Tasks 03-01 - Function Concepts & Business Modeling

Section 03: Functions as Business Models

A: Skills

Problem 1: Creating Tables and Graphs (x)

A streaming service charges according to the function \(S(h) = 10 + 2h\), where \(h\) is the number of hours watched per month and \(S(h)\) is the total cost in EUR.

  1. Create a table of values for \(h = 0, 5, 10, 15, 20, 25\) hours.

  2. Sketch the graph of \(S\) on graph paper or with graphing software.

  3. Explain what the number 10 in the function term means for a customer.

  4. A customer has a budget of at most 30 EUR per month. Determine how many hours per month the customer can watch.

  1. Table of values:
Hours \(h\) Cost \(S(h)\)
0 10 EUR
5 20 EUR
10 30 EUR
15 40 EUR
20 50 EUR
25 60 EUR

  1. Graph: plot the points from the table and connect them.

    • Points: \((0, 10)\), \((5, 20)\), \((10, 30)\), \((15, 40)\), \((20, 50)\), \((25, 60)\)
    • The graph is a straight line. It starts at \((0, 10)\) on the vertical axis and rises by 2 EUR per hour
    • Only \(h \geq 0\) makes sense, so the line starts at \(h = 0\)

  2. The number 10: \(S(0) = 10\). A customer who watches nothing still pays 10 EUR. The 10 EUR are the fixed monthly fee, the 2 EUR are the price per hour.

  3. Budget:

    • At most 30 EUR: \(10 + 2h \leq 30\)
    • \(2h \leq 20\)
    • \(h \leq 10\)
    • The customer can watch at most 10 hours per month. Check: \(S(10) = 30\) EUR ✓

Problem 2: Domain and Range (x)

Determine the domain and the range of each function.

  1. \(f(x) = \frac{3}{x - 4}\)

  2. \(g(x) = \sqrt{x + 5}\)

  3. \(h(x) = 3x - 8\)

  4. \(p(x) = \frac{1}{x^2 + 1}\) Hint: dividing the same positive number by a larger number gives a smaller result.

Hint: a number \(y\) belongs to the range if the equation \(f(x) = y\) has a solution \(x\) in the domain.

  1. \(f(x) = \frac{3}{x - 4}\)
    • Domain: \(x - 4 \neq 0\), so \(x \neq 4\). Domain: \(\mathbb{R} \setminus \{4\}\) or \((-\infty; 4) \cup (4; \infty)\)
    • Range: the numerator 3 is not 0, so the fraction is never 0. Every other \(y\) is an output: \(x = 4 + \frac{3}{y}\) gives \(f(x) = \frac{3}{3/y} = y\). Range: \(\mathbb{R} \setminus \{0\}\)
  2. \(g(x) = \sqrt{x + 5}\)
    • Domain: \(x + 5 \geq 0\), so \(x \geq -5\). Domain: \([-5; \infty)\)
    • Range: a square root is never negative, and \(g(-5) = 0\). Every \(y \geq 0\) is an output: \(x = y^2 - 5\) gives \(g(x) = \sqrt{y^2} = y\). Range: \([0; \infty)\)
  3. \(h(x) = 3x - 8\)
    • Domain: no restrictions for a linear function. Domain: \(\mathbb{R}\) or \((-\infty; \infty)\)
    • Range: every \(y\) is an output: \(x = \frac{y + 8}{3}\) gives \(h(x) = y\). Range: \(\mathbb{R}\) or \((-\infty; \infty)\)
  4. \(p(x) = \frac{1}{x^2 + 1}\)
    • Domain: \(x^2 + 1\) is never zero (always \(\geq 1\)). Domain: \(\mathbb{R}\) or \((-\infty; \infty)\)
    • Range: since \(x^2 \geq 0\), we have \(x^2 + 1 \geq 1\). So \(0 < \frac{1}{x^2 + 1} \leq 1\). The value 1 occurs at \(x = 0\): \(p(0) = 1\). Every \(y\) with \(0 < y \leq 1\) is an output: \(\frac{1}{x^2 + 1} = y \iff x^2 = \frac{1}{y} - 1 \geq 0\) has the solution \(x = \sqrt{\frac{1}{y} - 1}\). Range: \((0; 1]\)

Problem 3: Function or Not? (x)

Decide whether each relation is a function and give a reason. In c) to g), \(x\) is the input. Use the idea of the vertical line test: every vertical line \(x = c\) may meet the graph at most once.

  1. The table
\(x\) 1 2 3 4
\(y\) 6 6 9 11

  1. The pairs \((0, 2)\), \((1, 3)\), \((1, 4)\), \((2, 5)\)

  2. \(y = 2x^2 - 7\)

  3. \(y^2 = x + 4\)

  4. \(x = 3\)

  5. \(y = 4\)

  6. \(x^2 + y = 9\)

  7. Each invoice of a shop is assigned its amount in EUR.

  8. Each amount in EUR is assigned the invoices of the shop with this amount.

  1. Function. Each \(x\) has exactly one \(y\). That \(x = 1\) and \(x = 2\) have the same output 6 is allowed.

  2. Not a function. The input 1 has two outputs, 3 and 4.

  3. Function. Each \(x\) gives exactly one value \(2x^2 - 7\). Every vertical line meets the parabola once.

  4. Not a function. For \(x = 5\): \(y^2 = 9\), so \(y = 3\) or \(y = -3\). The vertical line \(x = 5\) meets the graph twice (a sideways parabola).

