Tasks 03-02 - Linear Functions & Economic Applications

Section 03: Functions as Business Models

A: Skills

Problem 1: Linear Function Forms (x)

Convert the following linear functions between different forms.

  1. Determine the equation of the line through the points \((3, 17)\) and \((7, 29)\) in slope-intercept form.

  2. Rewrite \(3x - 4y = 24\) in slope-intercept form.

  3. A line passes through \((5, 20)\) with slope \(m = -3\). Give its equation in point-slope form and in slope-intercept form.

  4. Determine the equation of the line parallel to \(y = 2x - 7\) that passes through the point \((4, 10)\).

  1. Finding the equation from two points:
    • Slope: \(m = \frac{29 - 17}{7 - 3} = \frac{12}{4} = 3\)
    • Using point-slope with \((3, 17)\): \(y - 17 = 3(x - 3)\)
    • Simplifying: \(y = 3x - 9 + 17 = 3x + 8\)
  2. Converting to slope-intercept form:
    • \(3x - 4y = 24\)
    • \(-4y = -3x + 24\)
    • \(y = \frac{3}{4}x - 6\)
  3. Line through \((5, 20)\) with slope \(-3\):
    • Point-slope: \(y - 20 = -3(x - 5)\)
    • Slope-intercept: \(y = -3x + 15 + 20 = -3x + 35\)
  4. Parallel line (same slope \(m = 2\)):
    • Through \((4, 10)\): \(y - 10 = 2(x - 4)\)
    • \(y = 2x - 8 + 10 = 2x + 2\)

Problem 2: Parallel and Perpendicular Lines (x)

In a) to d), decide whether the two lines are parallel, perpendicular or neither. Give a reason.

  1. \(y = 4x - 3\) and \(y = -0.25x + 6\)

  2. \(y = 1.5x + 2\) and \(3x - 2y = 10\)

  3. \(y = -5x + 2\) and \(y = 5x - 2\)

  4. \(y = \frac{2}{3}x + 1\) and \(3x + 2y = 4\)

  5. Determine the line through \((6, -1)\) that is perpendicular to \(y = 3x + 5\). Compute the intersection point of the two lines.

  6. Determine \(k\) so that the line \(y = kx - 5\) is perpendicular to \(y = 0.4x + 2\). Determine \(k\) so that it is parallel to \(2x + 5y = 1\).

  7. Two nail studios have monthly total costs of \(C_1(x) = 9.50x + 2300\) EUR and \(C_2(x) = 9.50x + 2750\) EUR for \(x\) appointments. Explain why the two graphs are parallel and what the vertical distance between them means. Decide whether there is a number of appointments at which both studios have the same costs.

  1. The slopes are \(4\) and \(-0.25\), and \(4 \cdot (-0.25) = -1\). Perpendicular.

  2. Solve the second equation for \(y\): \(-2y = -3x + 10\), so \(y = 1.5x - 5\). Both slopes are 1.5, the \(y\)-intercepts differ (2 and \(-5\)). Parallel.

  3. The slopes are \(-5\) and \(5\). They are not equal, and \((-5) \cdot 5 = -25 \neq -1\). Neither. Slopes with opposite signs are not enough for perpendicular lines: their product must be \(-1\).

  4. Solve the second equation for \(y\): \(2y = -3x + 4\), so \(y = -1.5x + 2\). \(\frac{2}{3} \cdot (-1.5) = -1\). Perpendicular.

  5. Perpendicular line: the slope \(m\) must satisfy \(3m = -1\), so \(m = -\frac{1}{3}\).

    • Point-slope form with \((6, -1)\): \(y + 1 = -\frac{1}{3}(x - 6)\)
    • \(y = -\frac{1}{3}x + 2 - 1 = -\frac{1}{3}x + 1\)

    Intersection: set the two terms equal and multiply by 3:

    • \(3x + 5 = -\frac{1}{3}x + 1 \iff 9x + 15 = -x + 3 \iff 10x = -12 \iff x = -1.2\)
    • \(y = 3 \cdot (-1.2) + 5 = 1.4\). Check: \(-\frac{1}{3} \cdot (-1.2) + 1 = 0.4 + 1 = 1.4\) ✓
    • Intersection point: \((-1.2, 1.4)\)

  6. Perpendicular: \(k \cdot 0.4 = -1\), so \(k = -2.5\).

    Parallel: \(2x + 5y = 1 \iff 5y = -2x + 1 \iff y = -0.4x + 0.2\). Same slope: \(k = -0.4\).

  7. Both costs rise by 9.50 EUR per appointment: the slopes (the variable costs per appointment) are equal, so the graphs are parallel.

    • Vertical distance: \(C_2(x) - C_1(x) = 2750 - 2300 = 450\) EUR for every \(x\). The second studio has 450 EUR higher fixed costs per month
    • Equal costs: \(9.50x + 2300 = 9.50x + 2750 \iff 2300 = 2750\), which is false. There is no such number of appointments. Parallel lines never meet, and the second studio is always 450 EUR more expensive

Problem 3: Slope and Intercept as Business Quantities (x)

In a) to d), give the slope and the \(y\)-intercept with their units and explain their meaning in the context. Decide whether the function is increasing or decreasing.

