enter image description hereThis 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??

  • $\begingroup$ 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) $\endgroup$
    – Henry
    Jan 7 '19 at 22:57

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 $p_i > p_i^0$

  • $\begingroup$ Thanks. Do you have any idea about the economical interpretation of this inequality? $\endgroup$ Jan 8 '19 at 9:10
  • 1
    $\begingroup$ @akd It is kind of difficult to say without more context, but probably the author is going after some form of elasticity discussion $\endgroup$
    – caverac
    Jan 8 '19 at 9:20

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