta P)(Q + \Delta Q) \\
\frac{R'}{R} &amp; = 1 + \frac{\Delta Q}{Q} + \frac{\Delta P}{P} + \frac{\Delta P \Delta Q}{PQ} \\
\end{align}</span></p>
<p>Meanwhile,
<span class="math-container">\begin{align}
\epsilon_D &amp; \equiv \frac{\Delta Q / Q}{\Delta P / P} \leqslant 0 \\
\frac{\Delta Q}{Q} &amp; = \frac{\Delta P}{P} \epsilon_D
\end{align}</span></p>
<p>Substituting:</p>
<p><span class="math-container">\begin{align}
\frac{R'}{R} &amp; = 1 + \frac{\Delta P}{P}(\epsilon_D+1)+(\frac{\Delta P}{P})^2\epsilon_D
\end{align}</span></p>
<p>Suppose the price decreased and the revenue also decreased. The textbooks say this means demand is inelastic. What do maths say about <span class="math-container">$\epsilon_D$</span>?</p>
<p><span class="math-container">\begin{align}
-1 &lt; \frac{\Delta P}{P} &amp; &lt; 0 \\
\frac{R'}{R} &amp; &lt; 1 \\
\frac{\Delta P}{P}(\epsilon_D+1)+(\frac{\Delta P}{P})^2\epsilon_D &amp; &lt; 0 \\
\epsilon_D+1+\frac{\Delta P}{P}\epsilon_D &amp; &gt; 0 \\
\epsilon_D(1+\frac{\Delta P}{P}) &amp; &gt; -1 \\
\epsilon_D &amp; &gt; -\frac{1}{1+\frac{\Delta P}{P}} \\
\epsilon_D &amp; &gt; -\frac{P}{P+\Delta P} \neq -1
\end{align}</span></p>
<p>Suppose the price changed from 11 to 10. This means the elasticity could have any value that's greater than -1.1. If the elasticity is -1.05, which means demand is technically elastic, the revenue would still decrease in this scenario.</p>
<p>Notice that such discrepancy is reconciled when the changes are small:</p>
<p><span class="math-container">\begin{align}
\lim_{\Delta P \rightarrow 0} (-\frac{P}{P+\Delta P}) = -1
\end{align}</span></p>
<p>What does this say about real world?</p>
<ul>
<li>We will never have a smooth, true-to-reality mathematical expression of the demand curve irl.</li>
<li>We will never change the price by an infinitesimal amount irl.</li>
<li>We likely don't care if the total revenue stays exactly the same. As long as we have a good idea of which direction it's going.</li>
<li>If the demand is very elastic, the total revenue increases when you decrease the price. Vice versa.</li>
<li>When the demand elasticity is roughly -1, the total revenue will stay roughly the same when you change the price.</li>
</ul>

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2023-09-23T03:11:13.880Z
2023-09-23T03:11:13.880Z
Vibius
<p>A question in relation to Price Elasticity of Demand.</p>

<p>My understanding is that when demand is elastic, a price reduction will increase
total revenue.</p>

<p>If the price of a product changes from \$100 to \$80 (20% decrease) and quantity demanded increases from 1000 to 1240 (24% increase), PED = 24/-20 = -1.2 (elastic).</p>

<p>Total revenue at the original price (\$100) and quantity demanded (1000) = \$100,000 Total Revenue after the price change is \$80 x 1240 = \$99,200 so total revenue has actually fallen</p>

<p>Can anyone reconcile this with the point that when demand is elastic a price reduction should increase total revenue?</p>

CC BY-SA 4.0
2019-10-25T10:56:31.053Z
If any answer does what you want, please consider ticking it as correct. A reputation of 1 is enough to do it. I remind you this because newcomers often forget to do so. See [What should I do when someone answers my question?](https://economics.stackexchange.com/help/someone-answers) That being said, welcome on [EC](https://economics.stackexchange.com/).
keepAlive
<elasticity>
Price Elasticity of Demand - elastic demand and impact of price reductions on total revenue
Gary
# Price Elasticity of Demand - elastic demand and impact of price reductions on total revenue

(Asked by: username_0 on 2019-10-25)

A question in relation to Price Elasticity of Demand.

