1
$\begingroup$

Given a linear demand curve $q(p) = a - bp$, how would one find the price elasticity of demand at $p = 0$? The quantity that would be demanded is given: $Q_0 = a$

The formula for the price elasticity of demand then dictates:

$\epsilon_{p=0} = \frac{dq}{dp} *\frac {p}{q} = -b * \frac{p}{a} = -b * \frac{0}{a}= 0$,

this would mean perfectly inelastic. Thus, no matter the change in price there should be no change in quantity. This does not seem to make sense as at this point demand should be the most elastic. Where did I go wrong?

As an example one could think of a free good (e.g. straws), the demand for which can be observed. Then a levy is imposed and the new demand is being observed, allowing one to derive a linear demand function. How would one find the price elasticity for straws when they are free ($p=0$)?

| improve this question | |
$\endgroup$
  • $\begingroup$ Remember that elasticity measures percentual rates of change. How does q change per unit of percentual change of p? If p=0, then any percentual change of p results in no change at all because 0 multiplied by any number is 0. The elasticity is not just 0 at p=0, it is also continuous at p=0. If you compute the elasticity for small but positive prices, and then take the limit as p goes to 0, the limit is 0. $\endgroup$ – brunosalcedo Jun 11 at 17:35
  • $\begingroup$ Alright, yes that makes a lot of sense, I don't know why I got confused. I took the limits myself but I thought I must be doing something wrong. Thank you very much!! $\endgroup$ – Pycruncher Jun 11 at 18:42
0
$\begingroup$

The Answer can be found in the comments by user brunosalcedo:

"Remember that elasticity measures percentual rates of change. How does q change per unit of percentual change of p? If $p=0$, then any percentual change of p results in no change at all because 0 multiplied by any number is 0. The elasticity is not just 0 at p=0, it is also continuous at $p=0$. If you compute the elasticity for small but positive prices, and then take the limit as p goes to 0, the limit is 0."

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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