  1. A fruit press (a company that presses apples into juice) has weekly total costs of \(C(x) = 1850 + 0.65x\) EUR for \(x\) litres of apple juice.

  2. A zoo sells \(Q_d(p) = 900 - 25p\) tickets per day at a ticket price of \(p\) EUR. Also determine the maximum price.

  3. A tractor has the value \(V(t) = 26000 - 3250t\) EUR, \(t\) years after its purchase. Also determine after how many years it is fully depreciated.

  4. The asparagus farms of a region offer \(Q_s(p) = 60p - 300\) kg of asparagus per day at a price of \(p\) EUR per kg. Also explain why the \(y\)-intercept has no meaning here and determine the price above which the farms offer asparagus.

  5. A car-sharing provider charges a monthly fee of 9 EUR and 0.35 EUR per minute of driving. A student writes the monthly costs for \(t\) minutes as \(C(t) = 9t + 0.35\). Find the mistake, correct the term and compute the costs for 120 minutes in one month.

  1. Slope 0.65 EUR per litre: each additional litre of juice adds 0.65 EUR to the costs (the variable costs per litre). \(y\)-intercept 1,850 EUR: the fixed costs per week, paid even when no juice is pressed. Increasing.

  2. Slope \(-25\) tickets per EUR: each EUR more on the ticket price costs 25 sold tickets per day. \(y\)-intercept 900 tickets: the market saturation, the number of tickets demanded at a price of 0 EUR. Decreasing.

    Maximum price: \(900 - 25p = 0 \iff p = 36\) EUR. At 36 EUR or more, nobody buys a ticket.

  3. Slope \(-3{,}250\) EUR per year: the tractor loses 3,250 EUR of value per year (the depreciation per year). \(y\)-intercept 26,000 EUR: the purchase price. Decreasing.

    Fully depreciated: \(26000 - 3250t = 0 \iff t = \frac{26000}{3250} = 8\) years (the useful life).

  4. Slope 60 kg per EUR: for each EUR more per kg, the farms offer 60 kg more per day. \(y\)-intercept \(-300\) kg: a negative amount of asparagus cannot be offered. At a price of 0 EUR nobody offers anything, so the intercept only fixes the position of the line. Increasing.

    Supply starts: \(60p - 300 = 0 \iff p = 5\) EUR per kg. The farms offer asparagus only at prices above 5 EUR per kg.

  5. Slope and intercept are swapped. The fee of 9 EUR is paid once per month (the intercept), the 0.35 EUR are paid for each minute (the slope):

    \[C(t) = 0.35t + 9\]

    \(C(120) = 0.35 \cdot 120 + 9 = 42 + 9 = 51\) EUR. The wrong term would give \(9 \cdot 120 + 0.35 = 1{,}080.35\) EUR.

Problem 4: Linear Models from Data (x)

  1. A new branch of the public library counts its registered readers at the end of each month.
Month \(m\) 1 2 3 4
Readers 1,460 1,535 1,610 1,685

Check that the data follow a linear trend. Determine a linear model \(N(m)\) with the point-slope form. Forecast the number of readers in month 12 and determine the first month in which the library has more than 2,000 readers.

  1. A website has 2,000, 2,200, 2,420 and 2,662 visitors in months 1 to 4. Decide whether a linear model fits these data and give a reason.

  2. An electrician charges 139 EUR for a job of 2 hours and 244 EUR for a job of 5 hours. The price is a linear function \(E(h)\) of the working time \(h\) in hours. Determine \(E(h)\) and interpret the slope and the \(y\)-intercept. A customer paid 419 EUR. Determine how long the job took.

  1. Linear trend: the differences are \(1535 - 1460 = 75\), \(1610 - 1535 = 75\) and \(1685 - 1610 = 75\). The number grows by the same amount each month, so a linear model fits with slope 75 readers per month.

    • Point-slope form with \((1, 1460)\): \(N - 1460 = 75(m - 1)\)
    • \(N(m) = 75m - 75 + 1460 = 75m + 1385\)
    • Forecast: \(N(12) = 900 + 1385 = 2285\) readers
    • More than 2,000: \(75m + 1385 > 2000 \iff 75m > 615 \iff m > 8.2\). The first such month is month 9. Check: \(N(8) = 1985\) and \(N(9) = 2060\) ✓

  2. The differences are \(200\), \(220\) and \(242\). They are not constant, so a linear model does not fit. Instead, each value is 1.1 times the one before (\(\frac{2200}{2000} = \frac{2420}{2200} = \frac{2662}{2420} = 1.1\)): the visitors grow by 10 % per month, which is exponential growth.

