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

Revision sheet: marginal analysis (201-103)

Q: What is the difference between marginal cost and average cost?

Marginal cost is the derivative of the total cost: it gives approximately what the next unit would cost. Average cost is the total cost divided by the quantity: it gives what each unit already produced costs on average. The two coincide at a single point, the one where the average cost is minimal.

Q: Why does marginal revenue have double the slope of the demand?

Because revenue is price times quantity. With a demand of the form a minus b times q, the revenue is aq minus bq squared, whose derivative is a minus two bq. Economically, selling one more unit forces the price down on all the units already sold, which costs twice as fast.

Q: How do you find the quantity that maximizes profit?

Set marginal revenue equal to marginal cost, which amounts to setting the derivative of the profit to zero, then check that the second derivative is negative. The quantity obtained is always below the one that maximizes revenue, as soon as producing one more unit costs something.

Q: Does a fixed cost change the optimal quantity?

No. A constant disappears on differentiation, so the optimality condition does not see it. The fixed cost lowers the maximum profit by the same amount and brings the two break-even points closer together, but the quantity to produce stays exactly the same. It affects the survival of the firm, not its production decision.

Q: What does the inflection point of a production function mean?

It is the place where the second derivative vanishes while changing sign, that is, where the curve stops accelerating and starts to slow down. Total output still rises there, but the marginal product reaches its maximum: this is the entry into diminishing returns.

This sheet covers the chapter that sets the differential calculus of the profile with mathematics apart from that of Natural Science: marginal analysis. The derivatives are the same, the vocabulary is not, and it is the vocabulary that costs the marks. It gives the eight places where a correct derivative answers the wrong question.

It is written for Social Science students at CEGEP taking course 201-103, differential calculus. The set of corrected exercises on the same chapter goes back over each move, from the minimum average cost to the reading of an economic statement.

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The thread of the chapter

In economics, MARGINAL means DERIVATIVE and AVERAGE means QUOTIENT. Almost every mistake in the chapter comes from using one in place of the other, or from optimizing the wrong quantity.

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

The essentials

Marginal, average: two words, two operations

• MARGINAL means DERIVATIVE: $C'(q)$ is the marginal cost, $R'(q)$ the marginal revenue, $Q'(L)$ the marginal product.
• AVERAGE means QUOTIENT: $\overline{C}(q)=\dfrac{C(q)}{q}$ . Never use one in place of the other.
• $C'(q)$ APPROXIMATES $C(q+1)-C(q)$ without being equal to it; for $C(q)=aq^{2}+bq+c$ , the gap is exactly $a$ , whatever the quantity.
• The MINIMUM average cost is found by setting $\overline{C}'(q)=\dfrac{C'(q)q-C(q)}{q^{2}}$ equal to zero, which gives the condition $C'(q)=\overline{C}(q)$ .

As long as the marginal cost (orange) is below the average cost (blue), it pulls the average down; above it, it pushes it up. The two curves therefore cross at the MINIMUM of the average cost, at

q=\sqrt{300}\approx 17.3

On an exam paper, writing “marginal cost, that is $C'(q)$ ” on the very first line earns the vocabulary mark and removes any ambiguity afterwards.

Demand, revenue and profit

• Linear demand $p=a-bq$ : revenue $R(q)=aq-bq^{2}$ and marginal revenue $R'(q)=a-2bq$ , with DOUBLE the slope of the demand.
• Marginal revenue is therefore BELOW the price: selling one more unit forces the price down on ALL the units.
• MAXIMUM PROFIT: $P'=R'-C'=0$ , so $R'(q)=C'(q)$ , with $P''<0$ to confirm that it is a maximum.
• The quantity of maximum profit is ALWAYS below the quantity of maximum revenue as soon as marginal cost is positive.

The optimal quantity is read at the crossing of

R'

and

C'

(

q^{*}\approx 13.3

); the PRICE is then read on the demand line, straight above:

73.3

dollars, not the height of the crossing.

The BREAK-EVEN POINTS are the roots of $P(q)=0$ : between them, and only between them, the firm makes money.

