1
$\begingroup$

I have attempted to solve this problem and would be grateful if someone looked over my solution

5% increase in price results in 5% decrease in total revenue. I need to determine price elasticity of demand. I have done calculations but I am not sure whether the way I have solved it is true.

$$TR=PQ\mapsto Q=\frac{TR}{P}\rightarrow Q=\frac{0.8TR}{1.2 P}=2/3$$ And then I wrote $$PE=\frac{\frac{8}{12}}{\frac{12}{10}}=5/9$$

But this solution doesn't appear in the test answer options. Can someone please check it and tell what is wrong?

Thanks in advance!

$\endgroup$
  • $\begingroup$ Perhaps a hint for future questions: You will get better answers when your questions are as general as possible. Also, then there are of more usefulness for future visitors. Here, the general question could've been on how to compute the demand-elasticity given the revenue-elasticity. $\endgroup$ – FooBar Apr 9 '15 at 15:06
1
$\begingroup$

Let's denote the elasticity of $X$ to $Y$ at $Y_0$ with $\eta_{X, Y}(Y_0)$ and solve for the demand ($Q$) elasticity in terms of the revenue elasticity ($R$).

We know that the elasticity is always given by

$$ \eta_{X,Y} = \frac{d X(Y)}{dY}\frac{Y}{X}$$

or equivalently (showing the equivalence is a useful exercise),

$$ \eta_{X,Y} = \frac{d \log X(Y))}{d \log Y}$$

Now,

$$ R(p) = p Q(p) \\ \eta_{R, p}(p) = R'(p) \frac{p}{pQ(p)}\\ \eta_{R, p}(p) = (Q(p) + pQ'(p_0) ) \frac{p}{pQ(p)} \\ \eta_{R, p}(p) = 1 + \frac{pQ'(p_0)}{Q} \\ \eta_{R, p}(p) = 1 + \eta_{Q, p}(p) \\ $$

Given that the left-hand-side elasticity is $-1$ at the initial price $p_0$, the right-hand-side elasticity must be $-2$:

$$-1 = 1 + \eta_{Q, p}(p_0) \Rightarrow \eta_{Q, p}(p_0) = -2$$

Your mistake

You are being sloppy with the mathematics. First you state

$$ Q = \frac{TR}{P}$$

But then you claim

$$ Q = \frac{0.8TR}{1.2P}$$

These two equations just cannot be true at the same time.

Perhaps you want to look at initial $TR_0$, at total revenue after price change $TR_1$, and then state that $TR_1/TR_0 = 0.95$.

$\endgroup$
  • $\begingroup$ How did you derive the second line? $\endgroup$ – Frank Booth Apr 9 '15 at 15:00
  • $\begingroup$ @orkhan471 I've added some general statements about elasticities. $\endgroup$ – FooBar Apr 9 '15 at 15:03

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.