  3. Slope: \(m = \frac{244 - 139}{5 - 2} = \frac{105}{3} = 35\) EUR per hour.

    • Point-slope form with \((2, 139)\): \(E - 139 = 35(h - 2)\)
    • \(E(h) = 35h - 70 + 139 = 35h + 69\)
    • The slope: each working hour costs 35 EUR. The \(y\)-intercept: a fixed fee of 69 EUR for each job (e.g. for the drive to the customer), paid however short the job is
    • \(35h + 69 = 419 \iff 35h = 350 \iff h = 10\). The job took 10 hours

Problem 5: Linear Depreciation (x)

A physiotherapy practice buys an ultrasound therapy device for 13,200 EUR. It depreciates the device linearly (the value falls by the same amount every year) over a useful life of 8 years. Let \(t\) be the time in years after the purchase.

  1. Determine the depreciation per year and the value function \(V(t)\).

  2. Give the domain and the range of \(V\) in this context.

  3. Compute \(V(3)\) and interpret the result.

  4. Determine when the device is worth 4,950 EUR.

  5. A driving school bought a car for 27,000 EUR and depreciates it linearly. Three years later, the car is worth 18,000 EUR. Determine the depreciation per year and the useful life of the car.

  1. Depreciation per year: \(\frac{13200}{8} = 1650\) EUR. Value function: \(V(t) = 13200 - 1650t\)

  2. Domain: \([0; 8]\) (from the purchase to the end of the useful life). \(V\) falls, so the range lies between \(V(8) = 0\) and \(V(0) = 13200\): range \([0; 13200]\) EUR.

  3. \(V(3) = 13200 - 4950 = 8250\). Three years after the purchase, the device is still worth 8,250 EUR.

  4. \(13200 - 1650t = 4950 \iff 1650t = 8250 \iff t = 5\). After 5 years.

  5. The car lost \(27000 - 18000 = 9000\) EUR in 3 years, i.e. \(\frac{9000}{3} = 3000\) EUR per year. \(V(t) = 27000 - 3000t\), and \(V(t) = 0\) for \(t = \frac{27000}{3000} = 9\). Useful life: 9 years.

B: Set Up & Check

Problem 6: From Text to Functions (xx)

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

  1. “A climbing hall (an indoor climbing gym) has found: at an entrance price of 0 EUR, 540 people per day would come (market saturation). At a price of 18 EUR or more, nobody comes (maximum price). The demand is linear.” Give the demand function \(Q_d(p)\).

    Check: How many visitors come per day at a price of 12 EUR?

  2. “A self-storage company rents out a storage unit of 4 m² for 58 EUR per month. For each additional square metre, the price rises by 11 EUR per month.” Give the monthly price \(p(a)\) of a unit with \(a\) square metres, \(a \geq 4\).

    Check: Does a unit of 10 m² cost 124 EUR per month?

  3. “A workshop sells bike lights for 19 EUR each. Material and assembly cost 11 EUR per light, and the fixed costs are 6,000 EUR per month. The contribution margin per light is … and the break-even quantity is …”

    Check: Does the workshop make a profit with 700 lights per month?

  4. “A tailoring shop buys an embroidery machine for 14,400 EUR. Its value falls by 12.5 % of the purchase price per year.” Give the value function \(V(t)\) and its domain in this context.

    Check: Is the machine worth 0 EUR after 8 years?

  5. “On a weekly market, the demand for goat cheese is given by \(p = 28 - 0.02Q\) (\(p\) in EUR per kg, \(Q\) in kg per week) and the supply by \(Q_s = 25p - 250\). The market is in equilibrium when supply equals demand.” Write the equilibrium condition as an equation for the price \(p\).

    Check: Is 22 EUR per kg the equilibrium price?

  1. Linear demand \(Q_d(p) = a - bp\). The market saturation is the quantity at a price of 0 EUR, so \(a = 540\). At the maximum price the demand is 0: \(540 - 18b = 0 \iff b = 30\).

    \(Q_d(p) = 540 - 30p\) for \(0 \leq p \leq 18\).

    Check: \(Q_d(12) = 540 - 360 = 180\) visitors per day.

    Typical mistake: \(Q_d(p) = 540 - 18p\), with the maximum price as the slope. Then \(Q_d(18) = 216\), although nobody comes at 18 EUR. The slope is market saturation divided by maximum price: \(\frac{540}{18} = 30\) visitors fewer per EUR.

  2. The price rises by 11 EUR per m², starting from the point \((4, 58)\). Point-slope form:

    \(p(a) - 58 = 11(a - 4) \iff p(a) = 11a - 44 + 58 = 11a + 14\) EUR.

    Check: yes, \(p(10) = 110 + 14 = 124\) EUR.

    Typical mistake: \(p(a) = 58 + 11a\). This adds 11 EUR for every square metre, although the 58 EUR already pay for the first 4 m². Then \(p(4) = 102 \neq 58\), and 10 m² would cost 168 EUR.

  3. Contribution margin: \(19 - 11 = 8\) EUR per light. Break-even quantity: \(\frac{6000}{8} = 750\) lights per month.

    Check: no. 700 lights bring contribution margins of \(700 \cdot 8 = 5{,}600\) EUR, which do not cover the fixed costs of 6,000 EUR: a loss of 400 EUR.

