Cost, revenue and monopoly in the Mathematik (CBK) exam at WU

The first of the nine topics in the syllabus of Mathematik in the CBK at WU Vienna is called "Lineare und quadratische Funktionen" (linear and quadratic functions); the syllabus does not list its subtopics.¹ The textbook by Birgit Rudloff, who leads the course, and Achim Zeileis has a first chapter with the same title, and part of it deals with a firm's costs, revenue and profit. Three questions come out of that: from what quantity does revenue cover the costs, at what quantity and price does a monopolist earn the most, and what is that profit.

To answer them you turn the wording of a task into a cost function and a revenue function, form the profit, solve a linear or quadratic equation and find the vertex of a parabola. No derivatives are needed. The easy things to mix up are price and quantity, and revenue and profit. German terms are given in brackets.

Cost, revenue and break-even: Mathematik (CBK)

A firm producing $x$ units has costs in two parts. Fixed costs $F$ (Fixkosten), such as rent, are there even at zero units. Variable costs rise by the same amount $v$ with every unit, the variable cost per unit. Together they give the linear cost function $C(x)=F+v\cdot x$ (Kostenfunktion $K(x)$). If every unit sells at a fixed price $p$, revenue (Erlös) is $R(x)=p\cdot x$, and profit (Gewinn), written $\pi$ here so it does not clash with the price $p$, is what is left after costs:

$$\pi (x)=R(x)-C(x)=(p-v)\cdot x-F$$

The difference $p-v$ is the contribution margin (Deckungsbeitrag), what each unit sold contributes towards the fixed costs. The break-even quantity (Gewinnschwelle) is where profit is exactly zero, $x=F/(p-v)$. With fixed costs of 240 MU (monetary units, Geldeinheiten or GE in German tasks), variable costs of 5 MU a unit and a price of 8 MU, each unit contributes 3 MU, and from 80 units on the firm no longer makes a loss.

Forget the fixed costs and break-even lands at zero. Divide by the price instead of the contribution margin and you are treating every unit as free to make. Round down when the question asks for a minimum quantity and you end up just inside the loss: if break-even falls between two whole numbers, the larger one is the answer.

The monopoly with linear demand: Mathematik (CBK)

A monopolist is the only seller in its market. It sets the price itself, but the higher the price, the fewer units it sells. In the linear market model demand depends linearly on the price, and there are two equivalent ways to write it: as a demand function giving the quantity at each price (Nachfragefunktion), for example $x(p)=60-2p$, or solved for the price as an inverse demand function giving the price at each quantity (Preis-Absatz-Funktion), here $p(x)=30-0.5x$. The price at which nobody buys any more is the choke price (Prohibitivpreis), here 30 MU.

With the inverse demand function, revenue $R(x)=p(x)\cdot x$ is a downward-opening parabola in the quantity, and after subtracting the linear costs so is profit. It is largest at the vertex: for a parabola $f(x)=\alpha x^2+\beta x+\gamma$ the vertex sits at $x=-\beta /(2\alpha )$. With the demand function you do the same in the price instead: $\pi (p)=(p-v)\cdot x(p)-F$ is a parabola in $p$. Derivatives are only the third topic in the syllabus and give the same result; here the vertex formula is enough.¹

An example: $p(x)=30-0.5x$ and $C(x)=160+6x$. Revenue is $R(x)=30x-0.5x^2$, profit $\pi (x)=-0.5x^2+24x-160$. Profit peaks at $x=24$ units, at the price $p(24)=18$ MU, and is then $\pi (24)=128$ MU. Working in the price gives the same: $\pi (p)=(p-6)(60-2p)-160$ has its vertex at $p=18$. Revenue on its own would peak at $x=30$, at 450 MU, but only 110 MU of profit remain there.

The profit-maximising quantity lies left of the revenue-maximising one because every extra unit costs something: beyond 24 units, one more unit adds less revenue than it costs. The monopolist makes a profit only between the zeros of the profit function, where the revenue curve crosses the cost line. Multiplying $\pi (x)=0$ by $-2$ gives $x^2-48x+320=0$, and the quadratic formula gives $x=(48\pm \sqrt{48^2-4\cdot 320})/2=24\pm 16$: 8, the break-even quantity, and 40, the upper break-even point (Gewinngrenze), beyond which the firm makes a loss again.

