# Economics: Elasticity Differential Equation Question

1. Aug 23, 2013

### Raze

Hi. Let me preface this by saying that I know nothing about economics. However, I learned a little bit about the concept of price elasticity of demand (that for something really elastic, if price goes up a little, demand will go down a lot), and I came across an equation relating price, demand, and elasticity. It was this:

dP/P = E*dD/D​

where P is price, E is elasticity, and D is demand, with P and D > 0.

In solving this, I get the following (unless I completely forgot how Calculus II works):

ln|P| = E*ln|D| + constant

P = DE + constant

So, I'm assuming that with an elastic good, if price goes up, demand will go down (I believe that is the Law of Demand). But this means that E has to be negative, does it not? Based on that equation, the only way I can see Price going up and Demand going down is if E is negative.

And if E = 0, then a good that is completely inelastic will have a constant price and a constant demand? (P = 1 + constant)? But then if E = 1, then it's P = D + constant, so what does an elasticity with 1 mean?

And if E > 0, then if price goes up, demand goes up? What kind of good would be like that? "Status" goods?

Anyway, would someone who is familiar with economics correct me or verify and/or elaborate on what I seem to get here? Thanks!

2. Aug 23, 2013

### vela

Staff Emeritus
I think you have E on the wrong side. The price elasticity of demand is given by
$$\varepsilon = \frac{dQ/Q}{dP/P}$$ where Q is the quantity demanded and P is the price.

You're assuming the elasticity is constant, which it probably isn't. You also made a mistake in your last step. You should've gotten $P = cD^E$ where $c$ is a constant. The correct relationship is what you got except with the price and quantity variables exchanged, $Q = cP^\varepsilon$, again assuming the elasticity is constant.

Yes, the elasticity is typically negative.

Starting from the correct equation, you would get Q = constant, which makes sense: regardless of what the price does, the quantity demanded stays constant for a perfectly inelastic good.

3. Aug 27, 2013

### Raze

First of all, thanks for your response!

Ah, that would be a pretty big mistake :) .

Yeah, when I e'd both sides I sort of screwed up what the plus sign meant due to bad handwriting on my scratch paper. Correct me if I'm wrong, but you integrate, and get $f(x) + c$, but when you 'e' it, it should be $e^{f(x) + c}$, not $e^{f(x)} + {e^c}$, which in this case would be $e^{(ln|D|^E + c)}$ = $e^{ln|D^E|}e^{c}$

which is,
$ce^{lnD^E}$ = $cD^E$, correct?

Of course, as you said, it should have been $Q = cP^E$ (why do they use Q for demand? Odd choice. Oh, wait, nevermind. dD/D is kind of awkward I guess, and Lord help you if you use D as a differential operator.)

In any case, if E is not a constant, you wouldn't still use this form of an equation? Say, E(Q,P,R...) = some quotient of partial derivatives, I guess?

Well, if I did the arithmetic right, then I should have $Q = cP^E$? So if Price goes up and demand goes down, E once again has to be negative. What a coincidence that my bad math gave a similar relationship between price and demand. Hmmm...

That actually makes more sense to me this way.

So, $Q = cP^0 => Q = c$, a constant. Demand doesn't change.

Now, if E = 1, you'd have Q = cP, which is just a straight line of slope c, right? (but then you'd have demand going up as price goes up). But what does that mean, if anything at all?

As far as the critical points here, I understand when E < 0. That's pretty much any normal good where demand goes down as price goes up. As for E = 0, again, that makes sense. It's a constant, so demand doesn't change.

But I'm not clear on any significance for E = 1, and also, when would revenue be maximized? I suppose with a revenue function I could find that just using the maxima finding technique from calculus, but I'm not sure from this equation.

Thanks again for all your work here.

4. Aug 28, 2013

### vela

Staff Emeritus
Correct.

Yes. It's simply that the integration is more difficult if the elasticity isn't constant.

It's a reversal of the law of demand. As you said in your original post, it would correspond to a status good, one where the higher price makes it more desirable for whatever reason.

Revenue won't have a maximum because you can always increase revenue by selling one more item. Increasing production, however, tends to increase costs, so there's some quantity at which you can maximize profit.