  5. Not a function. The graph is a vertical line. The input 3 has infinitely many outputs (every \(y\)), and every other input has none.

  6. Function. Every \(x\) has the output 4. The graph is a horizontal line, and every vertical line meets it once.

  7. Function. Solve for \(y\): \(y = 9 - x^2\). Each \(x\) gives exactly one \(y\).

  8. Function. Each invoice has exactly one amount.

  9. Not a function. Two invoices can have the same amount, e.g. two customers who each pay 19.99 EUR. Then one input has two outputs.

Problem 4: Function Notation and Function Values (x)

Given \(f(x) = x^2 + 4x - 5\) and \(g(x) = \frac{2x - 1}{x + 3}\):

  1. Compute \(f(0)\), \(f(-5)\), \(f(1.5)\), \(g(2)\) and \(g(-1)\). Explain why \(g(-3)\) does not exist.

  2. Compute and simplify \(f(a - 1)\) and \(f(x + h) - f(x)\).

  3. Determine all \(x\) with \(f(x) = 7\) and all \(x\) with \(g(x) = 1\).

  4. A shoe repair shop has monthly total costs of \(C(x) = 1900 + 22x\) EUR for \(x\) repairs. Compute \(C(150)\) and interpret the result. Determine the \(x\) with \(C(x) = 4540\). Explain the meaning of \(C(151) - C(150)\).

  1. Function values:
    • \(f(0) = 0 + 0 - 5 = -5\)
    • \(f(-5) = 25 - 20 - 5 = 0\)
    • \(f(1.5) = 2.25 + 6 - 5 = 3.25\)
    • \(g(2) = \frac{4 - 1}{2 + 3} = \frac{3}{5} = 0.6\)
    • \(g(-1) = \frac{-2 - 1}{-1 + 3} = \frac{-3}{2} = -1.5\)
    • \(g(-3)\): the denominator is \(-3 + 3 = 0\), and we cannot divide by zero. \(-3\) does not belong to the domain of \(g\)
  2. Replace every \(x\) by the new input, in brackets:
    • \(f(a - 1) = (a - 1)^2 + 4(a - 1) - 5 = a^2 - 2a + 1 + 4a - 4 - 5 = a^2 + 2a - 8\)
    • \(f(x + h) = (x + h)^2 + 4(x + h) - 5 = x^2 + 2xh + h^2 + 4x + 4h - 5\)
    • \(f(x + h) - f(x) = x^2 + 2xh + h^2 + 4x + 4h - 5 - (x^2 + 4x - 5) = 2xh + h^2 + 4h\)
  3. Equations:
    • \(x^2 + 4x - 5 = 7 \iff x^2 + 4x - 12 = 0 \iff (x + 6)(x - 2) = 0\)
    • \(x = -6\) or \(x = 2\). Check: \(f(2) = 4 + 8 - 5 = 7\) ✓ and \(f(-6) = 36 - 24 - 5 = 7\) ✓
    • \(g(x) = 1\) with \(x \neq -3\): \(2x - 1 = x + 3 \iff x = 4\). Check: \(g(4) = \frac{7}{7} = 1\) ✓
  4. Repair shop:
    • \(C(150) = 1900 + 22 \cdot 150 = 1900 + 3300 = 5200\). With 150 repairs per month, the total costs are 5,200 EUR
    • \(1900 + 22x = 4540 \iff 22x = 2640 \iff x = 120\) repairs
    • \(C(151) - C(150) = 22\) EUR: the costs of one more repair (the variable costs per repair)

Problem 5: Domain and Range in Context (xx)

Each function describes a business situation. Give the domain mathematically (all inputs for which the term is defined) and in this context. Then determine the range in this context.

  1. A beekeeper sells about \(H(a) = 40 + 20\sqrt{a}\) jars of honey per week in their farm shop when they spend \(a\) hundred EUR per week on advertising. They spend at most 900 EUR per week on advertising.

  2. An online language school has about \(U(t) = 2000 \cdot \ln(t + 1)\) registered users \(t\) months after its launch. The model is meant for the first two years. Round to whole users. Hint: like \(\log_{10}\), \(\ln\) grows.

  3. A tanker truck delivers 12,000 litres of heating oil. Its pump moves \(v\) litres per minute, so emptying the tank takes \(T(v) = \frac{12000}{v}\) minutes. The pump can be set to any rate from 200 to 800 litres per minute. Hint: dividing the same positive number by a larger number gives a smaller result.

Hint: first decide whether the function grows or falls on the domain in the context. Then the range lies between the values at the two ends.

