Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: elasticity and economic optimization (201-103)

This sheet covers the second chapter specific to calculus in the profile “with mathematics”: elasticity, and the optimizations that follow from it. An elasticity is a derivative in disguise, but the disguise changes everything you are allowed to say about it, and that is where papers lose their marks.

It is written for Social Science students at CEGEP taking the course 201-103, differential calculus. The corrected exercise set for the same chapter goes back over every move, from the elastic range to the revenue from a tax.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Elasticity is a ratio of RELATIVE changes: unitless, negative for a decreasing demand, and different at every point of the same line. It is its ABSOLUTE VALUE that decides whether raising the price pays or costs.

This chapter is part of 201-103, Differential Calculus for Social Science

The essentials

What an elasticity is, and what it is not

• $E=\dfrac{dq}{dp}\times \dfrac{p}{q}$ : a ratio of RELATIVE changes, so unitless, and negative for a decreasing demand.
• On a LINEAR demand, the slope $\dfrac{dq}{dp}$ is constant and the elasticity varies at every point: $-1$ at the midpoint, infinite near the maximum price, zero near the zero price.
• $\left|E\right|<1$ : INELASTIC demand. $\left|E\right|=1$ : unit elastic. $\left|E\right|>1$ : ELASTIC.
• ARC elasticity between two points: $\dfrac{\Delta q/\overline{q}}{\Delta p/\overline{p}}$ , with the AVERAGES in the denominator, so that the result does not depend on the direction of travel.

The three points are on the same line, so they have the same slope: it is the ratio

\dfrac{p}{q}

that changes, and it takes the elasticity from very large to almost zero.

Elasticity is a logarithmic derivative: it answers “how many PERCENT”, never “how many units”.

What elasticity decides

• $\dfrac{dR}{dp}=q\left(1+E\right)$ : in the INELASTIC range, raising the price INCREASES revenue; in the ELASTIC range, it lowers it.
• Total revenue is therefore greatest exactly where $E=-1$ .
• $R'(q)=p\left(1+\dfrac{1}{E}\right)$ : marginal revenue is NEGATIVE in the inelastic range, so the profit optimum is ALWAYS in the elastic range.
• Elasticity is larger in absolute value in the LONG run than in the short run, because adjusting takes time.

The revenue column rises, peaks at

1{,}250

and falls back, and the elasticity equals

-1

on exactly that row.

These two formulas spare you any case-by-case reasoning: read the range, deduce the direction of the change.

The other elasticities, and the two techniques of the chapter

• INCOME elasticity: negative for an inferior good, between $0$ and $1$ for a necessity, greater than $1$ for a luxury good.
• CROSS-PRICE elasticity: POSITIVE for substitutes, NEGATIVE for complements, zero for unrelated goods.
• CONSTRAINED OPTIMIZATION, in four moves: identify the objective, write the constraint, substitute to be left with one variable WITH ITS DOMAIN, differentiate and check, endpoints included.
• RELATED RATES: $\dfrac{dC}{dt}=C'(q)\times \dfrac{dq}{dt}$ . Profit can fall while production rises, as soon as $q$ goes past the optimum.

Income elasticity sorts goods: negative, the good is INFERIOR (people buy less as they get richer); between

0

and

1

, a necessity; beyond

1

, a luxury.

Unit tax $t$ on a linear demand with constant marginal cost: the optimal price rises by only $0.5t$ , and the tax revenue $T(t)$ eventually decreases.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Confusing the slope with the elasticity

2 marks, and every conclusion about the ranges

What not to write

“The demand is $q=400-5p$ : the elasticity is $-5$ everywhere.”

What to write

“ $-5$ is the SLOPE $\dfrac{dq}{dp}$ . The elasticity is $-5\times\dfrac{p}{q}$ : $-0.33$ at $p=20$ and $-1.11$ at $p=42$ .”

On the demand

q=400-5p

, the slope is

-5

at every point. The elasticity multiplies this slope by

\frac{p}{q}

-0.33

20

dollars,

-1.11

42

dollars.

