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Can unit elasticity be anywhere else on the demand curve other than the midpoint?

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    $\begingroup$ I find this question confusing... It's not hard to define a curve that extends infinitely. Where is the midpoint of such a curve? $\endgroup$ – dismalscience Oct 31 '15 at 23:22
  • $\begingroup$ Demand curve is normally NOT positive sloped, (other then Giffen or Veblen). Therefore it can not be infinite long. $\endgroup$ – Ben Oct 31 '15 at 23:47
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    $\begingroup$ Curves are not all straight, and some have asymptotic properties. You do know that, right? $\endgroup$ – dismalscience Oct 31 '15 at 23:52
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When the demand function is linear, $q = a-bp$, the only point were elasticity is unity is located in the midpoint of the demand curve (straight line). This is geometrical.

The demand line will cross the vertical $p$-axis at $p=a/b$ and the horizontal $q$ axis at $q=a$. For unitary elasticity (in absolute terms) we want

$$\frac{|dq/dp|}{q}\cdot p = \frac{bp}{a-bp}=1 \implies p=\frac {a}{2b} \implies q=\frac a2$$

So at unitary elasticity the corresponding price lies in the middle of the feasible price domain, and the corresponding quantity lies in the middle of the feasible quantity domain.

The related diagram is

enter image description here

The length $[AB]$ is equal to the length $[CD]$, and the length $[BC]$ is equal to the length $[DE]$. But this implies that the triangles $[ABC]$ and $[CDE]$ have two of their sides equal, so necessarily they will have also their third side equal. So $[AC] = [CE]$, which implies that the point of unitary elasticity $C$ is the midpoint of the demand line. Obviously, no other point can have unitary elasticity here.

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No, that is only true in the linear case. For a simple counterexample consider $$ D(p) = 1 - \sqrt{p}. $$ $$ \epsilon(p) = \frac{d D(p)}{dp} \cdot \frac{p}{D(p)} = -\frac{1}{2 \cdot \sqrt{p}} \cdot \frac{p}{1 - \sqrt{p}} = \frac{1}{2} \cdot \frac{\sqrt{p}}{1 - \sqrt{p}} $$ \begin{eqnarray*} \epsilon(p) & = & 1 \\ \\ \frac{1}{2} \cdot \frac{\sqrt{p}}{1 - \sqrt{p}} & = & 1 \\ \\ \sqrt{p} & = & 2 - 2 \cdot \sqrt{p} \\ \\ \sqrt{p} & = & \frac{2}{3} \\ \\ p & = & \frac{4}{9}. \end{eqnarray*} This is not in the middle of the demand curve in any sense.

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  • $\begingroup$ $D(p) = 1 - p^2$ would probably have been even simpler. $\endgroup$ – Giskard Oct 31 '15 at 23:51
  • $\begingroup$ All I want to know is in the linear case. Thanks. $\endgroup$ – Ben Nov 1 '15 at 0:14

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