Production, labour and returns

• $Q(L)$ gives the output as a function of labour; $Q'(L)$ is the marginal product, what the next worker brings in.
• The INFLECTION POINT of $Q$ , where $Q''(L)=0$ with a change of sign, marks the entry into diminishing returns.
• It is there that the MARGINAL product is maximal, not the total output.
• Maximizing output is not maximizing profit: you keep hiring as long as $pQ'(L)>w$ , where $w$ is the wage.

With

Q(L)=-\frac{L^{3}}{3}+12L^{2}

: the inflection of

Q

(at

L=12

) lies straight below the MAXIMUM of the marginal product

Q'

; the maximum of

Q

comes later, at

L=24

, where

Q'

vanishes.

A FIXED COST changes no derivative: it lowers the profit without moving the optimal quantity, and it narrows the interval of profitability.

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 marginal cost and average cost

2 points, and the rest of the question

What not to write

“ $C(q)=q^{2}+20q+300$ : the marginal cost for $q=10$ is $\dfrac{C(10)}{10}=\dfrac{600}{10}=60$ .”

What to write

“The AVERAGE cost is $60$ ; the MARGINAL cost is $C'(10)=2(10)+20=40$ . The first is a quotient, the second a derivative.”

At the point

q=10

: the TANGENT (green) has slope

C'(10)=40

, the marginal cost; the segment from the origin (red) has slope

\frac{600}{10}=60

, the average cost. Two lines, two numbers.

Why: The two numbers answer two different questions: how much each unit produced costs on average, and how much the NEXT unit would cost.

2. Taking marginal revenue for the price

2 points, and a wrong optimal quantity

What not to write

“The price is $p=100-2q$ , so the marginal revenue is too.”

What to write

“ $R(q)=pq=100q-2q^{2}$ , so $R'(q)=100-4q$ : the slope is DOUBLE and the marginal revenue vanishes at $q=25$ while the price vanishes at $q=50$ .”

Both lines start from the same point on the price axis, but the marginal revenue line reaches zero at

q=25

, half as far as the demand.

Why: Selling one more unit brings in the price of that unit MINUS the price cut granted on all the previous ones. Marginal revenue is therefore always below the demand curve.

3. Maximizing revenue instead of profit

the whole question, the quantity being almost doubled

What not to write

“ $R'(q)=0$ gives $q=25$ : this is the quantity that maximizes profit.”

What to write

“ $R'(q)=0$ gives the maximum revenue. Profit is maximal when $R'(q)=C'(q)$ , here $100-4q=2q+20$ , that is $q=\dfrac{40}{3}\approx 13.3$ .”

Revenue peaks at

q=25

(red), but at this quantity the profit is NEGATIVE:

-175

dollars. Profit peaks earlier, at

q\approx 13.3

(green), where

R'=C'

Why: Producing beyond $q^{*}$ still brings in revenue, but each additional unit costs more than it brings in. The quantity of maximum profit is always the smaller of the two.

4. Believing that a fixed cost moves the optimal quantity

1.5 points, and a wrong economic conclusion

What not to write

“The rent goes up by $100$ dollars: we have to produce more to absorb it.”

What to write

“A fixed cost changes NO derivative: $q^{*}$ stays $13.3$ . The maximum profit goes from $233$ to $133$ dollars, and the interval of profitability narrows from $[4.5, 22.2]$ to $[6.7, 20]$ .”

The two parabolas have exactly the same vertex abscissa; only the height changes, and with it the two points where the profit vanishes.

Why: The optimality condition involves only margins, that is, derivatives. A constant disappears on differentiation: it moves the curve down, never left or right.

5. Forgetting the second derivative

1 rigour mark, systematically deducted

What not to write

“ $P'(q)=0$ at $q=13.3$ : so this is the maximum profit.”

What to write

“ $P''(q)=-6<0$ for every $q$ : the function is concave, so this critical point is indeed a MAXIMUM.”

Why: A critical point can be a minimum, which really does happen on average cost functions. One line is enough to rule it out, and it is in the marking scheme of every optimization question.

6. Confusing maximum output and maximum profit

the whole decision question

What not to write

“Output is maximal at $L=12$ : this is the number of employees to hire.”

What to write

“You hire as long as the value of the marginal product exceeds the wage, that is, as long as $pQ'(L)>w$ . The last profitable employee comes well before the maximum output.”