    Typical mistake: \(\frac{6000}{19} \approx 316\) lights, dividing the fixed costs by the price. The price also has to pay for material and assembly. Only the contribution margin of 8 EUR per light is left to cover the fixed costs.

  4. 12.5 % of the purchase price are \(0.125 \cdot 14400 = 1800\) EUR, the same amount every year:

    \(V(t) = 14400 - 1800t\). The value cannot be negative: \(14400 - 1800t \geq 0 \iff t \leq 8\). Domain: \([0; 8]\).

    Check: yes, \(V(8) = 14400 - 14400 = 0\) EUR.

    Typical mistake: \(V(t) = 14400 \cdot 0.875^t\). This takes 12.5 % of the current value every year (exponential decay). Then the value never reaches 0, e.g. \(V(8) \approx 4{,}948\) EUR. “Of the purchase price” means the same amount every year, so the model is linear.

  5. First write the demand as a quantity that depends on the price:

    \(p = 28 - 0.02Q \iff 0.02Q = 28 - p \iff Q_d = 1400 - 50p\).

    Equilibrium, supply equals demand: \(1400 - 50p = 25p - 250\).

    Check: yes. \(Q_d(22) = 1400 - 1100 = 300\) kg and \(Q_s(22) = 550 - 250 = 300\) kg ✓

    Typical mistake: \(28 - 0.02Q = 25p - 250\). The left side is a price, the right side a quantity. Write both functions as quantities that depend on the price (or both as prices that depend on the quantity) before setting them equal.

Problem 7: One Paragraph, All Conditions (xx)

On the regional market for firewood, the households would buy 2,800 m³ (cubic metres) per month at a price of 0 EUR (market saturation). At a price of 140 EUR per m³ or more, they buy no firewood (maximum price). The demand is linear. The forest owners only offer firewood at prices above 30 EUR per m³, and for every EUR above 30 EUR they offer 25 m³ more per month. One of them, a small forestry company, has fixed costs of 2,000 EUR per month and variable costs (cutting, splitting, delivery) of 45 EUR per m³. It can deliver at most 160 m³ per month. The company wants to know at which market price it breaks even when it sells 100 m³ per month. The city council considers a price cap of 60 EUR per m³ (firewood may not be sold at a higher price).

List the demand function and the supply function (with \(p\) the price in EUR per m³), the condition for the market equilibrium, the company’s total costs \(C(x)\) and revenue \(R(x)\) for \(x\) m³ per month at the market price \(p\), the domain of \(x\) in this context, the company’s break-even condition and the condition for a shortage under the price cap. Do not solve anything.

Check: Does the price cap of 60 EUR per m³ lead to a shortage? Which of your conditions shows it?

  1. Demand: market saturation 2,800 m³, maximum price 140 EUR, so the slope is \(\frac{2800}{140} = 20\) m³ per EUR: \(Q_d(p) = 2800 - 20p\) for \(0 \leq p \leq 140\)
  2. Supply: the line passes through \((30, 0)\) with slope 25: \(Q_s(p) = 25(p - 30) = 25p - 750\) for \(p \geq 30\)
  3. Market equilibrium: \(2800 - 20p = 25p - 750\)
  4. Company: \(C(x) = 2000 + 45x\) EUR and \(R(x) = p \cdot x\) EUR
  5. Domain: \(0 \leq x \leq 160\)
  6. Break-even at 100 m³: \(R(100) = C(100)\), i.e. \(100p = 2000 + 45 \cdot 100\)
  7. Shortage under the price cap: \(Q_d(60) > Q_s(60)\)

Check: \(Q_d(60) = 2800 - 1200 = 1600\) m³ and \(Q_s(60) = 1500 - 750 = 750\) m³. Condition 7 holds: the households want 850 m³ more than the forest owners offer. The price cap leads to a shortage.

Typical mistake: \(Q_s(p) = 30 + 25p\) or \(Q_s(p) = 25p\). The 30 EUR is the price at which the supply starts, i.e. the point \((30, 0)\) on the line, not a quantity. The point-slope form with \((30, 0)\) gives \(25p - 750\).

C: Applied

Problem 8: Strawberry Market in a Harvest Region (xx)

During the strawberry harvest, the farms of a region sell their strawberries at stands along the roads. At a price of \(p\) EUR per kg, the customers buy \(Q_d(p) = 3600 - 150p\) kg per day. At a price of 8 EUR per kg, the farms together offer 1,600 kg per day. For each additional EUR per kg, they offer 250 kg more.

  1. Explain the meaning of the numbers 3600 and 150 in \(Q_d(p)\). Determine the maximum price.

  2. Set up the supply function \(Q_s(p)\) with the point-slope form. Determine the price below which the farms offer no strawberries. Give the prices for which both functions make sense in this context.

(for verification: \(Q_s(p) = 250p - 400\))

  1. Determine the market equilibrium. Compute the farms’ total revenue per day at this price.

(for verification: 10 EUR per kg)

  1. At the start of the season, the stands ask 9 EUR per kg. Decide whether there is a shortage or a surplus, compute its size and explain in which direction the price will move.