Why: The slope compares ABSOLUTE changes, the elasticity RELATIVE ones. The factor $\dfrac{p}{q}$ changes all along the line, from very large to almost zero.

2. Losing the sign of the elasticity

0.5 mark per occurrence, and more on a cross-price elasticity

What not to write

“The elasticity is $2$ : the demand is elastic.”

What to write

“The price elasticity of a decreasing demand is NEGATIVE: $E=-2$ . It is its ABSOLUTE VALUE, $2$ , that exceeds $1$ and makes the demand elastic.”

Why: On the price elasticity, the sign is always the same and forgetting it is minor. On the income and cross-price elasticities, the SIGN carries all the information.

3. Believing that a price rise always increases revenue

the whole decision question

What not to write

“We raise the price from $50$ to $60$ dollars: revenue will increase.”

What to write

“At $p=50$ , $E=-1$ : it is the peak. Beyond it, we enter the ELASTIC range and revenue FALLS, from $1{,}250$ to $1{,}200$ dollars.”

Revenue rises as long as the demand is inelastic (blue) and peaks at

p=50

, where

E=-1

. Going from

50

60

dollars moves into the elastic range (orange): revenue drops from

1{,}250

1{,}200

Why: $\dfrac{dR}{dp}=q(1+E)$ changes sign exactly at $E=-1$ . Beyond that, the quantity lost costs more than the price gained brings in.

4. Looking for the profit optimum in the inelastic range

1.5 marks, and an economically absurd result

What not to write

“The optimal price is $p=30$ , where the demand is inelastic: we sell a lot.”

What to write

“In the inelastic range, $R'(q)=p\left(1+\dfrac{1}{E}\right)$ is NEGATIVE: each extra unit sold reduces revenue. The profit optimum is ALWAYS in the elastic range.”

Above

p=40

(

E=-1

), the elastic range (orange); below it, the inelastic range (blue), where marginal revenue is negative. The optimum

p=42

is elastic;

p=30

cannot be.

Why: If marginal revenue is negative and marginal cost positive, producing less increases profit on both sides at once: so you cannot be at the optimum there.

5. Computing the arc elasticity on the starting point

1 mark, and two different answers for the same arc

What not to write

“From $p=40$ to $p=50$ , $q$ goes from $30$ to $25$ : $E=\dfrac{-5/30}{10/40}=-0.67$ .”

What to write

“The arc elasticity uses the AVERAGES: $E=\dfrac{-5/27.5}{10/45}=-0.82$ . That way the result does not depend on the direction of travel.”

Why: With the starting point in the denominator, going from $40$ to $50$ and back from $50$ to $40$ give two different elasticities, which no descriptive quantity should do.

6. Getting the sign of the cross-price elasticity wrong

1 mark, and the economic interpretation of the question

What not to write

“The cross-price elasticity of tea with respect to the price of coffee is $-0.4$ : they are substitutes.”

What to write

“A NEGATIVE cross-price elasticity means COMPLEMENTS. For substitutes, a rise in the price of coffee would increase the demand for tea, so the cross-price elasticity would be POSITIVE.”

Why: Here the sign is all the information: the absolute value only gives the strength of the link, never its nature.

7. Optimizing under a constraint without writing the domain

1 mark for rigour, systematically deducted

What not to write

“After substitution, $A(x)=x(30-x)$ is maximal at $x=15$ .”

What to write

“The constraint imposes $0<x<30$ , and $x=15$ lies in it: I can conclude. Without this domain, a negative root could pass for a solution.”

A(x)=x(30-x)

exists for every

x

, but the situation only makes sense for

0<x<30

(green). Outside this domain (red), the area would be negative: the substitution erased that constraint.

Why: Substituting turns a geometric or economic problem into a plain function and makes the positivity constraints disappear. Writing them back is the only way to keep them.

8. Believing that the revenue from a tax grows with the rate

1.5 marks, and the conclusion of the tax question

What not to write

“We double the unit tax: revenue will double.”

