Marginal cost, elasticity and optimisation in Mathematik (CBK) at WU

The third of the nine topics in the syllabus of Mathematik in the CBK at WU Vienna is differential calculus; the syllabus does not list its subtopics.¹ In the textbook by Birgit Rudloff, who leads the course, and Achim Zeileis, the third chapter has the same title. Its later sections ask what one more unit costs, how strongly demand reacts to the price, and at what quantity profit is largest.

The easy things to mix up are marginal cost and average cost, an elasticity and its reciprocal, and the minima of different cost curves. According to the syllabus, help is available in the forum on Canvas and at the walk-in tutorial, where you work through exercises with a tutor on hand.¹ German terms are given in brackets.

Marginal cost and average cost: Mathematik (CBK)

A cost function $K(x)$ gives the cost, in MU (monetary units, Geldeinheiten or GE in German tasks), of producing $x$ units. Its derivative $K^{\prime }(x)$ is the marginal cost (Grenzkosten): roughly how much the cost goes up when one more unit is made. Our example is

$$K(x)=0.25x^3-3x^2+19x+60,K^{\prime }(x)=0.75x^2-6x+19$$

with fixed costs of 60 MU that are there even at zero units.

Producing 8 units costs $K(8)=148$ MU, and marginal cost there is $K^{\prime }(8)=48-48+19=19$ MU. What the ninth unit really adds is $K(9)-K(8)=170.25-148=22.25$ MU. Marginal cost estimates that increase along the tangent, and since the curve keeps getting steeper here, the estimate comes out lower.

The usual mix-up is with average cost (Stückkosten), $k(x)=K(x)/x$, the cost that falls on each unit on average: at 8 units, $148/8=18.5$ MU. In a question asking for marginal cost, that is the wrong option sitting right next to the correct 19 MU.

Relative rate of change and elasticity: Mathematik (CBK)

Marginal cost measures a change in absolute terms (MU per unit). The relative rate of change (relative Änderungsrate) $f^{\prime }(x)/f(x)$ measures it against the size of the quantity itself. For the demand $x(p)=120-4p$, with the price $p$ in MU and the quantity $x$ in units, $x^{\prime }(p)=-4$. At a price of 20, 40 units are demanded, and the relative rate of change is $-4/40=-0.1$: at $p=20$, one more MU on the price cuts demand by 10%.

The price elasticity of demand (Preiselastizität) tells you by roughly what percentage the quantity changes when the price rises by 1%. It is the relative rate of change times the price:

$$\epsilon (p)=x^{\prime }(p)\cdot \frac{p}{x(p)}$$

At a price of 20, $\epsilon =20\cdot (-0.1)=-2$: 1% dearer means about 2% less demand. The minus sign belongs to the answer, because demand falls when the price rises; 2 without the sign is right only if the question asks for the absolute value. The fraction the wrong way up, $x/(x^{\prime }\cdot p)$, gives the reciprocal, here −0.5.

Along the demand line the elasticity changes: $\epsilon (p)=-4p/(120-4p)$ is −0.5 at a price of 10, −1 at 15 and −2 at 20. If its absolute value is below 1, demand is called inelastic; above 1, elastic. That decides what happens to revenue $p\cdot x(p)$: 800 MU at a price of 10, 900 MU at 15, and 800 MU again at 20. When demand is inelastic, a 1% higher price loses less than 1% of the quantity, so revenue $p\cdot x$ rises; when it is elastic, revenue falls. It is largest at the price 15, where $|\epsilon |=1$. That is half the choke price (Prohibitivpreis) of 30, at which nobody buys any more; with linear demand the revenue peak always sits there.

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): demand x(p) = 120 − 4p, prohibitive price 30 MU.

ε(p) = x′(p) · p / x(p) = −4p / (120 − 4p), E(p) = p · x(p) = 120p − 4p²

price p	10	15	20
quantity x(p)	80	60	40
ε(p)	−0.5	−1	−2
revenue E(p)	800	900	800

On a straight line the percentage is exact: if p = 10 rises by 1% to 10.1, x falls from 80 to 79.6, that is by 0.5%. Revenue is largest where E′(p) = 120 − 8p = 0, at p = 15, and exactly there ε = −60 / 60 = −1. The marker during the animation calculates with the whole price it names.

Curve sketching for the cost function: Mathematik (CBK)

Back to the cost function. Its second derivative $K^{\prime \prime }(x)=1.5x-6$ describes the curvature: below 4 units it is negative, so $K$ is concave and rises more and more slowly; above 4 units $K$ is convex and rises faster and faster. The inflection point is at $x=4$. There $K^{\prime \prime }$, the derivative of marginal cost, is 0 and turns from negative to positive, so marginal cost is lowest there, $K^{\prime }(4)=12-24+19=7$ MU. Hence $K^{\prime }(x)>0$ everywhere, and $K$ rises throughout.