  1. Honey:
    • Mathematically: the root needs \(a \geq 0\), so \([0; \infty)\)
    • In this context: 900 EUR are 9 hundred EUR, so \(0 \leq a \leq 9\), i.e. \([0; 9]\)
    • \(\sqrt{a}\) grows with \(a\), so \(H\) grows. \(H(0) = 40\) and \(H(9) = 40 + 20 \cdot 3 = 100\)
    • Range: \([40; 100]\). Without advertising the shop sells about 40 jars per week, with the full budget about 100 jars
  2. Language school:
    • Mathematically: the logarithm needs \(t + 1 > 0\), so \(t > -1\), i.e. \((-1; \infty)\)
    • In this context: the first two years, so \([0; 24]\)
    • \(\ln\) grows, so \(U\) grows. \(U(0) = 2000 \cdot \ln(1) = 0\) and \(U(24) = 2000 \cdot \ln(25) \approx 6437.75\)
    • Range: about \([0; 6438]\) users
  3. Heating oil:
    • Mathematically: the denominator must not be zero, so \(v \neq 0\), i.e. \(\mathbb{R} \setminus \{0\}\)
    • In this context: \([200; 800]\)
    • The larger \(v\), the smaller \(\frac{12000}{v}\), e.g. \(\frac{12000}{300} = 40 > \frac{12000}{400} = 30\). So \(T\) falls. \(T(200) = 60\) and \(T(800) = 15\)
    • Range: \([15; 60]\) minutes. The faster the pump, the shorter the delivery

B: Set Up & Check

Problem 6: From Text to Functions (xx)

Translate each statement into a function or a domain, or complete it. Then answer the check question: it tells you whether your translation works.

  1. “A cosmetics start-up fills \(x\) jars of face cream per month. Its filling machine can fill at most 1,500 jars per month.” Give the domain of \(x\).

    Check: Do \(x = 1500\) and \(x = 750.5\) belong to the domain?

  2. “The monthly fee of a music school consists of a basic fee of 25 EUR for the practice rooms and 18 EUR for each lesson.”

    Check: A parent pays 115 EUR for one month. How many lessons did the child have?

  3. “A riding stable has 2,400 kg of hay in stock and feeds 150 kg per day to its horses. No new delivery arrives, and the stock cannot be negative.”

    Check: Does \(t = 20\) days belong to the domain?

  4. “A catering firm (a company that delivers meals for events) has total costs of \(C(x) = 1150 + 7.40x\) EUR per day for \(x\) meals. \(C(0)\) means …”

    Check: Is \(C(0) = 0\), since no meal is cooked?

  5. “A smoothie stand at a train station makes a weekly profit of \(P(x) = 2.80x - 1260\) EUR with \(x\) smoothies. It can sell at most 900 smoothies per week. The range of the profit function in this context is …”

    Check: Can the stand make a weekly profit of 1,500 EUR?

  1. \(x\) is a whole number with \(0 \leq x \leq 1500\), i.e. the domain is \(\{0, 1, 2, \ldots, 1500\}\).

    Check: \(x = 1500\) belongs to the domain, since “at most” includes 1,500. \(x = 750.5\) does not, since the start-up only fills whole jars.

    Typical mistake: \(x < 1500\). “At most” includes the boundary, so the sign is \(\leq\).

  2. With \(n\) the number of lessons per month: \(F(n) = 25 + 18n\) EUR, where \(n\) is a whole number with \(n \geq 0\).

    Check: \(25 + 18n = 115 \iff 18n = 90 \iff n = 5\) lessons.

    Typical mistake: \(F(n) = (25 + 18)n = 43n\). The basic fee is paid once per month, not for each lesson. The check shows the mistake: \(\frac{115}{43} \approx 2.67\) is not a whole number of lessons.

  3. Stock after \(t\) days: \(S(t) = 2400 - 150t\) kg. The stock cannot be negative:

    \(2400 - 150t \geq 0 \iff 150t \leq 2400 \iff t \leq 16\)

    Domain: \([0; 16]\). The stock falls from \(S(0) = 2400\) kg to \(S(16) = 0\) kg, so the range is \([0; 2400]\).

    Check: no, \(S(20) = 2400 - 3000 = -600 < 0\).

    Typical mistake: the domain \(t \geq 0\) only. “The stock cannot be negative” is a condition on the output \(S(t)\), and it gives the upper bound for the input \(t\).

  4. \(C(0) = 1150 + 7.40 \cdot 0 = 1150\) EUR: the costs per day when the firm cooks no meal, i.e. its fixed costs (e.g. rent, salaries, vans).

    Check: no, \(C(0) = 1{,}150\) EUR.

    Typical mistake: “\(C(0) = 0\), since nothing is produced.” The 0 is the input (0 meals), not the output. Fixed costs arise even without production.

  5. The domain is \(\{0, 1, \ldots, 900\}\). Each smoothie adds 2.80 EUR, so \(P\) grows with \(x\). The smallest value is \(P(0) = -1260\) EUR (a loss equal to the fixed costs), the largest is \(P(900) = 2520 - 1260 = 1260\) EUR.

