<p>Revenue <span class="math-container">$R(p)$</span>, which depends on price can be written as:
<span class="math-container">$$
R(p) = p D(p),
$$</span>
where <span class="math-container">$D(p)$</span> is the demand at price <span class="math-container">$p$</span>. Then taking the derivative with respect to <span class="math-container">$p$</span> gives:
<span class="math-container">$$
\begin{align*}
R'(p) &amp;= D(p) + p D'(p) \\
&amp;= D(p) \left(1 + p \frac{D'(p)}{D(p)}\right)\\
&amp;= D(p) \left(1 + \varepsilon^D_p\right)
\end{align*}
$$</span>
Here <span class="math-container">$\varepsilon^D_p$</span> is the price elasticity of demand. Notice that the sign of <span class="math-container">$R'(p)$</span> is entirely determined by the sign of <span class="math-container">$1 + \varepsilon^D_p$</span>.</p>
<ul>
<li>If demand is inelastic then <span class="math-container">$\varepsilon^D_p \ge -1$</span> so <span class="math-container">$R'(p) \ge 0$</span>.</li>
<li>If demand is elastic then <span class="math-container">$\varepsilon^D_p &lt; -1$</span> so <span class="math-container">$R'(p) &lt; 0$</span>.</li>
</ul>
<p>Intuitively, revenue is price times demand. Under inelastic demand, quantity will decrease but not so fast (proportionally as price increases, so total revenue will increase. If demand is elastic, then demand will decrease faster (proportionally) than the price increases, so revenue will drop.</p>

CC BY-SA 4.0
2022-03-22T16:01:12.740Z
2022-03-22T16:01:12.740Z
tdm
<p>I am learning about the total revenue test for AP Microeconomics. I know that demand curves tend to have different price elasticities when calculating using different points (one interval that is elastic, one interval that is unitary, and one interval that is inelastic).</p>
<p>However, the total revenue test seems to determine if the demand for a product is elastic or inelastic. In this case, won't we get different results from looking at total revenue when raising prices beginning at different initial prices?</p>
<p>For the same curve:</p>
<ol>
<li><p>Starting at a point within the elastic interval and raising prices will decrease total revenue -&gt; elastic demand</p>
</li>
<li><p>Starting at a point within the inelastic interval and raising prices will increase total revenue -&gt; inelastic demand</p>
</li>
</ol>
<p>How does this work?</p>

<microeconomics>
How to interpret total revenue test for demand curves with non-constant price elasticity?
Andy
# How to interpret total revenue test for demand curves with non-constant price elasticity?

(Asked by: username_0 on 2022-03-22)

I am learning about the total revenue test for AP Microeconomics. I know that demand curves tend to have different price elasticities when calculating using different points (one interval that is elastic, one interval that is unitary, and one interval that is inelastic).

However, the total revenue test seems to determine if the demand for a product is elastic or inelastic. In this case, won't we get different results from looking at total revenue when raising prices beginning at different initial prices?

For the same curve:

1. Starting at a point within the elastic interval and raising prices will decrease total revenue -> elastic demand
2. Starting at a point within the inelastic interval and raising prices will increase total revenue -> inelastic demand

How does this work?

**Tags:** microeconomics

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## 1 answers

### Answer by username_1 on 2022-03-22. Score: 5 | Accepted answer

Revenue $R(p)$, which depends on price can be written as:
$$
R(p) = p D(p),
$$
where $D(p)$ is the demand at price $p$. Then taking the derivative with respect to $p$ gives:
$$
\begin{align\*}
R'(p) &= D(p) + p D'(p) \\
&= D(p) \left(1 + p \frac{D'(p)}{D(p)}\right)\\
&= D(p) \left(1 + \varepsilon^D\_p\right)
\end{align\*}
$$
Here $\varepsilon^D\_p$ is the price elasticity of demand. Notice that the sign of $R'(p)$ is entirely determined by the sign of $1 + \varepsilon^D\_p$.

* If demand is inelastic then $\varepsilon^D\_p \ge -1$ so $R'(p) \ge 0$.
* If demand is elastic then $\varepsilon^D\_p < -1$ so $R'(p) < 0$.

Intuitively, revenue is price times demand. Under inelastic demand, quantity will decrease but not so fast (proportionally as price increases, so total revenue will increase. If demand is elastic, then demand will decrease faster (proportionally) than the price increases, so revenue will drop.


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