# Find quantity from price elasticity

I have $P_1 = 24$, $Q_1 = 800000$, $P_2 = 32$ and price elasticity $e = -8$ and need to find the function $P(Q)$.

I guess I need to find $Q_2$ by

$$\frac{(Q_2-Q_1)/Q_1}{-(P_2-P_1)/P_1} = -8 \Leftrightarrow Q_2 = \frac{8800000}{3}$$

But does it make sense that the quantity will be so much higher by a price increment and thus not follow the law of demand? I know that $e = -8$ is a high price elasticity but this is quite extreme I think.

• You probably have an error in your equation. It's hard to be sure without seeing the context in your textbook (this could be an example intented to illustrate the concept of Giffen goods). But if that's not the case then the left hand side should be $[(Q_2-Q_1)/Q_1] / [(P_2-P_1)/P_1]$ (i.e. with no negative sign). – Ubiquitous Mar 9 '15 at 13:18
• But without the negative sign I get $Q_2 = -4000000/3$. Wouldn't it be a bad example if the quantity becomes negative? – Jamgreen Mar 9 '15 at 13:51
• Yes, and I now notice that you said the example concerns the potato market, which is the classic example of a Giffen good. I'd recommend that you read about such goods (if you haven't already) to see why price and quantity move in the same direction. – Ubiquitous Mar 9 '15 at 18:34

The good is not a giffen good. These goods have a positive elasticity of demand. Yours has a negative one, by assumption. It cannot yield a giffen good at all.

The error is in the negative sign that you (and also Thomas's answer) assumed.

The formula for the elasticity is:

$$e_d = \frac{\Delta Q}{\Delta P}\frac{P}{Q}$$

The negativeness of the elasticity is given by the sign of the derivative (a price increase leads to a fall in quantity demanded; the contrary in a Giffen good).

In your example, assuming as starting point $Q_1$ and $P_1$ (you could alternative use the end point, or the arc elasticity; see here), you get:

$$-8 = \frac{Q_2 - 800,000}{12}\frac{24}{800,000}$$

Which yields

$$Q_2 = -2,400,000$$

This is, in effect, a negative quantity. To see why this is the case, rewrite the elasticity function as follows:

$$e_d \frac{\Delta P}{P_1} = \frac{\Delta Q}{Q_1}$$

This is, the percentage change in the quantity is equivalent to the percentage change in prices times the elasticity.

Replacing the numbers, you get:

$$-4 = \frac{\Delta Q}{Q_1}$$

This is, the quantity falls by 400% percent!. That is why it becomes negative (from 800,000 to -2,400,000).

The origin of the problem is the massive change in price (50%) together with the massive elasticity of demand (-8), which gives the total change in quantity ($50\% \times -8 = -400\%$). A less dramatic combination would yield a positive $Q_2$.

You have that price elasticity is defined by:

$$\varepsilon_{d} = -\frac{\partial Q}{\partial P}\frac{P}{Q} \approx -\frac{P}{Q}\frac{\Delta Q}{\Delta P}$$

Initially we have $P = P_{1}$ and $Q=Q_{1}$ and after the perturbation we have $\Delta P = (P_{2} - P_{1})$ and $\Delta Q = (Q_{2} - Q_{1})$, therefore:

$$-\frac{P_{1}}{Q_{1}}\frac{Q_{2}-Q_{1}}{P_{2}-P_{1}}=-8 \implies Q_{2}-Q_{1}=\frac{8Q_{1}(P_{2}-P_{1})}{P_{1}}$$

And therefore:

$$Q_{2}=Q_{1}+\frac{8Q_{1}(P_{2}-P_{1})}{P_{1}}=\frac{8800000}{3}$$

This is so large, partly because the price elasticity is so large, but also probably because of the error in the finite difference approximation method over such large changes in $P$.

• Thank you. But does it make sense that the quantity increases even when the price increases? My textbook problem is about potatoes. – Jamgreen Mar 9 '15 at 10:31
• @Jamgreen Well in the case of potatoes it doesn't seem to make sense; although as $\varepsilon_{d} < 0$ it does make sense mathematically. Does your textbook give the formula for $\varepsilon_{d}$, sometimes they define it without the negative coefficient? Usually this sort of behaviour occurs with Giffen Goods. – Thomas Russell Mar 9 '15 at 10:34