What to write

“The optimal price rises by $0.5t$ and the quantity falls: the revenue $T(t)=t\,q(t)$ goes from $370$ to $720$ dollars, that is $1.95$ times, and it eventually DECREASES beyond a tax of $38$ dollars.”

The curve rises, peaks at a tax of

38

dollars and falls back to zero: beyond the peak, raising the rate lowers the revenue.

Why: The tax acts twice: it takes more per unit and it reduces the number of units. The product of the two is a parabola, so it has a maximum.

Which method to choose

Which range are we in, and what should we do with the price?

Compute $E$ at the price in question, then read its ABSOLUTE VALUE.

If $\left|E\right|<1$ , INELASTIC demand → raising the price INCREASES revenue; the profit optimum is not here

Example: $E=-0.33$ at $p=20$ on $q=400-5p$
If $\left|E\right|=1$ , unit elasticity → REVENUE is greatest exactly at this point

Example: $p=40$ on $q=400-5p$
If $\left|E\right|>1$ , ELASTIC demand → raising the price LOWERS revenue, and this is where the profit optimum lies

Example: $p=42$ , where $E=-1.11$
If you only have TWO observations, with no function → ARC elasticity, with the averages in the denominator

Example: from $p=40$ to $p=50$ , $E\approx -0.82$

A permanent consistency check: the profit-maximizing price is always HIGHER than the revenue-maximizing price as soon as producing costs something.

Which elasticity does the question ask for?

Look at what varies and what reacts, then read the expected sign.

If the PRICE of the good varies and its own demand reacts → PRICE elasticity, negative for a normal demand

Example: $E=-1.11$
If consumers' INCOME varies → INCOME elasticity: negative for an inferior good, greater than $1$ for a luxury good

Example: airline tickets, $E=1.8$
If the price of ANOTHER good varies → CROSS-PRICE elasticity: positive for substitutes, negative for complements

Example: gasoline and cars, $E<0$
If a quantity varies OVER TIME → it is no longer an elasticity but RELATED RATES: $\dfrac{dq}{dt}=\dfrac{dq}{dp}\times\dfrac{dp}{dt}$

Example: a price rising by $0.50$ dollar per month

If the question does not say what varies, it is almost always the price elasticity: it is the only one that is named without a qualifier.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Computing and interpreting an elasticity

When to use it: As soon as the question asks for the elasticity at a given price, or for the range we are in.

1 Write the demand function and compute $\dfrac{dq}{dp}$ .
2 Compute $q$ at the price asked, then $E=\dfrac{dq}{dp}\times\dfrac{p}{q}$ , with two decimals and its SIGN.
3 Name the range from $\left|E\right|$ .
4 Translate into plain language: “a $1$ % rise in the price changes the quantity by $\left|E\right|$ %”.
5 Conclude on the effect of a price rise on REVENUE.

Concluding sentence

“At $p=42$ dollars, $E=-1.11$ : the demand is ELASTIC. A $1$ % rise in the price reduces the quantity demanded by $1.11$ %, so a price rise would LOWER revenue.”

The trap: Giving the elasticity without saying what it implies: half of the marks are in the interpretation sentence.

Marking: 1 mark for the derivative, 1 mark for the calculation with its sign, 1 mark for the range, 1 mark for the interpretation in percentages.

Writing up a constrained optimization

When to use it: When the question gives an objective and a relation between two variables.

1 Name the variables and write the OBJECTIVE function.
2 Write the CONSTRAINT, then solve it for one variable in terms of the other.
3 Substitute, and write the DOMAIN that the constraint imposes on the remaining variable.
4 Differentiate, set to zero, reject the roots outside the domain, and check the nature of the extremum.
5 Also compare the values at the ENDPOINTS when the domain is closed, then answer in the units of the question.

Concluding sentence

“On the domain $0<x<30$ imposed by the constraint, the objective function is maximal at $x=15$ , where the second derivative is negative; the dimensions sought are therefore $15$ and $15$ metres.”