  2. The farmers’ association guarantees its farms a price of 12 EUR per kg. At this price, it buys all strawberries that the customers do not buy and sells them to a jam factory for 3 EUR per kg. Compute the quantity the association buys per day and its daily loss. Compute the farms’ total revenue per day and its change compared with c) in percent, rounded to one decimal place.

a) \(Q_d(0) = 3600\) is the market saturation: at a price of 0 EUR the customers would take 3,600 kg per day. The 150 is the price sensitivity: for each EUR per kg that the price rises, the customers buy 150 kg less per day.

Maximum price, where the demand is 0:

\[3600 - 150p = 0 \iff 150p = 3600 \iff p = 24\]

At 24 EUR per kg or more, nobody buys strawberries.

b) The supply line passes through \((8, 1600)\) and rises by 250 kg per EUR:

\[Q_s(p) - 1600 = 250(p - 8) \iff Q_s(p) = 250p - 2000 + 1600 = 250p - 400 \quad \checkmark\]

No supply: \(250p - 400 = 0 \iff p = 1.60\). Below 1.60 EUR per kg, the farms offer no strawberries.

Both functions make sense where both quantities are not negative: \(Q_s(p) \geq 0\) for \(p \geq 1.60\) and \(Q_d(p) \geq 0\) for \(p \leq 24\). So \(1.60 \leq p \leq 24\), i.e. \([1.60; 24]\).

c) Supply equals demand:

\[3600 - 150p = 250p - 400 \iff 4000 = 400p \iff p^* = 10\]

\[Q^* = Q_d(10) = 3600 - 1500 = 2100 \qquad \text{Check: } Q_s(10) = 2500 - 400 = 2100 \; \checkmark\]

The equilibrium price is 10 EUR per kg, and 2,100 kg are sold per day. The farms’ total revenue is \(10 \cdot 2100 = 21{,}000\) EUR per day.

d) At 9 EUR per kg:

  • Demand: \(Q_d(9) = 3600 - 1350 = 2250\) kg
  • Supply: \(Q_s(9) = 2250 - 400 = 1850\) kg

The customers want \(2250 - 1850 = 400\) kg more than the farms offer: a shortage of 400 kg per day. The price lies below the equilibrium price. The strawberries sell out, so the stands can ask more, and the price rises towards 10 EUR per kg.

e) At 12 EUR per kg:

  • Demand: \(Q_d(12) = 3600 - 1800 = 1800\) kg
  • Supply: \(Q_s(12) = 3000 - 400 = 2600\) kg

The association buys the surplus of \(2600 - 1800 = 800\) kg per day. It pays \(800 \cdot 12 = 9{,}600\) EUR and gets \(800 \cdot 3 = 2{,}400\) EUR from the jam factory: a loss of 7,200 EUR per day.

The farms sell all 2,600 kg at 12 EUR: \(2600 \cdot 12 = 31{,}200\) EUR per day (21,600 EUR from the customers and 9,600 EUR from the association). Compared with c):

\[\frac{31200 - 21000}{21000} = \frac{10200}{21000} \approx 0.4857\]

The farms’ revenue rises by about 48.6 %. The customers buy 300 kg less than in the equilibrium and pay 2 EUR more per kg.

Problem 9: Beeswax Wraps from a Start-Up (xx)

A start-up sells sets of beeswax wraps (cotton cloths coated with beeswax that replace cling film). Its fixed costs for rent, a machine lease and insurance are 12,000 EUR per month. Material and labour cost 8 EUR per set, and a set sells for 20 EUR. Let \(x\) be the number of sets per month.

  1. Give terms for the total costs \(C(x)\) and the revenue \(R(x)\) in EUR. Show that the profit function is \(P(x) = 12x - 12000\). Compute the contribution margin per set and explain its meaning.

  2. Determine the break-even quantity and the revenue at the break-even point.

(for verification: 1,000 sets)

  1. The start-up wants a monthly profit of at least 6,000 EUR. Determine how many sets it must sell. It plans to sell 1,500 sets per month. Give the margin of safety (the number of sets sold above the break-even quantity) in sets and in percent of the planned sales, rounded to one decimal place.

  2. A supplier offers a faster coating machine. Its lease raises the fixed costs by 3,000 EUR per month, and the variable costs fall to 6.50 EUR per set. Give the new profit function, the new contribution margin and the new break-even quantity in whole sets.

(for verification: \(P_{\text{new}}(x) = 13.5x - 15000\))

  1. Determine for which monthly sales the new machine brings the higher profit. Decide with a reason whether the start-up should lease it at its planned sales of 1,500 sets.

a) Fixed costs 12,000 EUR, variable costs 8 EUR per set:

\[C(x) = 12000 + 8x, \qquad R(x) = 20x\]

\[P(x) = R(x) - C(x) = 20x - (12000 + 8x) = 20x - 12000 - 8x = 12x - 12000 \quad \checkmark\]

Contribution margin: \(20 - 8 = 12\) EUR per set. Each set sold contributes 12 EUR to cover the fixed costs. Once the fixed costs are covered, each further set adds 12 EUR to the profit. The contribution margin is the slope of \(P\).

b) Break-even: \(P(x) = 0 \iff 12x = 12000 \iff x = 1000\) sets per month, i.e. fixed costs divided by the contribution margin.

