2
$\begingroup$

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

$\endgroup$
3
  • 2
    $\begingroup$ I find this question confusing... It's not hard to define a curve that extends infinitely. Where is the midpoint of such a curve? $\endgroup$ Oct 31, 2015 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, 2015 at 23:47
  • 2
    $\begingroup$ Curves are not all straight, and some have asymptotic properties. You do know that, right? $\endgroup$ Oct 31, 2015 at 23:52

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ $D(p) = 1 - p^2$ would probably have been even simpler. $\endgroup$
    – Giskard
    Oct 31, 2015 at 23:51
  • $\begingroup$ All I want to know is in the linear case. Thanks. $\endgroup$
    – Ben
    Nov 1, 2015 at 0:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.