Why: The maximum of $Q$ is the point where the next worker no longer adds anything; well before it, he already adds less than he costs.

7. Believing that the derivative gives exactly the cost of the next unit

0.5 point, and more if the question is about the approximation

What not to write

“ $C'(10)=40$ : the eleventh unit will cost exactly $40$ dollars.”

What to write

“ $C(11)-C(10)=41$ : the derivative APPROXIMATES this cost. For a quadratic cost $aq^{2}+bq+c$ , the gap is exactly $a$ , here $1$ dollar.”

Over one unit, the curve rises by

41

(green) and the tangent by only

40

(red):

C'(10)

APPROXIMATES the cost of the eleventh unit. The gap,

1

dollar, is the coefficient

a

q^{2}

Why: The derivative is a limit: it gives the instantaneous rate, not the change over a whole unit. The gap is small and perfectly computable.

8. Mistaking the inflection point for a maximum

2 points, and the question on diminishing returns

What not to write

“ $Q''(L)=0$ at $L=12$ : output is maximal there.”

What to write

“At $L=12$ , it is the MARGINAL product $Q'(L)$ that is maximal: beyond it, each additional employee still adds output, but less and less.”

The curve still rises after the marked point: what changes there is the CURVATURE, and so the slope starts to decrease.

Why: $Q''=0$ sets the derivative of $Q'$ to zero, so it is $Q'$ that reaches an extremum, not $Q$ . The curve changes curvature and keeps rising.

Which method to choose

What is the economic statement asking?

Spot the WORD in the statement, then write the matching equation before any calculation.

If “marginal”, whatever follows → DIFFERENTIATE the matching function

Example: marginal cost is $C'(q)$
If “average” or “per unit” → DIVIDE the function by the quantity

Example: average cost is $\dfrac{C(q)}{q}$
If “maximize the profit” → set $R'(q)=C'(q)$ , or set $P'$ to zero, then check $P''<0$

Example: $100-4q=2q+20$
If “minimum average cost” → set $\overline{C}'(q)$ to zero, which amounts to $C'(q)=\overline{C}(q)$

Example: $q=\sqrt{300}\approx 17.3$
If “break-even point” or “from which quantity” → solve $P(q)=0$ , and give BOTH roots

Example: $q=4.5$ and $q=22.2$

If the word “marginal” and the word “average” both appear in the statement, the question is almost always about their meeting point, which is the minimum of the average cost.

Is this critical point a maximum?

Once $f'(q)=0$ is solved, pick the quickest test according to the form of $f$ .

Same equation

f'(q)=0

, two outcomes: a maximum if the curve opens downward, a minimum if it opens upward. On a closed interval, the extremum can be at the ENDPOINT, with no horizontal tangent.

If $f$ is a second-degree polynomial → the sign of the leading coefficient is enough: negative, it is a maximum

Example: $P(q)=-3q^{2}+80q-300$ , so a maximum
If $f''$ is easy to compute → SECOND derivative test: $f''(q^{*})<0$ for a maximum

Example: $P''=-6<0$
If $f''$ is heavy to write out → SIGN table of $f'$ around the critical point

Example: an average cost with a term in $\dfrac{1}{q}$
If the domain is a CLOSED interval → also compare the values at the ENDPOINTS, since the extremum may be reached there

Example: a production capacity limited to 20 units

In economics, the domain is almost always $q>0$ : a negative root is rejected in one line, and that line is in the marking scheme.

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.

Writing up an economic optimization

When to use it: As soon as the statement asks for the quantity that maximizes the profit or minimizes a cost.

1 Write down the functions of the statement and the DOMAIN, usually $q>0$ .
2 Build the function to optimize, profit or average cost, and say so.
3 Differentiate, set to zero, solve, then REJECT the roots outside the domain.
4 Check the nature of the extremum with the second derivative or a sign table.
5 Answer the question asked in the UNITS of the statement: quantity, price and profit.

Concluding sentence

“The profit $P(q)=-3q^{2}+80q-300$ is maximal for $q^{*}=\dfrac{40}{3}\approx 13.3$ units, since $P'(q^{*})=0$ and $P''(q)=-6<0$ . The corresponding price is $p=100-2q^{*}\approx 73.3$ dollars and the maximum profit $233.33$ dollars.”