The trap: Substituting without rewriting the domain: a negative root then becomes an acceptable answer, which no real-world problem allows.

Marking: 1 mark for the objective, 1 mark for the constraint, 1 mark for the domain, 2 marks for the solution, 1 mark for the endpoints.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sign of the price elasticity

On a decreasing demand, $E$ is NEGATIVE. A positive result almost always comes from a derivative whose sign was lost.

$E=+1.11$ on $q=400-5p$ : the slope $-5$ was written $+5$ .
The profit optimum is in the elastic range

Compute $E$ at the optimal price found: its absolute value must exceed $1$ . Otherwise, the optimum equation is set up wrongly.

$E(42)=-1.11$ : the absolute value exceeds $1$ , the result is consistent.
The profit-maximizing price exceeds the revenue-maximizing price

Compare the two: since producing costs something, the profit optimum is always at a higher price and a lower quantity.

$42$ dollars against $40$ : the order is the right one.
An arc elasticity does not depend on the direction

Redo the calculation in the other direction: with the averages in the denominator, you must get exactly the same number.

$-0.82$ in both directions, whereas the starting-point method would give $-0.67$ and then $-1.00$ .

The typical problem, taken apart

A cafeteria's pricing decision, all the way to the tax

The daily demand for meals is $q=400-5p$ , where $p$ is the price in dollars. The variable cost is $4$ dollars per meal.

We ask for the elasticity at $p=20$ , the price that maximizes revenue, the one that maximizes profit, and the effect of a unit tax of $2$ dollars.

Step 1

$\dfrac{dq}{dp}=-5$ ; at $p=20$ , $q=300$ , so $E=-5\times\dfrac{20}{300}\approx -0.33$ : INELASTIC demand.

Why

The slope is $-5$ everywhere, the elasticity is not: it is the ratio $\dfrac{p}{q}$ that makes it vary, and it equals $\dfrac{1}{15}$ here.

Step 2

$R(p)=p(400-5p)=400p-5p^{2}$ , so $R'(p)=400-10p=0$ at $p=40$ . Then $q=200$ and $E=-5\times\dfrac{40}{200}=-1$ .

Why

The peak of revenue falls exactly on unit elasticity: it is the quickest check of the chapter, and it is worth its mark.

Step 3

$P(p)=(p-4)(400-5p)=-5p^{2}+420p-1600$ , so $P'(p)=-10p+420=0$ at $p=42$ , with $P''=-10<0$ . Then $q=190$ and $P=7.220$ dollars.

Why

The profit-maximizing price exceeds the revenue-maximizing one, as always when producing costs something. The gap, here $2$ dollars, is half the variable cost.

Step 4

$E(42)=-5\times\dfrac{42}{190}\approx -1.11$ : the absolute value exceeds $1$ , the optimum is indeed in the ELASTIC range.

Why

This check is not decoration: an optimum that fell in the inelastic range would signal a sign error or an error in the equation.

Step 5

With a tax of $2$ dollars: $P(p)=(p-6)(400-5p)$ , so $P'(p)=-10p+430=0$ at $p=43$ . The price rises by $1$ dollar, that is $0.5t$ , and $q$ falls to $185$ .

Why

The seller passes only half of the tax on to the customer: the other half comes out of their margin. This is the result the question almost always looks for.

Step 6

Tax revenue: $T(2)=2\times 185=370$ dollars, and $T(4)=4\times 180=720$ dollars, that is $1.95$ times and not $2$ . $T(t)=t(190-2.5t)$ is maximal at $t=38$ dollars.

Why

The tax brings in more per unit and reduces the number of units: the product of the two is a parabola, so it eventually comes back down.

The conclusion, written out

“At $20$ dollars, the demand is inelastic and a price rise would increase revenue. Revenue is greatest at $40$ dollars and PROFIT at $42$ dollars, for $190$ meals and $7{,}220$ dollars per day. A unit tax of $2$ dollars brings the optimal price to $43$ dollars, that is, half of the tax passed on to the customer.”

