<p>Everything comes from the constraint</p>

<p><span class="math-container">$$
C(p, \gamma^0) \leq S(p, \gamma^0, z^0_k) \tag{C.17}
$$</span></p>

<p>By construction at <span class="math-container">$p_0$</span> these two things are equal:</p>

<p><span class="math-container">$$
C(p^0, \gamma^0) = S(p^0, \gamma^0, z^0_k) \tag{1}
$$</span></p>

<p>Consider a very small <em>positive</em> number <span class="math-container">$\delta p_i$</span>, and note that (I dropped the other components of <span class="math-container">$p$</span> since they remain constant)</p>

<p><span class="math-container">\begin{align}
C(p_i^0 - \delta p_i^0, \gamma^0) &amp;&lt;  S(p^0_i - \delta p_i^0, \gamma^0, z^0_k) \\ 
C(p_i^0 - \delta p_i^0, \gamma^0) - C(p_i^0, \gamma^0) &amp;&lt;  S(p^0_i - \delta p_i^0, \gamma^0, z^0_k) - C(p_i^0, \gamma^0)\\
C(p_i^0 - \delta p_i^0, \gamma^0) - C(p_i^0, \gamma^0) &amp;&lt;  S(p^0_i - \delta p_i^0, \gamma^0, z^0_k) - S(p_i^0, \gamma^0, z_k^0) \\
\frac{C(p_i^0 - \delta p_i^0, \gamma^0) - C(p_i^0, \gamma^0)}{\delta p_i^0} &amp;&lt;  \frac{S(p^0_i - \delta p_i^0, \gamma^0, z^0_k) - S(p_i^0, \gamma^0, z_k^0)}{\delta p_i^0}\\
\lim_{\delta p_i^0\to 0^+}\frac{C(p_i^0 - \delta p_i^0, \gamma^0) - C(p_i^0, \gamma^0)}{\delta p_i^0} &amp;&lt;  \lim_{\delta p_i^0\to 0^+} \frac{S(p^0_i - \delta p_i^0, \gamma^0, z^0_k) - S(p_i^0, \gamma^0, z_k^0)}{\delta p_i^0} \\
-\frac{\partial C}{\partial p_i} &amp;&lt; -\frac{\partial S}{\partial p_i}
\end{align}</span></p>

<p>or in other words,</p>

<p><span class="math-container">$$
\frac{\partial C}{\partial p_i} &gt; \frac{\partial S}{\partial p_i} ~~\mbox{for}~~ p_i &lt; p_i^0 \tag{2}
$$</span></p>

<p>or in plain English: <span class="math-container">$S$</span> is flatter than <span class="math-container">$C$</span> for <span class="math-container">$p_i &lt; p_i^0$</span>. You can repeat the same argument for <span class="math-container">$p_i &gt; p_i^0$</span></p>

CC BY-SA 4.0
2019-01-08T09:10:00.287Z
Thanks. Do you have any idea about the economical interpretation of this inequality?
Richardlion
CC BY-SA 4.0
2019-01-08T09:20:41.607Z
@akd It is kind of difficult to say without more context, but probably the author is going after some form of elasticity discussion
caverac
CC BY-SA 4.0
2019-01-07T22:38:27.280Z
2019-01-07T22:38:27.280Z
caverac
<p><a href="https://i.sstatic.net/JaEJO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JaEJO.png" alt="enter image description here"></a>This is an image of page 133 of Microeconomics; Gravell and Rees.In the fourth line there is a statement about the slope of those curves. Can anyone please explain me the reason??</p>

CC BY-SA 4.0
2019-01-07T22:57:27.853Z
I would have thought the diagram was clear enough (at least on the right hand side) where both curves are increasing and the $S$ curve is the higher curve.  So the $S$ curve must be steeper upwards when moving right from the tangent  (otherwise it would be lower than the $C$ curve) and the $C$ curve must be steeper downwards when moving left (otherwise it would be higher than the $S$ curve)
Henry
<microeconomics>
Short-run and Long-run cost against the price of the variable input
Richardlion
# Short-run and Long-run cost against the price of the variable input

(Asked by: username_0 on 2019-01-07)

[![enter image description here](https://i.sstatic.net/JaEJO.png)](https://i.sstatic.net/JaEJO.png)This is an image of page 133 of Microeconomics; Gravell and Rees.In the fourth line there is a statement about the slope of those curves. Can anyone please explain me the reason??

**Tags:** microeconomics

> **Comments:**
> **username_1**: I would have thought the diagram was clear enough (at least on the right hand side) where both curves are increasing and the $S$ curve is the higher curve.  So the $S$ curve must be steeper upwards when moving right from the tangent  (otherwise it would be lower than the $C$ curve) and the $C$ curve must be steeper downwards when moving left (otherwise it would be higher than the $S$ curve)  

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## 1 answers

### Answer by username_2 on 2019-01-07. Score: 1 | Accepted answer

Everything comes from the constraint

$$
C(p, \gamma^0) \leq S(p, \gamma^0, z^0\_k) \tag{C.17}
$$

By construction at $p\_0$ these two things are equal:

$$
C(p^0, \gamma^0) = S(p^0, \gamma^0, z^0\_k) \tag{1}
$$

Consider a very small *positive* number $\delta p\_i$, and note that (I dropped the other components of $p$ since they remain constant)

\begin{align}
C(p\_i^0 - \delta p\_i^0, \gamma^0) &< S(p^0\_i - \delta p\_i^0, \gamma^0, z^0\_k) \\
C(p\_i^0 - \delta p\_i^0, \gamma^0) - C(p\_i^0, \gamma^0) &< S(p^0\_i - \delta p\_i^0, \gamma^0, z^0\_k) - C(p\_i^0, \gamma^0)\\
C(p\_i^0 - \delta p\_i^0, \gamma^0) - C(p\_i^0, \gamma^0) &< S(p^0\_i - \delta p\_i^0, \gamma^0, z^0\_k) - S(p\_i^0, \gamma^0, z\_k^0) \\
\frac{C(p\_i^0 - \delta p\_i^0, \gamma^0) - C(p\_i^0, \gamma^0)}{\delta p\_i^0} &< \frac{S(p^0\_i - \delta p\_i^0, \gamma^0, z^0\_k) - S(p\_i^0, \gamma^0, z\_k^0)}{\delta p\_i^0}\\
\lim\_{\delta p\_i^0\to 0^+}\frac{C(p\_i^0 - \delta p\_i^0, \gamma^0) - C(p\_i^0, \gamma^0)}{\delta p\_i^0} &< \lim\_{\delta p\_i^0\to 0^+} \frac{S(p^0\_i - \delta p\_i^0, \gamma^0, z^0\_k) - S(p\_i^0, \gamma^0, z\_k^0)}{\delta p\_i^0} \\
-\frac{\partial C}{\partial p\_i} &< -\frac{\partial S}{\partial p\_i}
\end{align}

or in other words,

$$
\frac{\partial C}{\partial p\_i} > \frac{\partial S}{\partial p\_i} ~~\mbox{for}~~ p\_i < p\_i^0 \tag{2}
$$

or in plain English: $S$ is flatter than $C$ for $p\_i < p\_i^0$. You can repeat the same argument for $p\_i > p\_i^0$

> **Comments:**
> **username_0**: Thanks. Do you have any idea about the economical interpretation of this inequality?  
> **username_2**: @akd It is kind of difficult to say without more context, but probably the author is going after some form of elasticity discussion  

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