    Range in this context: from \(-1{,}260\) EUR to \(1{,}260\) EUR in steps of 2.80 EUR, i.e. within \([-1260; 1260]\).

    Check: no, 1,500 EUR lies above 1,260 EUR, outside the range.

    Typical mistake: the range \(\mathbb{R}\) (the range of the term without the context) or \([0; 1260]\). A profit can be negative, and a negative profit is a loss.

Problem 7: One Paragraph, All Conditions (xx)

An escape room business (groups of players solve puzzles to get out of a themed room) has 2 rooms. Each room can host at most 6 games per day, and the business opens on 25 days per month. A group pays 96 EUR per game. Each game causes 8 EUR for cleaning and resetting the room. The fixed costs for rent, staff and the website are 6,800 EUR per month. The owner wants a monthly profit of at least 12,000 EUR. Their bank only grants a loan for a third room if the monthly revenue is above 24,500 EUR.

List the total costs \(C(x)\), the revenue \(R(x)\) and the profit \(P(x)\) with \(x\) the number of games per month, the domain in this context and every condition as a mathematical statement. Do not solve anything.

Check: Which of your conditions rules out 240 games per month?

  1. Games: \(x\) is a whole number
  2. Capacity: 2 rooms with 6 games each on 25 days, so \(0 \leq x \leq 2 \cdot 6 \cdot 25 = 300\). Domain: \(\{0, 1, \ldots, 300\}\)
  3. Total costs: \(C(x) = 6800 + 8x\) EUR
  4. Revenue: \(R(x) = 96x\) EUR
  5. Profit: \(P(x) = R(x) - C(x) = 96x - (6800 + 8x) = 88x - 6800\) EUR
  6. Profit target: \(88x - 6800 \geq 12000\)
  7. Bank: \(96x > 24500\)

Check: for \(x = 240\) the profit is \(88 \cdot 240 - 6800 = 21120 - 6800 = 14{,}320\) EUR \(\geq 12{,}000\) EUR, so condition 6 holds. The revenue is \(96 \cdot 240 = 23{,}040\) EUR, not above 24,500 EUR. Condition 7 (the bank) rules out 240 games.

Typical mistake: the capacity \(x \leq 12\) (or \(x \leq 6\)). The 2 rooms can each host 6 games on each of the 25 days, so the capacity is 300 games per month.

C: Applied

Problem 8: Celebration Cakes of a Patisserie (xx)

A patisserie (a pastry shop) bakes celebration cakes to order. Rent and utilities cost 2,500 EUR per month. Each cake needs ingredients for 12 EUR and labour for 8 EUR. The patisserie sells a cake for 45 EUR. Its capacity (the maximum possible output) is 200 cakes per month. Let \(x\) be the number of cakes per month.

  1. Give terms for the total costs \(C(x)\) and the revenue \(R(x)\) in EUR. Show that the profit function is \(P(x) = 25x - 2500\). Give the domain of these functions in this context.

(for verification: \(C(x) = 2500 + 20x\))

  1. Create a table with \(C(x)\), \(R(x)\) and \(P(x)\) for \(x = 0, 50, 100, 150, 200\). Interpret the values \(P(0)\) and \(P(50)\).

  2. Sketch the graphs of \(C\) and \(R\) in one coordinate system. Mark the break-even point in your sketch and compute it exactly.

(for verification: 100 cakes)

  1. Determine the range of the profit function in this context. Give the largest possible monthly profit and explain why it occurs at the capacity.

  2. Rent and utilities rise by 10 %. Determine the new break-even point. The owner wants a monthly profit of at least 1,500 EUR. Determine how many cakes per month the patisserie must now sell.

a) Fixed costs 2,500 EUR, variable costs \(12 + 8 = 20\) EUR per cake:

\[C(x) = 2500 + 20x, \qquad R(x) = 45x\]

\[P(x) = R(x) - C(x) = 45x - (2500 + 20x) = 45x - 2500 - 20x = 25x - 2500 \quad \checkmark\]

Domain in this context: whole numbers of cakes up to the capacity, i.e. \(\{0, 1, 2, \ldots, 200\}\).

b)

Cakes \(x\) \(C(x)\) \(R(x)\) \(P(x)\)
0 2,500 EUR 0 EUR \(-2{,}500\) EUR
50 3,500 EUR 2,250 EUR \(-1{,}250\) EUR
100 4,500 EUR 4,500 EUR 0 EUR
150 5,500 EUR 6,750 EUR 1,250 EUR
200 6,500 EUR 9,000 EUR 2,500 EUR

\(P(0) = -2500\): without any cakes, the patisserie loses its fixed costs of 2,500 EUR. \(P(50) = -1250\): with 50 cakes it still makes a loss of 1,250 EUR. Each cake reduces the loss by 25 EUR.

c) Both graphs are straight lines. \(C\) runs from \((0, 2500)\) to \((200, 6500)\), \(R\) from \((0, 0)\) to \((200, 9000)\). They meet at \((100, 4500)\): the break-even point.