The trap: Giving the quantity and stopping there: the statement almost always asks for the PRICE and the PROFIT as well, and each is worth a mark.

Marking: 1 mark for the function set up, 2 marks for the differentiation and the solving, 1 mark for the second derivative, 2 marks for the answers with units.

Showing that the average cost is minimal when it equals the marginal cost

When to use it: When the statement asks for the quantity of minimum average cost, or for a proof of this property.

1 Write $\overline{C}(q)=\dfrac{C(q)}{q}$ on the domain $q>0$ .
2 Differentiate by the quotient rule: $\overline{C}'(q)=\dfrac{C'(q)q-C(q)}{q^{2}}$ .
3 Set the NUMERATOR to zero, the denominator never vanishing on the domain, which gives $C'(q)=\dfrac{C(q)}{q}$ .
4 Solve, reject the negative root, and check with a sign table that it is a minimum.

Concluding sentence

“The average cost is minimal for $q=\sqrt{300}\approx 17.3$ units, and it equals $\overline{C}\approx 54.64$ dollars there, exactly the value of the marginal cost at that point.”

The trap: Setting the whole fraction to zero instead of its numerator: the equation $\dfrac{A}{q^{2}}=0$ is solved by setting $A$ to zero, and $q=0$ is outside the domain.

Marking: 1 mark for the derivative of the quotient, 1 mark for setting the numerator to zero, 1 mark for the solving, 1 mark for the nature of the extremum.

Check before you hand in

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

The quantity of maximum profit is below that of maximum revenue

Compare $q^{*}$ with the root of $R'(q)=0$ : as soon as the marginal cost is positive, $q^{*}$ is the smaller.

$13.3$ against $25$ : the order is right. The reverse signals an equation set up wrongly.
The optimal price lies in the upper half of the demand

Substitute $q^{*}$ into $p=a-bq$ : the price obtained must exceed half the maximum price.

$p=100-2(13.3)\approx 73.3$ , well above $50$ .
Marginal cost equals average cost at its minimum

At the point of minimum average cost, compute both and check that they coincide. This is the quickest check in the chapter.

$\overline{C}(17.32)=54.64$ and $C'(17.32)=54.64$ : the root is right.
The profit is positive between the two break-even points

Test a quantity lying between the two roots of $P(q)=0$ : the profit must be positive there, and negative beyond.

$P(13.3)=233>0$ between $4.5$ and $22.2$ , $P(25)=-175<0$ beyond.

The typical problem, taken apart

A firm from start to finish, from marginal cost to break-even points

The demand is $p=100-2q$ and the total cost $C(q)=q^{2}+20q+300$ , where $q$ is in units and the amounts in dollars.

The question asks for the quantity that maximizes the profit, the corresponding price and profit, the quantity of minimum average cost, and the break-even points.

Step 1

$R(q)=pq=100q-2q^{2}$ , so $R'(q)=100-4q$ ; and $C'(q)=2q+20$ .

Why

Marginal revenue has double the slope of the demand: writing it down right away avoids mixing up $p$ and $R'$ in the rest of the question.

Step 2

$P(q)=R-C=-3q^{2}+80q-300$ , so $P'(q)=-6q+80=0$ at $q^{*}=\dfrac{40}{3}\approx 13.33$ . $P''(q)=-6<0$ : maximum.

Why

The same quantity comes from $R'(q)=C'(q)$ , that is $100-4q=2q+20$ : the two routes must give the same number, which checks the calculation.

Step 3

$p=100-2(13.33)\approx 73.33$ dollars and $P(13.33)\approx 233.33$ dollars.

Why

The question asks for three numbers, not one. The quantity alone is worth only a third of the marks.

Step 4

$\overline{C}(q)=q+20+\dfrac{300}{q}$ , so $\overline{C}'(q)=1-\dfrac{300}{q^{2}}=0$ at $q=\sqrt{300}\approx 17.32$ .

Why

Immediate check: $C'(17.32)=54.64$ and $\overline{C}(17.32)=54.64$ , so the equality announced by the course holds.

Step 5

Break-even points: $-3q^{2}+80q-300=0$ gives $q=\dfrac{80\pm\sqrt{2800}}{6}$ , that is $q\approx 4.51$ and $q\approx 22.15$ .