Price takers and the shutdown point: Mathematik (CBK)

Under perfect competition (vollständige Konkurrenz) the firm is a price taker: the market price $p$ is given, and it only chooses the quantity. An extreme value lies where the first derivative is 0, and it is a maximum if the second derivative is negative there. For profit (Gewinn) $G(x)=p\cdot x-K(x)$ that means $G^{\prime }(x)=p-K^{\prime }(x)=0$: price equals marginal cost. Since $G^{\prime \prime }(x)=-K^{\prime \prime }(x)$ is negative only to the right of the inflection point, the crossing has to lie on the rising branch of marginal cost.

At a market price of 34 MU that means $0.75x^2-6x+19=34$, so $x^2-8x-20=0$, with the solutions 10 and −2. The negative one drops out, and the firm produces 10 units. Costs are then $K(10)=200$ MU and profit is $34\cdot 10-200=140$ MU.

In the short run producing pays even at a loss, as long as the price covers the variable costs, because the fixed costs have to be paid anyway. Average variable cost $k_v(x)=(K(x)-60)/x=0.25x^2-3x+19$ is a parabola with its vertex at $x=3/(2\cdot 0.25)=6$ and $k_v(6)=10$ MU. The quantity 6 is the shutdown point (Betriebsminimum); the price there, 10 MU, is the short-run price floor, because below it no price covers the variable costs. There $K^{\prime }(6)=27-36+19=10$ as well: when the next unit costs exactly the average, the average stops falling.

The typical mistake is the minimum of the wrong curve. Average total cost $K(x)/x$ includes the fixed costs; its minimum is higher. The quantity there is the Betriebsoptimum, the firm's most efficient scale, and its cost level is the long-run price floor. The minimum of marginal cost, 7 MU at 4 units, is not a price floor. Also check whether the price or the quantity is asked for.

Pricing for a monopolist: Mathematik (CBK)

A monopolist chooses the price itself; how much it sells at that price is up to demand. Here demand is given as the inverse demand function (Preis-Absatz-Funktion) $p(x)=35-x$: the same kind of relation as a demand function $x(p)$, only solved for the price. Revenue (Erlös) is $E(x)=35x-x^2$, and its derivative $E^{\prime }(x)=35-2x$ is marginal revenue (Grenzerlös), what one more unit adds to revenue. It lies below the price, because every extra unit lowers the price on all units.

With the same costs $K(x)$, profit is largest where marginal revenue equals marginal cost: $35-2x=0.75x^2-6x+19$, so $0.75x^2-4x-16=0$ with the positive solution $x=8$. To the left of it each further unit adds more revenue than it costs; to the right, less. At 8 units marginal revenue and marginal cost are both 19 MU, while the price on the inverse demand line is well above them at $p(8)=27$ MU. Revenue is $E(8)=216$ MU, and after the costs of $K(8)=148$ MU a profit of 68 MU is left.

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): x in units, amounts in MU, price-demand function p(x) = 35 − x.

Revenue: E(x) = p(x) · x = 35x − x², marginal revenue E′(x) = 35 − 2x. Cost: K(x) = 0.25x³ − 3x² + 19x + 60, marginal cost K′(x) = 0.75x² − 6x + 19.

Equally steep: 35 − 2x = 0.75x² − 6x + 19, so 0.75x² − 4x − 16 = 0, x = (4 ± 8) / 1.5, that is x = 8 (the other solution is negative). There E′(8) = K′(8) = 19.

p(8) = 27, E(8) = 216, K(8) = 148, profit 216 − 148 = 68. Profit is positive between the zeros x ≈ 3.04 and x ≈ 11.71 (found numerically); the area between the curves is shaded there. The tangents carry the slopes at the quantity the marker shows, to one decimal.

The same question without derivatives, with linear costs and the vertex formula, is the subject of our article on cost, revenue and monopoly.

Practice questions on marginal cost and elasticity: Mathematik (CBK)

The exam has 20 multiple-choice questions with 5 options each, exactly one of them right. A right answer earns 1 point. Your first 3 wrong answers cost nothing, every further one costs 0.25 points, and you pass with 11 points.¹

The three questions below are our own, written in that format; none comes from a WU exam or from the textbook. Every wrong option is a typical mistake, and each solution shows the working and how to rule options out.