Exactly: \(R(x) = C(x) \iff 45x = 2500 + 20x \iff 25x = 2500 \iff x = 100\).

The break-even point is at 100 cakes, with revenue and total costs of 4,500 EUR each. Below 100 cakes the patisserie makes a loss, above 100 cakes a profit.

d) Each cake adds 25 EUR to the profit, so \(P\) grows with \(x\). The smallest value is \(P(0) = -2500\), the largest is \(P(200) = 25 \cdot 200 - 2500 = 2500\).

Range in this context: from \(-2{,}500\) EUR to \(2{,}500\) EUR in steps of 25 EUR, i.e. within \([-2500; 2500]\).

The largest possible monthly profit is 2,500 EUR with 200 cakes. Since the profit grows with every cake, more cakes always mean more profit, and the capacity stops the output at 200 cakes.

e) New rent and utilities: \(2500 \cdot 1.10 = 2750\) EUR. New profit function: \(P_{\text{new}}(x) = 25x - 2750\).

New break-even point: \(25x - 2750 = 0 \iff x = 110\) cakes.

Profit target:

\[25x - 2750 \geq 1500 \iff 25x \geq 4250 \iff x \geq 170\]

Together with the capacity, the patisserie must sell 170 to 200 cakes per month. Check: \(P_{\text{new}}(170) = 4250 - 2750 = 1500\) EUR ✓

Problem 9: Weekly Production Plan of an Electronics Manufacturer (xx)

A small electronics manufacturer produces tablets and smartphones. Each tablet needs 3 hours of assembly and 2 hours of testing. Each smartphone needs 2 hours of assembly and 1 hour of testing. Per week, the factory has 120 hours of assembly time and 70 hours of testing time. A tablet sells for 300 EUR and costs 180 EUR to produce. A smartphone sells for 200 EUR and costs 110 EUR to produce. Let \(t\) be the number of tablets and \(s\) the number of smartphones per week.

  1. Give the conditions that \(t\) and \(s\) must meet as inequalities.

  2. Give terms for the weekly revenue and the weekly profit in EUR. Fixed costs are not considered.

(for verification: profit \(120t + 90s\) EUR)

  1. Check whether the plans “30 tablets and 20 smartphones” and “24 tablets and 20 smartphones” are possible. Compute the profit of each possible plan.

  2. The manufacturer produces only tablets. Give the domain of the profit function \(P(t) = 120t\) in this context and its range.

  3. Determine the plan that uses the assembly time and the testing time completely. Compute its profit and compare it with the largest profit from d).

(for verification: 20 tablets)

a)

  • Assembly time: \(3t + 2s \leq 120\)
  • Testing time: \(2t + s \leq 70\)
  • \(t\) and \(s\) are whole numbers with \(t \geq 0\) and \(s \geq 0\)

b) Revenue: \(300t + 200s\) EUR. Profit per tablet: \(300 - 180 = 120\) EUR, per smartphone: \(200 - 110 = 90\) EUR. Profit: \(120t + 90s\) EUR.

c)

  • 30 tablets and 20 smartphones: assembly \(3 \cdot 30 + 2 \cdot 20 = 130\) hours \(> 120\) hours, testing \(2 \cdot 30 + 20 = 80\) hours \(> 70\) hours. Not possible, both conditions are violated
  • 24 tablets and 20 smartphones: assembly \(72 + 40 = 112 \leq 120\) ✓, testing \(48 + 20 = 68 \leq 70\) ✓. Possible, with a profit of \(120 \cdot 24 + 90 \cdot 20 = 2880 + 1800 = 4{,}680\) EUR

d) With \(s = 0\): assembly \(3t \leq 120 \iff t \leq 40\), testing \(2t \leq 70 \iff t \leq 35\). Both conditions must hold, so \(t \leq 35\). The testing time is the limit.

Domain: \(\{0, 1, 2, \ldots, 35\}\).

Each tablet adds 120 EUR, so \(P\) grows with \(t\). Range: \(\{0, 120, 240, \ldots, 4200\}\), i.e. from 0 EUR to \(P(35) = 4{,}200\) EUR in steps of 120 EUR.

e) Both times are used completely:

\[\begin{cases} 3t + 2s = 120 \\ 2t + s = 70 \end{cases}\]

From the second equation: \(s = 70 - 2t\). Substitute into the first:

\[3t + 2(70 - 2t) = 120 \iff 3t + 140 - 4t = 120 \iff t = 20\]

Then \(s = 70 - 40 = 30\). Check: \(3 \cdot 20 + 2 \cdot 30 = 120\) ✓ and \(2 \cdot 20 + 30 = 70\) ✓

The plan is 20 tablets and 30 smartphones per week, with a profit of \(120 \cdot 20 + 90 \cdot 30 = 2400 + 2700 = 5{,}100\) EUR. That is 900 EUR more than the 4,200 EUR with tablets only.