Why

Between these two quantities, and only between them, the firm is profitable. Giving only one root costs half the question.

Step 6

With the fixed cost raised to $400$ : $q^{*}$ stays $13.33$ , the maximum profit drops to $133.33$ , and the break-even points tighten to $6.67$ and $20$ .

Why

This last calculation isolates what a fixed cost really does: it has no influence on the production decision, only on the survival of the firm.

The conclusion, written out

“Profit is maximal for $q^{*}\approx 13.33$ units sold at $73.33$ dollars, for a profit of $233.33$ dollars. The average cost, for its part, is minimal for $q\approx 17.32$ units, and the firm is profitable between $4.51$ and $22.15$ units.”

The classic mistake on this problem: Setting $R'(q)$ to zero and answering $q=25$ : this is the quantity of maximum revenue, where the profit is only $-175$ dollars.

Learn by heart

• MARGINAL means DERIVATIVE, AVERAGE means QUOTIENT. Neither word is ever used in place of the other.
• Demand $p=a-bq$ : $R'(q)=a-2bq$ , with DOUBLE the slope, and marginal revenue is always below the price.
• MAXIMUM PROFIT: $R'(q)=C'(q)$ with $P''<0$ . The quantity obtained is always BELOW that of maximum revenue.
• MINIMUM AVERAGE COST: $C'(q)=\overline{C}(q)$ , and it is also the check on the result.
• A FIXED COST does not move $q^{*}$ : it lowers the profit and narrows the interval of profitability.
• BREAK-EVEN POINTS: the TWO roots of $P(q)=0$ . Between them, the firm makes money.
• $Q''(L)=0$ marks the maximum of the MARGINAL product, not of the total output.

Frequently asked questions

What is the difference between marginal cost and average cost?

Marginal cost is the derivative of the total cost: it gives approximately what the next unit would cost. Average cost is the total cost divided by the quantity: it gives what each unit already produced costs on average. The two coincide at a single point, the one where the average cost is minimal.

Why does marginal revenue have double the slope of the demand?

Because revenue is price times quantity. With a demand of the form a minus b times q, the revenue is aq minus bq squared, whose derivative is a minus two bq. Economically, selling one more unit forces the price down on all the units already sold, which costs twice as fast.

How do you find the quantity that maximizes profit?

Set marginal revenue equal to marginal cost, which amounts to setting the derivative of the profit to zero, then check that the second derivative is negative. The quantity obtained is always below the one that maximizes revenue, as soon as producing one more unit costs something.

Does a fixed cost change the optimal quantity?

No. A constant disappears on differentiation, so the optimality condition does not see it. The fixed cost lowers the maximum profit by the same amount and brings the two break-even points closer together, but the quantity to produce stays exactly the same. It affects the survival of the firm, not its production decision.

What does the inflection point of a production function mean?

It is the place where the second derivative vanishes while changing sign, that is, where the curve stops accelerating and starts to slow down. Total output still rises there, but the marginal product reaches its maximum: this is the entry into diminishing returns.

Practise it

Corrected exercises: Marginal analysis, 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

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

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Revision sheet: marginal analysis (201-103)

The essentials

Marginal, average: two words, two operations

Demand, revenue and profit

Production, labour and returns

The mistakes that cost marks

1. Confusing marginal cost and average cost

2. Taking marginal revenue for the price

3. Maximizing revenue instead of profit

4. Believing that a fixed cost moves the optimal quantity

5. Forgetting the second derivative

6. Confusing maximum output and maximum profit

7. Believing that the derivative gives exactly the cost of the next unit

8. Mistaking the inflection point for a maximum

Which method to choose

What is the economic statement asking?

Is this critical point a maximum?

How the answer is expected to be written

Writing up an economic optimization

Showing that the average cost is minimal when it equals the marginal cost

Check before you hand in

The typical problem, taken apart

A firm from start to finish, from marginal cost to break-even points

Learn by heart

Frequently asked questions

What is the difference between marginal cost and average cost?

Why does marginal revenue have double the slope of the demand?

How do you find the quantity that maximizes profit?

Does a fixed cost change the optimal quantity?

What does the inflection point of a production function mean?

Corrected exercises: Marginal analysis, 201-103

See also

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