There is a quick check for the maximum. With linear demand and linear costs, the profit-maximising price lies exactly halfway between the choke price and the variable cost per unit, here $(30+6)/2=18$. If your price does not match, you have made a mistake somewhere. Before calculating, also note what is being asked: quantity, price and profit at the maximum are three different numbers, and with five options any of them can appear as a wrong answer.

Our calculation

Numbers from this article’s example, chosen by unipass for the article.

R(x) = p(x) · x = 30x − 0.5x²

π(x) = (p(x) − 6) · x − 160 = −0.5x² + 24x − 160

The filled rectangle above the unit cost is the contribution margin (p − 6) · x. At x = 24 and p = 18: 12 · 24 = 288, profit 288 − 160 = 128.

The whole rectangle is the revenue p · x. At x = 30 and p = 15: 15 · 30 = 450, profit 450 − 160 − 6 · 30 = 110.

Check: (30 + 6) / 2 = 18 and (30 + 0) / 2 = 15. The largest rectangle under a falling line has its corner at the line’s midpoint.

Practice questions on cost and monopoly: 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

We wrote these three questions for this article in that format; the stories, numbers and wording are our own, not questions from WU or the textbook. Every wrong option comes from a typical mistake. Work each one out first, then open the solution: it shows the working, the typical mistake, and how to rule options out before you have finished calculating.

Practice question 1Linear and quadratic functions

A screen-printing shop prints tote bags. Its fixed costs are 500 MU (monetary units) a week; each bag costs it another 3.50 MU and sells for 8 MU. What is the smallest number of bags the shop must sell per week to avoid a loss?

1. a)62
2. b)63
3. c)111
4. d)112
5. e)143

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Cost C(x) = 500 + 3.5x, revenue R(x) = 8x, with x bags per week.
2. Profit π(x) = R(x) − C(x) = 8x − 500 − 3.5x = 4.5x − 500.
3. π(x) = 0 at x = 500 / 4.5 ≈ 111.1.
4. Only whole bags are sold. At 111 the shop is 0.50 MU short; at 112 it makes 4 MU. The answer is 112.

a)

Wrong. The variable cost forgotten and rounded down: 500 / 8 = 62.5.

0 or −0.25

b)

Wrong. The variable cost forgotten: 500 / 8 = 62.5, rounded up.

0 or −0.25

c)

Wrong. Correct calculation, rounded down: 111 bags bring in 111 · 4.50 = 499.50 MU, 0.50 MU short.

0 or −0.25

d)

Right. Break-even at 500 / 4.5 ≈ 111.1, rounded up to the next whole bag.

+1

e)

Wrong. Divided by the variable cost per unit instead of the contribution margin: 500 / 3.5 ≈ 142.9, rounded up.

0 or −0.25

–

Left blank.

0

Typical mistake. Rounding down. 111.1 is the quantity at which profit is exactly zero; the whole number below it still makes a loss. When the question asks for a minimum quantity, round up.

Ruling out before you finish. Each bag contributes 8 − 3.50 = 4.50 MU towards the fixed costs. 100 bags cover 450 MU, 120 bags cover 540 MU, so the answer lies between 100 and 120. That rules out options a, b and e 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.

Practice question 2Linear and quadratic functions

A manufacturer is the only supplier of a spare part. To sell x units it can charge the price p(x) = 90 − 0.5x, in MU (monetary units). Its costs are C(x) = 400 + 10x. At what quantity is its profit highest?

1. a)50
2. b)80
3. c)90
4. d)100
5. e)160

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Revenue R(x) = p(x) · x = 90x − 0.5x².
2. Profit π(x) = 90x − 0.5x² − (400 + 10x) = −0.5x² + 80x − 400.
3. The parabola opens downwards; its vertex is at x = −80 / (2 · (−0.5)) = 80.
4. Check: the price is then p(80) = 50 MU, the profit π(80) = 2,800 MU.

a)

Wrong. The price instead of the quantity: at the profit maximum p(80) = 50.

0 or −0.25

b)

Right. Vertex of the profit parabola π(x) = −0.5x² + 80x − 400: x = −80 / (2 · (−0.5)) = 80.

+1

c)

Wrong. Revenue maximised instead of profit: vertex of R(x) = 90x − 0.5x².

0 or −0.25

d)

Wrong. The variable costs added instead of subtracted: π(x) = −0.5x² + 100x − 400, vertex at 100.

0 or −0.25

e)

Wrong. The price set equal to the variable cost of 10 MU per unit, as if it only had to cover that: 90 − 0.5x = 10.