Problem 10: Customers of a Grocery Delivery Service (xx)

A new grocery delivery service starts in a town with 6,000 households. It models the share of households that have ordered at least once by

\[A(t) = \frac{100t}{t + 8}\]

percent, \(t\) weeks after the start. The model is meant for the first year (\(t\) up to 52 weeks).

  1. Compute \(A(0)\), \(A(8)\) and \(A(24)\) and interpret the values.

  2. Give the domain of \(A\) mathematically and in this context. Show that \(A(t) = 100 - \frac{800}{t + 8}\) and explain with this form why the share always stays below 100 %.

  3. Determine the range of \(A\) in this context, in percent rounded to two decimal places. Hint: dividing the same positive number by a larger number gives a smaller result.

(for verification: the largest value is about 86.67 %)

  1. Give a term for the number of households \(N(t)\) that have ordered at least once. Determine when 5,000 households have ordered.

(for verification: \(N(t) = \frac{6000t}{t + 8}\))

  1. The owner claims that 95 % of the households will have ordered within the first year. Decide with the result of c) whether the claim is right. Determine when the model reaches 95 %.

a)

  • \(A(0) = \frac{0}{8} = 0\): at the start, no household has ordered yet
  • \(A(8) = \frac{800}{16} = 50\): after 8 weeks, half of the households have ordered
  • \(A(24) = \frac{2400}{32} = 75\): after 24 weeks, three quarters have ordered

The first 50 % take 8 weeks, the next 25 % take 16 more weeks: the growth slows down.

b) Mathematically: \(t + 8 \neq 0\), so \(t \neq -8\), i.e. \(\mathbb{R} \setminus \{-8\}\). In this context: \([0; 52]\).

Common denominator:

\[100 - \frac{800}{t + 8} = \frac{100(t + 8) - 800}{t + 8} = \frac{100t + 800 - 800}{t + 8} = \frac{100t}{t + 8} = A(t) \quad \checkmark\]

For \(t \geq 0\), the denominator \(t + 8\) is positive, so \(\frac{800}{t + 8} > 0\). \(A(t)\) is 100 minus a positive number and therefore always below 100 %.

c) As \(t\) grows, the denominator \(t + 8\) grows, and 800 divided by a larger number gives a smaller result, e.g. \(\frac{800}{10} = 80 > \frac{800}{20} = 40\). So \(\frac{800}{t + 8}\) gets smaller and \(A\) grows. The smallest value is \(A(0) = 0\), the largest is

\[A(52) = \frac{5200}{60} \approx 86.67\]

Range in this context: \([0; 86.67]\) percent (rounded).

d) 1 % of 6,000 households are 60 households:

\[N(t) = 60 \cdot A(t) = \frac{6000t}{t + 8}\]

5,000 households, with \(t + 8 \neq 0\):

\[\frac{6000t}{t + 8} = 5000 \iff 6000t = 5000(t + 8) \iff 6000t = 5000t + 40000 \iff t = 40\]

After 40 weeks, 5,000 households have ordered. \(t = 40\) lies in \([0; 52]\). Check: \(N(40) = \frac{240000}{48} = 5000\) ✓

e) 95 % is not in the range \([0; 86.67]\) from c). Within the first year, at most about 86.67 % of the households order. The claim is wrong.

\[\frac{100t}{t + 8} = 95 \iff 100t = 95t + 760 \iff 5t = 760 \iff t = 152\]

The model reaches 95 % only after 152 weeks, i.e. after almost three years (\(\frac{152}{52} \approx 2.9\)), if it still holds for that long.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Standing Desks from a Carpentry Workshop [21 pts. total]

A carpentry workshop builds height-adjustable standing desks for offices. Its fixed costs, which arise even when no desk is built, amount to 29,000 EUR per month. Each additional desk costs a little more than the one before, because larger outputs need overtime pay and express deliveries of oak wood. For an output of \(x\) QU per month, the workshop models its variable costs (they grow with the output) by \((x^2 + 34x)\) MU. It sells every desk for 480 EUR. Its capacity (the maximum possible output) is 1,300 desks per month.

(QU = quantity units, MU = monetary units. 1 QU = 100 desks, 1 MU = 1,000 EUR. Total costs = fixed costs + variable costs. Revenue = quantity × price. Profit = revenue − total costs.)

Part a)

Give a term for the total costs \(C(x)\) in MU and the domain of the cost function in this context. Compute \(C(4)\) and interpret the value in EUR. [3 pts.]

(for verification: \(C(x) = x^2 + 34x + 29\))

Total costs. Fixed costs: 29,000 EUR \(= 29\) MU. Total costs = fixed costs + variable costs:

\[C(x) = x^2 + 34x + 29\]

\(C(0) = 29\) MU are the fixed costs: the workshop pays them even when it builds no desk.

Domain. The output cannot be negative, and the capacity is 1,300 desks \(= 13\) QU:

\[D = [0; 13]\]

Since only whole desks are built, \(x\) is a multiple of 0.01 QU (1 desk).