The classic mistake on this problem: Answering $40$ dollars to the profit question: that is the revenue-maximizing price, where profit is only $7{,}200$ dollars.

Learn by heart

• $E=\dfrac{dq}{dp}\times\dfrac{p}{q}$ : unitless, negative, and VARIABLE along a line whose slope, by contrast, is constant.
• $\left|E\right|<1$ inelastic, $=1$ unit elastic, $>1$ elastic. REVENUE is greatest at $E=-1$ .
• $\dfrac{dR}{dp}=q(1+E)$ : raising the price pays in the inelastic range, costs in the elastic range.
• The PROFIT optimum is always in the ELASTIC range, because marginal revenue is positive there.
• ARC elasticity: the AVERAGES in the denominator, so that the result does not depend on the direction of travel.
• Cross-price POSITIVE for substitutes, NEGATIVE for complements. Income elasticity greater than $1$ for a luxury good.
• Unit tax $t$ : the optimal price rises by $0.5t$ , and the tax revenue $T(t)$ eventually decreases.

Frequently asked questions

What is the difference between the slope and the elasticity of a demand?

The slope compares absolute changes, in units per dollar, and stays constant on a line. The elasticity compares relative changes, in percent per percent, and equals the slope multiplied by the price divided by the quantity. It therefore changes at every point of the same line, from very large near the maximum price to almost zero near a price of zero.

When does a price increase raise revenue?

When the demand is inelastic, that is, when the absolute value of the elasticity is less than one. The quantity lost is then proportionally smaller than the price gained. As soon as the absolute value exceeds one, the effect reverses, and revenue is greatest exactly where the elasticity equals minus one.

Why is the profit optimum always in the elastic range?

Because marginal revenue is negative in the inelastic range: selling one more unit lowers total revenue. If producing also costs something, producing less would raise profit on both sides. So you cannot be at the optimum while you are in the inelastic range.

How do you compute an arc elasticity between two points?

Divide the relative change in quantity by the relative change in price, putting in the denominators the averages of the two quantities and of the two prices rather than the starting values. You then get the same number whether you go from the first point to the second or the reverse, which the starting-point method does not guarantee.

Who pays a unit tax on a product?

Both sides. With a linear demand and a constant marginal cost, the optimal price rises by exactly half of the tax: the customer pays half and the seller the other half, out of their margin. Tax revenue first rises, then falls, because the quantity sold shrinks as the rate goes up.

Practise it

Corrected exercises: Elasticity and economic optimization, 201-103

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Marginal analysis

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-103-elasticity-optimization. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a 201-103 tutor in Montreal?

Get in touch for a first session. Calculus in the mathematics profile is the course that opens the limited-enrolment admissions to business and economics programs, and it is won or lost in the first weeks.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: elasticity and economic optimization (201-103)

The essentials

What an elasticity is, and what it is not

What elasticity decides

The other elasticities, and the two techniques of the chapter

The mistakes that cost marks

1. Confusing the slope with the elasticity

2. Losing the sign of the elasticity

3. Believing that a price rise always increases revenue

4. Looking for the profit optimum in the inelastic range

5. Computing the arc elasticity on the starting point

6. Getting the sign of the cross-price elasticity wrong

7. Optimizing under a constraint without writing the domain

8. Believing that the revenue from a tax grows with the rate

Which method to choose

Which range are we in, and what should we do with the price?

Which elasticity does the question ask for?

How the answer is expected to be written

Computing and interpreting an elasticity

Writing up a constrained optimization

Check before you hand in

The typical problem, taken apart

A cafeteria's pricing decision, all the way to the tax

Learn by heart

Frequently asked questions

What is the difference between the slope and the elasticity of a demand?

When does a price increase raise revenue?

Why is the profit optimum always in the elastic range?

How do you compute an arc elasticity between two points?

Who pays a unit tax on a product?

Corrected exercises: Elasticity and economic optimization, 201-103

See also

Looking for a 201-103 tutor in Montreal?