My understanding is that when demand is elastic, a price reduction will increase
total revenue.

If the price of a product changes from \$100 to \$80 (20% decrease) and quantity demanded increases from 1000 to 1240 (24% increase), PED = 24/-20 = -1.2 (elastic).

Total revenue at the original price (\$100) and quantity demanded (1000) = \$100,000 Total Revenue after the price change is \$80 x 1240 = \$99,200 so total revenue has actually fallen

Can anyone reconcile this with the point that when demand is elastic a price reduction should increase total revenue?

**Tags:** elasticity

> **Comments:**
> **username_1**: If any answer does what you want, please consider ticking it as correct. A reputation of 1 is enough to do it. I remind you this because newcomers often forget to do so. See [What should I do when someone answers my question?](https://economics.stackexchange.com/help/someone-answers) That being said, welcome on [EC](https://economics.stackexchange.com/).  

---

## 3 answers

### Answer by username_2 on 2019-10-25. Score: 2

Like anything derived from calculus, the statement is guaranteed to hold true ***only for infinitesimal change***. Unfortunately, since I don't know the entire demand function, I can't say much more than this.

A clear example of why this wouldn't work: starting at the same point \$100 and Q = 1000, we can decrease price to zero and total revenue would go down.

> **Comments:**
> **username_3**: "we can decrease price to zero and total revenue would go down" regardless of the increase in quantity sold.  

---

### Answer by username_1 on 2019-10-25. Score: 1

> My understanding is that when demand is elastic, a price reduction will increase total revenue.

Some readers may want to know that an elastic demand can either be [*unitary elastic*](https://sites.google.com/site/economicsbasics/unitary-elastic-demand) ($E\_d = −1$) or [*relatively elastic*](https://sites.google.com/site/economicsbasics/relatively-elastic-demand) ($−∞ < E\_d < −1$). When unitary elastic, the above statement is not true, strictly speaking.

---

> Can anyone reconcile this with the point that when demand is elastic a price reduction should increase total revenue?

Actually, to be logically compatible with your "point" (in which the observed change in total revenue is the objective reference of accuracy), your elasticity should be smaller than $1$ (in absolute value). Which you do not get since you are using *an approximation* formulae, not good enough in this case. Indeed, as outlined in the other answer, this approximation formulae is only good for infinitesimal changes.

That being explained, note that the [mathematical construct behind elasticities](https://en.wikipedia.org/wiki/Elasticity_(economics)#Mathematical_construct) is based on continuous sets. A better approximation is then

$E\_d = \frac{\ln(1240/1000)}{\ln(80/100)} = \frac{.215}{-.223} = -.964$

so $−1 < E\_d < 0$, which caracterizes a [*relatively inelastic demand*](https://sites.google.com/site/economicsbasics/relatively-inelastic-demand). In this case, as summarized on [wikipedia](https://en.wikipedia.org/wiki/Price_elasticity_of_demand#Effect_on_total_revenue), when the price goes down, the total revenu decreases... ***which is what you see.*** Put differently, the "reconciliation" you are searching lies in the fact that you are actually ***not*** dealing with an elastic demand.

---

***NB***: The better approximation *here* lies in computing variations as $\left[\ln\frac{a}{b}\right]$ instead of $\left[-1 + \frac{a}{b}\right]$. See [this answer of mine](https://economics.stackexchange.com/questions/23790/using-an-elasticity-to-evaluate-changes-in-a-dependent-variable/23792#23792) for explanations.