\(C(4)\): \(C(4) = 16 + 136 + 29 = 181\) MU \(= 181{,}000\) EUR. With 400 desks per month, the total costs are 181,000 EUR.

Grading: 1 Pt. \(C(x)\) with the fixed costs converted to MU, 1 Pt. domain \([0; 13]\) with the capacity converted to QU, 1 Pt. \(C(4)\) in EUR with its meaning.

Part b)

Give a term for the revenue function \(R(x)\) in MU. Show that the profit function is given by \(P(x) = -x^2 + 14x - 29\). [3 pts.]

Revenue. 1 QU = 100 desks at 480 EUR each \(= 48{,}000\) EUR \(= 48\) MU:

\[R(x) = 48x\]

Profit:

\[P(x) = R(x) - C(x) = 48x - (x^2 + 34x + 29) = -x^2 + 14x - 29 \quad \checkmark\]

Grading: 1 Pt. \(R(x) = 48x\) with the conversion to MU, 1 Pt. \(P = R - C\), 1 Pt. simplifying with the sign change.

Part c)

Determine the ranges of \(C\) and \(R\) in this context. Compute the profit at full capacity in EUR and explain what it means for the workshop. [3 pts.]

Range of \(C\). For \(x \geq 0\), both \(x^2\) and \(34x\) grow with \(x\), so \(C\) grows. The smallest value is \(C(0) = 29\), the largest is \(C(13) = 169 + 442 + 29 = 640\):

\[[29; 640] \text{ MU, i.e. from } 29{,}000 \text{ EUR to } 640{,}000 \text{ EUR}\]

Range of \(R\). \(R\) grows with \(x\): \(R(0) = 0\), \(R(13) = 624\), so \([0; 624]\) MU.

Profit at full capacity:

\[P(13) = R(13) - C(13) = 624 - 640 = -16 \text{ MU} = -16{,}000 \text{ EUR}\]

(Check with b): \(-169 + 182 - 29 = -16\) ✓) At full capacity the workshop makes a loss of 16,000 EUR per month: the total costs grow faster than the revenue, so building as many desks as possible does not pay.

Grading: 1 Pt. range of \(C\) with a reason, 1 Pt. range of \(R\), 1 Pt. loss of 16,000 EUR with its meaning.

Part d)

Determine the (lower) break-even point and the profit limit (upper break-even point) with the quadratic formula. Give both exactly and in QU, rounded to two decimal places. [5 pts.]

(for verification: \(x_1 = 7 - 2\sqrt{5} \approx 2.53\))

Zero form. At a break-even point the profit is zero, \(P(x) = 0\). Multiply by \(-1\):

\[-x^2 + 14x - 29 = 0 \;\big|\cdot(-1) \iff x^2 - 14x + 29 = 0\]

Discriminant: \(\Delta = (-14)^2 - 4 \cdot 1 \cdot 29 = 196 - 116 = 80 > 0\). So there are two zeros. 80 is not a perfect square, so the quadratic formula is needed:

\[x_{1,2} = \frac{14 \pm \sqrt{80}}{2} = \frac{14 \pm 4\sqrt{5}}{2} = 7 \pm 2\sqrt{5}\]

\[x_1 = 7 - 2\sqrt{5} \approx 2.53, \qquad x_2 = 7 + 2\sqrt{5} \approx 11.47\]

Check \(x_1\) with the perfect square formula: \((7 - 2\sqrt{5})^2 - 14(7 - 2\sqrt{5}) + 29 = 69 - 28\sqrt{5} - 98 + 28\sqrt{5} + 29 = 0\) ✓

Both zeros lie in the domain \([0; 13]\). Break-even point: \(x_1 = 7 - 2\sqrt{5} \approx 2.53\) QU. Profit limit: \(x_2 = 7 + 2\sqrt{5} \approx 11.47\) QU.

Grading: 1 Pt. zero form, 1 Pt. discriminant, 1 Pt. quadratic formula, 1 Pt. exact and rounded values, 1 Pt. naming break-even point and profit limit.

Part e)

Decide in which output range the workshop makes a profit and state this profit zone as an interval. Give the smallest and the largest number of desks per month with which the workshop makes a profit. [3 pts.]

Test values below, between and above the zeros:

  • \(x = 1\): \(P(1) = -1 + 14 - 29 = -16 < 0\), a loss
  • \(x = 5\): \(P(5) = -25 + 70 - 29 = 16 > 0\), a profit
  • \(x = 13\): \(P(13) = -16 < 0\), a loss (part c)

Profit zone: \((7 - 2\sqrt{5}; 7 + 2\sqrt{5}) \approx (2.53; 11.47)\). The interval is open, since the profit is zero at both ends, and it lies inside the domain \([0; 13]\).

Whole desks. \(x_1 \approx 2.5279\) QU \(= 252.79\) desks and \(x_2 \approx 11.4721\) QU \(= 1147.21\) desks. Round the lower bound up and the upper bound down: the workshop makes a profit with 253 to 1,147 desks per month.