Revenue at the break-even point: \(R(1000) = 20{,}000\) EUR. Check: \(C(1000) = 12000 + 8000 = 20{,}000\) EUR ✓

c) Profit target:

\[12x - 12000 \geq 6000 \iff 12x \geq 18000 \iff x \geq 1500\]

The start-up must sell at least 1,500 sets per month.

Margin of safety at the planned sales: \(1500 - 1000 = 500\) sets, i.e. \(\frac{500}{1500} \approx 33.3\,\%\) of the planned sales. The sales may fall by a third before the start-up makes a loss.

d) New fixed costs \(12000 + 3000 = 15000\) EUR, new variable costs 6.50 EUR per set:

\[P_{\text{new}}(x) = 20x - (15000 + 6.5x) = 13.5x - 15000\]

New contribution margin: \(20 - 6.50 = 13.50\) EUR per set.

New break-even quantity: \(13.5x = 15000 \iff x = \frac{15000}{13.5} \approx 1111.1\). Only whole sets are sold, so round up: 1,112 sets. With 1,111 sets, \(P_{\text{new}}(1111) = 14998.50 - 15000 = -1.50\) EUR, still a small loss.

e) The new machine brings the higher profit if

\[13.5x - 15000 > 12x - 12000 \iff 1.5x > 3000 \iff x > 2000\]

With more than 2,000 sets per month, the new machine is better. At exactly 2,000 sets, both profits are \(12 \cdot 2000 - 12000 = 12{,}000\) EUR.

At the planned 1,500 sets: \(P(1500) = 6{,}000\) EUR and \(P_{\text{new}}(1500) = 20250 - 15000 = 5{,}250\) EUR. The start-up should not lease the machine: it would earn 750 EUR less per month. Each set saves 1.50 EUR, and the extra 3,000 EUR of fixed costs are only covered from \(\frac{3000}{1.5} = 2000\) sets on.

Problem 10: Equipment of a Landscaping Company (xx)

A landscaping company (it designs and maintains gardens) buys new equipment. It depreciates each item linearly: the book value (the value in the company’s accounts) falls by the same amount every year.

  • a mini excavator for 44,000 EUR, depreciation 5,500 EUR per year
  • three wood chippers (machines that shred branches) for 8,000 EUR each, depreciation 2,000 EUR per year each
  • planning software with computers for 15,000 EUR, depreciation 3,000 EUR per year

Let \(t\) be the time in years after the purchase.

  1. Give the value functions of the excavator, of one wood chipper and of the software. Determine the useful life of each item and give the domain of each value function in this context.

  2. Give a term for the total book value \(W(t)\) of all the equipment and explain why it only holds for \(0 \leq t \leq 4\). Compute \(W(3)\) and interpret the slope of \(W\).

(for verification: \(W(3) = 39500\))

  1. The company wants to sell the excavator when its book value has fallen to 40 % of the purchase price. Determine when this is the case.

  2. The company also owns a tipper truck (a truck whose loading area can be tilted), which it depreciates linearly as well. Two years after the purchase, its book value was 31,200 EUR, and six years after the purchase 20,800 EUR. Determine the purchase price and the useful life of the truck.

  3. The tax advisor suggests a different method for the excavator: its book value falls by 25 % of the current book value every year, so \(V(t) = 44000 \cdot 0.75^t\) EUR. Compute the book value after 3 years with this method and compare it with the linear method. Determine when the book value reaches 40 % of the purchase price with this method, in years rounded to two decimal places, and compare with c).

a) Value functions and useful lives (purchase price divided by the depreciation per year):

  • Excavator: \(V_E(t) = 44000 - 5500t\), useful life \(\frac{44000}{5500} = 8\) years, domain \([0; 8]\)
  • One wood chipper: \(V_W(t) = 8000 - 2000t\), useful life \(\frac{8000}{2000} = 4\) years, domain \([0; 4]\)
  • Software: \(V_S(t) = 15000 - 3000t\), useful life \(\frac{15000}{3000} = 5\) years, domain \([0; 5]\)

b) One excavator, three wood chippers and the software:

\[W(t) = (44000 - 5500t) + 3(8000 - 2000t) + (15000 - 3000t) = 83000 - 14500t\]

After 4 years, the wood chippers are fully depreciated. Their book value stays at 0, but the term \(8000 - 2000t\) would become negative. So \(W(t)\) only holds as long as all items are still being depreciated, i.e. for \(0 \leq t \leq 4\).

\[W(3) = 83000 - 43500 = 39{,}500 \text{ EUR}\]

Check: \(V_E(3) + 3V_W(3) + V_S(3) = 27500 + 3 \cdot 2000 + 6000 = 39{,}500\) EUR ✓

The slope \(-14{,}500\) EUR per year: all the equipment together loses 14,500 EUR of book value per year (\(5500 + 3 \cdot 2000 + 3000\)).

c) 40 % of 44,000 EUR are \(0.4 \cdot 44000 = 17600\) EUR:

\[44000 - 5500t = 17600 \iff 5500t = 26400 \iff t = 4.8\]

The company sells the excavator after 4.8 years.

d) Slope from the two book values:

\[m = \frac{20800 - 31200}{6 - 2} = \frac{-10400}{4} = -2600 \text{ EUR per year}\]

Point-slope form with \((2, 31200)\): \(V_T(t) - 31200 = -2600(t - 2) \iff V_T(t) = 36400 - 2600t\).