Practice question 1Differential calculus

Demand for a product is x(p) = 120 − 0.05p², with the price p in MU (monetary units) and the quantity x in units. What is the price elasticity of demand at the price p = 30 (rounded)?

1. a)−1.20
2. b)−0.83
3. c)−0.60
4. d)−0.04
5. e)1.20

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Quantity at the price 30: x(30) = 120 − 0.05 · 900 = 120 − 45 = 75 units.
2. Derivative: x′(p) = −0.05 · 2p = −0.1p, so x′(30) = −3.
3. Elasticity: ε = x′(30) · 30 / x(30) = −3 · 30 / 75 = −90 / 75 = −1.20.

a)

Right. ε = x′(p) · p / x(p) = −3 · 30 / 75 = −1.20. With x′(p) = −0.1p, x′(30) = −3, and x(30) = 120 − 45 = 75.

+1

b)

Wrong. The fraction upside down: x / (x′ · p) = 75 / (−90) ≈ −0.83, the reciprocal of the elasticity.

0 or −0.25

c)

Wrong. Differentiated wrongly: x′(p) = −0.05p instead of −0.1p, the exponent 2 not brought down as a factor; −1.5 · 30 / 75 = −0.60.

0 or −0.25

d)

Wrong. The relative rate of change x′ / x = −3 / 75 instead of the elasticity, without the factor p.

0 or −0.25

e)

Wrong. The sign forgotten: only the absolute value 3 · 30 / 75.

0 or −0.25

–

Left blank.

0

Typical mistake. The fraction upside down. The elasticity divides the percentage change in quantity by that in price, x′ · p / x. Working out x / (x′ · p) gives the reciprocal, −0.83. Like the right answer it is close to −1, which is why it looks believable.

Ruling out before you finish. Demand falls when the price rises, so the elasticity is negative; that rules out e. And ε = p · x′ / x is, at the price 30, thirty times the relative rate of change x′ / x = −3 / 75 = −0.04. Option d is exactly that rate, with the factor p missing, so d goes too. That leaves a, b and c.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 2Differential calculus

A firm has the cost function K(x) = 0.2x³ − 6x² + 80x + 180, with x in units and K in MU (monetary units). What is the short-run price floor, the price at the shutdown point (Betriebsminimum), rounded?

1. a)15.00 MU
2. b)20.00 MU
3. c)35.00 MU
4. d)46.36 MU
5. e)47.00 MU

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Average variable cost without the fixed costs of 180: k_v(x) = (K(x) − 180) / x = 0.2x² − 6x + 80.
2. Vertex of the parabola: x = 6 / (2 · 0.2) = 15 units.
3. Check: K′(15) = 0.6 · 225 − 12 · 15 + 80 = 135 − 180 + 80 = 35, so marginal cost cuts k_v there.
4. Price floor: k_v(15) = 0.2 · 225 − 90 + 80 = 45 − 90 + 80 = 35.00 MU.

a)

Wrong. The quantity at the shutdown point, 15 units, instead of the price.

0 or −0.25

b)

Wrong. The minimum of marginal cost instead of average variable cost: K′(x) = 0.6x² − 12x + 80 is lowest at x = 10, K′(10) = 20.

0 or −0.25

c)

Right. Average variable cost k_v(x) = 0.2x² − 6x + 80 is lowest at x = 15: k_v(15) = 45 − 90 + 80 = 35. As a check, K′(15) = 135 − 180 + 80 = 35.

+1

d)

Wrong. The minimum of average total cost k(x) = K(x) / x, fixed costs included, at x ≈ 16.63: the Betriebsoptimum, the long-run price floor.

0 or −0.25

e)

Wrong. Fixed costs included: average total cost at 15 units, k(15) = 35 + 180 / 15 = 47.

0 or −0.25

–

Left blank.

0

Typical mistake. The minimum of the wrong curve. In the short run only average variable cost counts, because the fixed costs are due even without production. Including them gives the cost level at the Betriebsoptimum (the most efficient scale), 46.36 MU, the long-run price floor. The minimum of marginal cost, 20 MU, is not a price floor.

Ruling out before you finish. At the shutdown point marginal cost cuts average variable cost, on the rising branch of marginal cost, above its lowest value K′(10) = 20 MU. So the price floor is above 20 MU; that rules out a (15) and b (20). Three are left.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 3Differential calculus

A monopolist has the inverse demand function (Preis-Absatz-Funktion) p(x) = 30 − 3x and the cost function K(x) = 0.5x² + 2x + 20, with x in units, p and K in MU (monetary units). At what price is its profit largest?