Check: 252 desks give \(P(2.52) \approx -0.0704\) MU, a loss of about 70 EUR, and 1,148 desks give the same loss.

Grading: 1 Pt. test values or sign argument, 1 Pt. open interval, 1 Pt. smallest and largest number of desks, rounded in the right direction.

Part f)

The workshop currently sells 750 desks per month. An office furniture retailer offers to buy another 350 desks per month at the same price. Compute the monthly profit in EUR without and with the order. Compute the additional total costs per additional desk and decide with a reason whether the workshop should accept the order. [4 pts.]

Without the order: 750 desks \(= 7.5\) QU.

\[P(7.5) = -56.25 + 105 - 29 = 19.75 \text{ MU} = 19{,}750 \text{ EUR}\]

With the order: 1,100 desks \(= 11\) QU, within the capacity of 13 QU.

\[P(11) = -121 + 154 - 29 = 4 \text{ MU} = 4{,}000 \text{ EUR}\]

Additional costs: \(C(11) = 121 + 374 + 29 = 524\) MU and \(C(7.5) = 56.25 + 255 + 29 = 340.25\) MU.

\[C(11) - C(7.5) = 183.75 \text{ MU} = 183{,}750 \text{ EUR}, \qquad \frac{183750}{350} = 525 \text{ EUR per additional desk}\]

Decision: each additional desk costs 525 EUR but brings only 480 EUR, a loss of 45 EUR per desk. The profit falls by \(350 \cdot 45 = 15{,}750\) EUR, from 19,750 EUR to 4,000 EUR. The workshop should not accept the order at this price, although 1,100 desks still lie in the profit zone.

Grading: 1 Pt. profit without the order in EUR, 1 Pt. profit with the order in EUR, 1 Pt. 525 EUR per additional desk, 1 Pt. decision with reason.

Calculator Practice

Problem 11: Calculator: Function Values in Table Mode (x)

Press MENU → 9 (Tabellen) and enter \(f(x) = 2x^2 + 3x - 2\) and \(g(x) = x + 2\). Press = and set Start \(= -1\), Ende \(= 2\) and Inkre \(= 0.5\).

  1. Give \(f(0)\), \(f(2)\), \(f(-1)\) and \(f(0.5)\) from the table.

  2. Give the \(x\)-value in your table at which \(f\) and \(g\) have the same value.

  3. Solve \(f(x) = g(x)\) by hand. Explain why the table shows only one of the solutions.

The table for \(x = -1, -0.5, 0, \ldots, 2\):

\(x\) \(-1\) \(-0.5\) \(0\) \(0.5\) \(1\) \(1.5\) \(2\)
\(f(x)\) \(-3\) \(-3\) \(-2\) \(0\) \(3\) \(7\) \(12\)
\(g(x)\) \(1\) \(1.5\) \(2\) \(2.5\) \(3\) \(3.5\) \(4\)

  1. \(f(0) = -2\), \(f(2) = 12\), \(f(-1) = -3\), \(f(0.5) = 0\) (so \(x = 0.5\) is a zero of \(f\))

  2. At \(x = 1\): \(f(1) = g(1) = 3\)

  3. \(2x^2 + 3x - 2 = x + 2 \iff 2x^2 + 2x - 4 = 0 \iff x^2 + x - 2 = 0 \iff (x + 2)(x - 1) = 0\)

    • Solutions: \(x = 1\) and \(x = -2\)
    • The table only starts at \(x = -1\), so it cannot show \(x = -2\). A table only shows the \(x\)-values you choose

Problem 12: Calculator: Value Tables (x)

Create a value table for \(f(x) = x^2 - x - 12\) from \(x = -4\) to \(x = 5\) with step 1.

  1. Give all function values.

  2. Give the \(x\)-values at which \(f(x) = 0\).

  3. Give the smallest value in your table and the \(x\)-values at which it occurs. Compute \(f(0.5)\) and explain what this shows about reading values from a table.

Press MENU → 9 (Tabellen), enter \(f(x)\), press = and set Start \(= -4\), Ende \(= 5\), Inkre \(= 1\).

  1. Value table for \(f(x) = x^2 - x - 12\):
\(x\) \(f(x)\)
\(-4\) \(8\)
\(-3\) \(0\)
\(-2\) \(-6\)
\(-1\) \(-10\)
\(0\) \(-12\)
\(1\) \(-12\)
\(2\) \(-10\)
\(3\) \(-6\)
\(4\) \(0\)
\(5\) \(8\)

  1. Zeros: \(f(x) = 0\) at \(x = -3\) and \(x = 4\). Check by factoring: \(x^2 - x - 12 = (x + 3)(x - 4)\) ✓

  2. The smallest value in the table is \(-12\), at \(x = 0\) and at \(x = 1\). But \(f(0.5) = 0.25 - 0.5 - 12 = -12.25\) is even smaller. A table only shows the values at the chosen \(x\). Between these \(x\)-values the function can take other values, so the smallest value in a table need not be the smallest value of the function.