> **Comments:**
> **username_1**: Any question @username_0 ?  
> **username_3**: The Wikipedia article you yourself link to says "*The demand for a good is said to be elastic (or relatively elastic) when its PED is greater than one*", which is the concept used by the OP.  
> **username_1**: @username_3 Sorry, I do not understand what you say. The PED of the OP is wrong. Correctly computed, it is lower than one *(in absolute value)*, which reconciles this with the change in total revenue she sees.  
> **username_3**: Seems I have misunderstood what you meant by "you do not precise how elastic your demand is".  
> **username_1**: @username_3 ok, is it the wording better now?  
> **username_3**: Actually this is how I understood it. But why are you pointing out that the statement is only true for relatively elastic demand, when the OP already says that it is only true for elastic demand and as the Wikipedia article you yourself link to says these are synonyms? (see the quote in the comment above)  
> **username_1**: @username_3 They are hyponyms of "elastic", but not synonyms: an elastic demand can either be unitary or relative. When unitary, the statement is not true, strictly speaking. Do you think that I should simply omit that? To me, it looks like the OP does not have this nuance in mind.  
> **username_3**: I don't think "elastic" is generally used in this way (i.e. $\geq 1$ rather than $>1$), the Wikipedia page certainly doesn't use it like this. There it is clearly treated as a synonym. Anyway, this is but a nuance.  

---

### Answer by username_4 on 2023-09-23. Score: 0

Like @username_2 said, I think the key thing to realise is that the elasticity is only well defined for small intervals. When you have a big change in price and quantity, the statement of "revenue increases when price decreases if demand is elastic" are no longer accurate / well defined.

We can see the problem mathematically. Let R, P, Q be the old revenue, price, and quantity respectively. Let R' be the new revenue, and $\Delta P$, $\Delta Q$ to be the change in price and quantity, respectively. Let $\epsilon\_D$ be the price elasticity of demand.

\begin{align}
R & \equiv P Q \\
R' & \equiv (P + \Delta P)(Q + \Delta Q) \\
\frac{R'}{R} & = 1 + \frac{\Delta Q}{Q} + \frac{\Delta P}{P} + \frac{\Delta P \Delta Q}{PQ} \\
\end{align}

Meanwhile,
\begin{align}
\epsilon\_D & \equiv \frac{\Delta Q / Q}{\Delta P / P} \leqslant 0 \\
\frac{\Delta Q}{Q} & = \frac{\Delta P}{P} \epsilon\_D
\end{align}

Substituting:

\begin{align}
\frac{R'}{R} & = 1 + \frac{\Delta P}{P}(\epsilon\_D+1)+(\frac{\Delta P}{P})^2\epsilon\_D
\end{align}

Suppose the price decreased and the revenue also decreased. The textbooks say this means demand is inelastic. What do maths say about $\epsilon\_D$?

\begin{align}
-1 < \frac{\Delta P}{P} & < 0 \\
\frac{R'}{R} & < 1 \\
\frac{\Delta P}{P}(\epsilon\_D+1)+(\frac{\Delta P}{P})^2\epsilon\_D & < 0 \\
\epsilon\_D+1+\frac{\Delta P}{P}\epsilon\_D & > 0 \\
\epsilon\_D(1+\frac{\Delta P}{P}) & > -1 \\
\epsilon\_D & > -\frac{1}{1+\frac{\Delta P}{P}} \\
\epsilon\_D & > -\frac{P}{P+\Delta P} \neq -1
\end{align}

Suppose the price changed from 11 to 10. This means the elasticity could have any value that's greater than -1.1. If the elasticity is -1.05, which means demand is technically elastic, the revenue would still decrease in this scenario.

Notice that such discrepancy is reconciled when the changes are small:

\begin{align}
\lim\_{\Delta P \rightarrow 0} (-\frac{P}{P+\Delta P}) = -1
\end{align}

What does this say about real world?

* We will never have a smooth, true-to-reality mathematical expression of the demand curve irl.
* We will never change the price by an infinitesimal amount irl.
* We likely don't care if the total revenue stays exactly the same. As long as we have a good idea of which direction it's going.
* If the demand is very elastic, the total revenue increases when you decrease the price. Vice versa.
* When the demand elasticity is roughly -1, the total revenue will stay roughly the same when you change the price.


---