1. a)4 MU
2. b)6 MU
3. c)9 MU
4. d)15 MU
5. e)18 MU

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Revenue E(x) = (30 − 3x) · x = 30x − 3x², marginal revenue E′(x) = 30 − 6x.
2. Marginal cost K′(x) = x + 2. From 30 − 6x = x + 2 it follows that 7x = 28, so x = 4 units.
3. Price on the inverse demand line: p(4) = 30 − 3 · 4 = 18 MU.

a)

Wrong. The profit-maximising quantity, 4 units, instead of the price.

0 or −0.25

b)

Wrong. Marginal cost at the optimum, K′(4) = 4 + 2 = 6, instead of the price on the inverse demand line.

0 or −0.25

c)

Wrong. Price set equal to marginal cost as under perfect competition: 30 − 3x = x + 2 gives x = 7 and p(7) = 9.

0 or −0.25

d)

Wrong. The revenue maximum instead of the profit maximum: E′(x) = 30 − 6x = 0 at x = 5, p(5) = 15.

0 or −0.25

e)

Right. Marginal revenue equals marginal cost: 30 − 6x = x + 2, so x = 4 and p(4) = 30 − 12 = 18. Profit is then 72 − 36 = 36 MU.

+1

–

Left blank.

0

Typical mistake. Price equal to marginal cost. That condition holds for a price taker, whose quantity does not move the price. A monopolist lowers the price on all units with every extra unit, so what counts is marginal revenue. With p = K′ you get 7 units at 9 MU, and profit would be 63 − 58.5 = 4.5 MU instead of 36 MU.

Ruling out before you finish. A price taker would set price equal to marginal cost: 30 − 3x = x + 2 gives x = 7 and the price 9 MU. The monopolist sells less, because his marginal revenue is below the price, and so charges more than 9 MU. That rules out a, b and c and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Calculator and time for optimisation questions in Mathematik (CBK)

WU does not publish which of the 20 questions covers which chapter; this article follows the textbook. There, "Marginale Änderung" (marginal change), "Relative Änderungsrate" and "Elastizitäten" belong to the section "Interpretationen der Ableitung" (interpretations of the derivative). Then come "Kurvendiskussion" (curve sketching) and "Anwendungen – Optimierung" (applications: optimisation), which also includes "Optimale Lagerhaltung" (optimal inventory), left out here.²

A calculator is allowed if it has no functions for differential calculus, integration or matrices, cannot solve linear systems and has no text memory.¹ The derivatives therefore happen on paper, and the quadratic equations from $p=K^{\prime }(x)$ or $E^{\prime }(x)=K^{\prime }(x)$ you solve with the quadratic formula.

Under WU's guideline for conducting exams, formula sheets and watches of any kind are not allowed unless the syllabus says otherwise.³ So you cannot bring your own formula sheet, and WU does not say whether a formula sheet is handed out. Keep in your head the derivative rules, $\epsilon =x^{\prime }\cdot p/x$ with its sign, and three conditions: price equals marginal cost for a price taker, marginal revenue equals marginal cost for a monopolist, and the minimum of average variable cost for the shutdown point.

The exam plan sets aside two hours for the whole exam; the syllabus does not give the writing time.⁴ With 20 questions, that leaves at most six minutes for each.

Next step after marginal cost and elasticity: Mathematik (CBK)

1. Do the three questions above before opening the solutions, and for each one note first which curve gives the answer: marginal cost, average variable cost, marginal revenue or elasticity.
2. If you got one wrong, find the mistake in the solution that matches your answer. From now on, write down the condition you mixed up before you calculate. For more practice, see the third chapter of the free textbook by Birgit Rudloff and Achim Zeileis.⁵
3. If taking derivatives is not yet solid, start with the article on derivative rules and the chain rule. Exam dates and registration are in the overview of the Mathematik (CBK) exam.

Sources

1. 1Syllabus 0003 Mathematik (LVP), Wintersemester 2026/27WU Wien, Vorlesungsverzeichnis · undated page · checked on 27 September 2026
2. 2Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften, Kapitel 3: Differentialrechnungmathe4wiwi.org · undated page · checked on 27 September 2026
3. 3Richtlinie zur Abhaltung von Präsenzprüfungen, 2026 (PDF)WU Wien · document dated 2 March 2026 · checked on 27 September 2026
4. 4Prüfungsplan Semestermitte WiSe 2026/27, Prüfungswoche November 2026 (PDF)WU Wien, Prüfungsorganisation · document dated 30 March 2026 · checked on 27 September 2026
5. 5Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften (Online-Buch, Version 2023)mathe4wiwi.org · undated page · checked on 26 September 2026