Purchase price: \(V_T(0) = 36{,}400\) EUR. Useful life: \(36400 - 2600t = 0 \iff t = \frac{36400}{2600} = 14\) years.

e) After 3 years: \(V(3) = 44000 \cdot 0.75^3 = 44000 \cdot 0.421875 = 18{,}562.50\) EUR. That is \(27500 - 18562.50 = 8{,}937.50\) EUR less than with the linear method: in the first years, the book value falls faster.

40 % of the purchase price:

\[44000 \cdot 0.75^t = 17600 \iff 0.75^t = 0.4 \iff t \cdot \ln(0.75) = \ln(0.4) \iff t = \frac{\ln(0.4)}{\ln(0.75)} \approx \frac{-0.9163}{-0.2877} \approx 3.19\]

With this method, the book value reaches 40 % after about 3.19 years, i.e. about 1.61 years earlier than with the linear method (4.8 years). It never reaches 0 EUR, since \(0.75^t > 0\).

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: Smart Door Locks from a Start-Up [22 pts. total]

A start-up has developed a smart door lock that opens with a smartphone app. At first, it is the only supplier of such locks in its region, so it sets the price itself. Its fixed costs (development, staff, rent) amount to 46,000 EUR per month. Material and assembly cost 100 EUR per lock (variable costs).

(QU = quantity units, MU = monetary units. 1 QU = 100 locks, 1 MU = 100 EUR, so a price of \(p\) MU per QU is a price of \(p\) EUR per lock. Total costs = fixed costs + variable costs. Revenue = quantity × price. Profit = revenue − total costs.)

Part a)

Determine the linear price-demand function \(p(x)\) (price in MU per QU at a demand of \(x\) QU per month) assuming a maximum price of 260 EUR per lock and a market saturation at 2,600 locks per month. Interpret the slope of \(p\) in EUR. Compute how many locks the start-up sells per month at a price of 195 EUR per lock. [4 pts.]

(for verification: \(p(x) = 260 - 10x\))

Units. A price of 260 EUR per lock means \(100 \cdot 260 = 26{,}000\) EUR \(= 260\) MU per QU. The market saturation is 2,600 locks \(= 26\) QU.

Linear price-demand function \(p(x) = c - dx\). At the maximum price the demand is 0, so \(p(0) = 260\) and \(c = 260\). At the market saturation the price is 0:

\[p(26) = 260 - 26d = 0 \iff d = 10 \qquad \Rightarrow \qquad p(x) = 260 - 10x, \quad 0 \leq x \leq 26\]

Slope \(-10\): for each additional 100 locks that the start-up wants to sell per month, it must lower the price by 10 EUR per lock.

Price of 195 EUR \(= 195\) MU per QU:

\[260 - 10x = 195 \iff 10x = 65 \iff x = 6.5 \text{ QU} = 650 \text{ locks per month}\]

Grading: 1 Pt. conversion of maximum price and market saturation to MU per QU and QU, 1 Pt. \(p(x)\), 1 Pt. slope interpreted in EUR per lock, 1 Pt. 650 locks.

Part b)

Give terms for the revenue function \(R(x)\) and the cost function \(C(x)\) in MU. Show that the profit function is given by \(P(x) = -10x^2 + 160x - 460\). [3 pts.]

Revenue = quantity × price:

\[R(x) = x \cdot p(x) = x(260 - 10x) = -10x^2 + 260x\]

Costs. Fixed costs 46,000 EUR \(= 460\) MU. Variable costs 100 EUR per lock \(= 100 \cdot 100 = 10{,}000\) EUR per QU \(= 100\) MU per QU:

\[C(x) = 100x + 460\]

Profit:

\[P(x) = R(x) - C(x) = -10x^2 + 260x - (100x + 460) = -10x^2 + 160x - 460 \quad \checkmark\]

Grading: 1 Pt. \(R(x) = x \cdot p(x)\), 1 Pt. \(C(x)\) with both conversions to MU, 1 Pt. \(P = R - C\) simplified with the sign change.