0 or −0.25

–

Left blank.

0

Typical mistake. Maximising revenue instead of profit. The vertex of R(x) is at 90. Because each extra unit costs 10 MU, the vertex of the profit lies further left, at 80.

Ruling out before you finish. If every unit costs something, the profit-maximising quantity lies to the left of the revenue maximum. Revenue peaks halfway between 0 and 180, the quantity at which the price reaches zero, so at 90. That rules out options c, d and e 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.

Practice question 3Linear and quadratic functions

A language school is the only one in town running a particular intensive course. At a price of p MU (monetary units), x(p) = 200 − 4p people sign up. It has fixed costs of 300 MU and costs of 8 MU per person. What is its highest possible profit?

1. a)1,464 MU
2. b)1,764 MU
3. c)2,004 MU
4. d)2,304 MU
5. e)2,436 MU

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Solve for the price: 4p = 200 − x, so p(x) = 50 − 0.25x. (Working in p with π(p) = (p − 8) · (200 − 4p) − 300 gives the same result.)
2. Profit π(x) = (50 − 0.25x) · x − 300 − 8x = −0.25x² + 42x − 300.
3. Vertex at x = −42 / (2 · (−0.25)) = 84 people, at the price p(84) = 29 MU.
4. π(84) = −1,764 + 3,528 − 300 = 1,464 MU.

a)

Right. π(84) = −0.25 · 84² + 42 · 84 − 300 = 1,464.

+1

b)

Wrong. The fixed costs of 300 MU not subtracted.

0 or −0.25

c)

Wrong. The demand read the wrong way round, as the price for a quantity, p(x) = 200 − 4x: π(x) = −4x² + 192x − 300, vertex at 24.

0 or −0.25

d)

Wrong. The demand read the wrong way round and the fixed costs forgotten: −4 · 24² + 192 · 24.

0 or −0.25

e)

Wrong. The revenue at the profit maximum, without subtracting the costs: 29 · 84.

0 or −0.25

–

Left blank.

0

Typical mistake. Reading the demand the wrong way round. x(p) = 200 − 4p gives the quantity at a price. Either solve it for the price, p(x) = 50 − 0.25x, or work in the price directly: π(p) = (p − 8) · (200 − 4p) − 300 has its vertex at p = 29, with the same profit. Reading 200 − 4x as the price for quantity x leads to 2,004 MU.

Ruling out before you finish. The largest revenue is at x = 100, at a price of 25 MU, and comes to 2,500 MU. Profit is always less than revenue minus the fixed costs, so it cannot reach 2,500 − 300 = 2,200 MU. That rules out options d and e and leaves three.

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.

WU's rules for the Mathematik (CBK) exam

WU does not publish which of the 20 questions covers which chapter. This article therefore follows the structure of the textbook, where costs and revenue (Kosten und Erlöse) and the linear market model for the monopoly (Lineares Marktmodell) are sections of the first chapter.²

You may use a calculator, but not one with functions for differential calculus, integration or matrices, not one that solves linear systems, and not one with a text memory. If German is not your first language you may bring a dictionary; the syllabus allows nothing else.¹ Any permitted model will do for this chapter: you need basic arithmetic, squares and square roots.

WU's guideline for conducting exams bans formula sheets and watches of any kind unless the syllabus says otherwise.³ So you cannot bring your own formula sheet, and WU does not say whether one is handed out. Learn the vertex formula and the break-even formula by heart.

The exam timetable allows two hours for the whole exam; the syllabus does not say how long you actually write.⁴ Spread over 20 questions, that is six minutes a question at most. If a question on cost or monopoly comes up, you want to solve it quickly, so that time is left for questions that take more calculation.

Next step after cost and monopoly: Mathematik (CBK)

1. Read the sections Kosten und Erlöse and Lineares Marktmodell (Monopol) in the free textbook by Birgit Rudloff and Achim Zeileis.⁵ Then do the three questions above without looking at the solutions first.
2. If you got one wrong, find the mistake in the solution that matches your answer. If it was an arithmetic slip in a monopoly question, make the halfway-price check a habit. If it was a mix-up, write down before calculating whether the question asks for a quantity, a price or a profit.
3. Arithmetic sequences are another section of the same chapter. Exam dates and how to register are in our 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 1: Lineare und quadratische Funktionenmathe4wiwi.org · undated page · checked on 26 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