Difficulty: Intermediate | Prerequisites: Basic differentiation, understanding of demand curves and revenue. This is an extra credit topic.
Elasticity of demand measures how sensitive consumer demand is to changes in price. It connects calculus (derivatives) to economics (pricing strategy). This topic appears as extra credit on the exam, so it is worth studying if you want to pick up additional points. It ties together derivatives, the demand equation, and revenue maximisation in one framework.
Elasticity E = −(p/q) · (dq/dp). If E > 1, demand is elastic (price-sensitive). If E < 1, demand is inelastic. Revenue is maximised when E = 1 (unit elastic).
Elasticity of demand (E)
A measure of how much the quantity demanded changes in response to a change in price. The formula is E = −(p/q) · (dq/dp). The negative sign makes E positive (since dq/dp is typically negative). Think of it as: a ratio that tells you whether a price increase will help or hurt total revenue.
Elastic demand (E > 1)
Demand is highly responsive to price changes. A small price increase causes a proportionally larger drop in quantity demanded. Raising the price decreases total revenue.
Inelastic demand (E < 1)
Demand is not very responsive to price changes. A price increase causes a proportionally smaller drop in quantity demanded. Raising the price increases total revenue.
Unit elastic (E = 1)
The percentage change in quantity exactly matches the percentage change in price. This is where revenue is maximised.
E = −(p/q) · (dq/dp)
You need the demand equation relating q (quantity) and p (price), then differentiate q with respect to p.
Revenue R = p · q. Revenue is maximised at the price where E = 1. This gives you a way to find the optimal price without writing out R(p) explicitly.
Worked example (extra credit problem 1): q = 342.5 / p^(0.5314)
dq/dp = 342.5 · (−0.5314) · p^(−1.5314) = −181.995 / p^(1.5314)
E = −(p/q) · (dq/dp) = −(p / [342.5/p^(0.5314)]) · (−181.995/p^(1.5314))
Simplify: E = (p^(1.5314) / 342.5) · (181.995 / p^(1.5314)) = 181.995 / 342.5 = 0.5314
E = 0.5314 < 1 for all prices. Demand is always inelastic. (This is because the exponent on p is less than 1.)
Worked example (extra credit problem 2): p = 90 − 10q, find elasticity at p = $50.
Solve for q: q = (90 − p)/10 = 9 − p/10
dq/dp = −1/10
At p = 50: q = 9 − 5 = 4
E = −(50/4)(−1/10) = 50/40 = 1.25
E > 1, so demand is elastic at p = $50.
Revenue maximisation (problem 2b-d):
Revenue is maximised at E = 1: −(p/q)(−1/10) = 1 gives p/(10q) = 1, so p = 10q.
Substitute into the demand equation: p = 90 − 10q and p = 10q, so 10q = 90 − 10q, 20q = 90, q = 4.5 (thousands).
Revenue-maximising price: p = 10(4.5) = $45.
Maximum revenue: R = pq = 45 · 4.5 = 202.5 (in thousands of dollars), or $202,500.
Students often forget the negative sign in the elasticity formula and end up with a negative E. The formula has a built-in negative to make E positive.
Elastic does not mean "demand drops to zero." It means the percentage change in quantity is larger than the percentage change in price.
Students sometimes confuse which direction is which: E > 1 is elastic (price-sensitive), E < 1 is inelastic. A mnemonic: "elastic stretches more" (bigger response).
⚠️ Revenue maximisation at E = 1 is a key result. Expect a question that asks you to find the price that maximises revenue using elasticity.
⚠️ The beef problem (extra credit 1) has constant elasticity. Recognise that when q = k/p^n, the elasticity is always n, regardless of price.
Fill in the blank: Demand is elastic when E is ______ than 1.
Greater.
True or False: Revenue is maximised when demand is inelastic.
False. Revenue is maximised when E = 1 (unit elastic).
Fill in the blank: The elasticity formula is E = −()/() · dq/dp.
p / q.
Q: The demand for McGuffins is p = 90 − 10q. Is demand elastic or inelastic at p = $50?
A: q = (90 − 50)/10 = 4, dq/dp = −1/10. E = −(50/4)(−1/10) = 1.25. Since E > 1, demand is elastic.
Q: At what price is revenue maximised for the McGuffin demand equation p = 90 − 10q?
A: Set E = 1. This gives p = 10q. Combined with p = 90 − 10q: 10q = 90 − 10q, q = 4.5, p = $45.
Q: What is the maximum revenue?
A: R = p · q = 45 · 4.5 = $202.50 (in thousands), or $202,500.
Q: For q = 342.5/p^(0.5314), is demand ever elastic?
A: E = 0.5314 for all prices (constant elasticity equals the exponent). Since 0.5314 < 1, demand is always inelastic.
Elasticity ties together the demand curve (from economics) with the derivative (from calculus). It connects to revenue maximisation, which is an optimisation problem at heart. The idea of marginal revenue equalling zero at maximum revenue is equivalent to E = 1.
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