Part c)

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: \(x_1 = 8 - 3\sqrt{2} \approx 3.76\))

Zero form. At a break-even point the profit is zero, \(P(x) = 0\). Divide by \(-10\):

\[-10x^2 + 160x - 460 = 0 \;\big|: (-10) \iff x^2 - 16x + 46 = 0\]

Discriminant: \(\Delta = (-16)^2 - 4 \cdot 1 \cdot 46 = 256 - 184 = 72 > 0\). So there are two zeros. 72 is not a perfect square, so the quadratic formula is needed. With \(\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}\):

\[x_{1,2} = \frac{16 \pm 6\sqrt{2}}{2} = 8 \pm 3\sqrt{2}\]

\[x_1 = 8 - 3\sqrt{2} \approx 3.76, \qquad x_2 = 8 + 3\sqrt{2} \approx 12.24\]

Check \(x_1\) with the perfect square formula: \((8 - 3\sqrt{2})^2 - 16(8 - 3\sqrt{2}) + 46 = 64 - 48\sqrt{2} + 18 - 128 + 48\sqrt{2} + 46 = 0\) ✓

Both zeros lie in the domain \([0; 26]\). Break-even point: \(x_1 = 8 - 3\sqrt{2} \approx 3.76\) QU. Profit limit: \(x_2 = 8 + 3\sqrt{2} \approx 12.24\) QU.

Grading: 1 Pt. zero form, 1 Pt. discriminant, 1 Pt. quadratic formula with \(\sqrt{72} = 6\sqrt{2}\), 1 Pt. exact and rounded values, 1 Pt. naming break-even point and profit limit.

Part d)

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

Test values below, between and above the zeros:

  • \(x = 1\): \(P(1) = -10 + 160 - 460 = -310 < 0\), a loss
  • \(x = 5\): \(P(5) = -250 + 800 - 460 = 90 > 0\), a profit
  • \(x = 13\): \(P(13) = -1690 + 2080 - 460 = -70 < 0\), a loss

Profit zone: \((8 - 3\sqrt{2}; 8 + 3\sqrt{2}) \approx (3.76; 12.24)\). The interval is open, since the profit is zero at both ends.

Whole locks. \(x_1 \approx 3.7574\) QU \(= 375.74\) locks and \(x_2 \approx 12.2426\) QU \(= 1224.26\) locks. Round the lower bound up and the upper bound down: the start-up makes a profit with 376 to 1,224 locks per month.

Check: 375 locks give \(P(3.75) = -0.625\) MU, a loss of 62.50 EUR, and 1,225 locks give the same loss.

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

Part e)

Two years later, several competitors sell similar locks, and the price now forms on the market. The demand in the region has not changed. At a price of \(p\) EUR per lock, all suppliers together offer \(Q_s(p) = 0.3p - 36\) QU per month.

Determine the price above which the suppliers offer locks. Give the demand as a function of the price, \(Q_d(p)\) in QU. Determine the market equilibrium and interpret it. [4 pts.]

(for verification: equilibrium price 155 EUR per lock)

Supply starts: \(0.3p - 36 > 0 \iff 0.3p > 36 \iff p > 120\). The suppliers offer locks only at prices above 120 EUR per lock.

Demand \(Q_d(p) = a - bp\) with the market saturation \(a = 26\) QU. At the maximum price of 260 EUR the demand is 0: \(26 - 260b = 0 \iff b = 0.1\).

\[Q_d(p) = 26 - 0.1p\]

(Or solve \(p = 260 - 10x\) for \(x\): \(x = 26 - 0.1p\).)

Equilibrium: supply equals demand.

\[26 - 0.1p = 0.3p - 36 \iff 62 = 0.4p \iff p^* = 155\]

\[Q^* = Q_d(155) = 26 - 15.5 = 10.5 \text{ QU} \qquad \text{Check: } Q_s(155) = 46.5 - 36 = 10.5 \; \checkmark\]

At 155 EUR per lock, the customers want to buy exactly as many locks as all suppliers offer: 1,050 locks per month. There is neither a shortage nor a surplus.

Grading: 1 Pt. price of 120 EUR, 1 Pt. \(Q_d(p)\), 1 Pt. equation and equilibrium price, 1 Pt. equilibrium quantity in locks with interpretation.

Part f)

The development costs are now paid off, so the start-up’s fixed costs fall to 22,000 EUR per month. It sells at the equilibrium price, and its variable costs stay at 100 EUR per lock. Compute the contribution margin per lock and the break-even quantity. The start-up expects to sell 40 % of the locks traded on the market. Decide whether it makes a profit and compute its monthly profit or loss in EUR. [3 pts.]

Contribution margin: \(155 - 100 = 55\) EUR per lock.

Break-even quantity: \(\frac{22000}{55} = 400\) locks per month.

Expected sales: 40 % of 1,050 locks \(= 0.4 \cdot 1050 = 420\) locks. That is more than 400 locks, so the start-up makes a profit:

\[420 \cdot 55 - 22000 = 23100 - 22000 = 1{,}100 \text{ EUR per month}\]

The margin is small: the start-up needs a market share of at least \(\frac{400}{1050} \approx 38.1\,\%\), and 20 locks fewer per month would already mean no profit.

Grading: 1 Pt. contribution margin, 1 Pt. break-even quantity, 1 Pt. 420 locks with the profit of 1,100 EUR and the decision.