Tasks 01-06 - Synthesis & Linear Equations

Integrating Mathematical Foundations

A: Skills

Problem 1: Multi-Technique Problems (xx)

Each item combines several techniques from Section 01.

  1. Solve \(\log_2(x^2 - 9) = \log_2(x - 3) + 3\).

Hint for a): write \(3\) as a logarithm to base 2, and check your solutions in the original equation.

  1. Factor \(x^4 - 17x^2 + 16\) completely, then evaluate it at \(x = \sqrt{2}\).

  2. Evaluate \(\frac{(x^3 - 8) \cdot \sqrt{x + 2}}{(x - 2) \cdot (x^2 + 2x + 4)}\) at \(x = 6\) and simplify the result.

  3. If \(2^x = 3\) and \(3^y = 4\), determine the value of \(6^{xy}\).

  4. Expand \((1 + \sqrt{2})^3\) and give the result in the form \(a + b\sqrt{2}\).

  1. Solve: \(\log_2(x^2 - 9) = \log_2(x - 3) + 3\)
    • Rewrite: \(\log_2(x^2 - 9) = \log_2(x - 3) + \log_2(8)\)
    • \(\log_2(x^2 - 9) = \log_2[8(x - 3)]\)
    • Therefore: \(x^2 - 9 = 8(x - 3)\)
    • \(x^2 - 9 = 8x - 24\)
    • \(x^2 - 8x + 15 = 0\)
    • \((x - 3)(x - 5) = 0\)
    • So \(x = 3\) or \(x = 5\)
    • Check \(x = 3\): \(\log_2(0)\) is undefined, so \(x = 3\) is not a solution (the logarithm of 0 is not defined)
    • Check \(x = 5\): \(\log_2(16) = \log_2(2) + 3 = 4 = 1 + 3\) ✓
    • Therefore: \(x = 5\)
  2. Factor: \(x^4 - 17x^2 + 16\)
    • Let \(u = x^2\): \(u^2 - 17u + 16 = (u-1)(u-16)\)
    • \(= (x^2 - 1)(x^2 - 16) = (x+1)(x-1)(x+4)(x-4)\)
    • At \(x = \sqrt{2}\): \((\sqrt{2}+1)(\sqrt{2}-1)(\sqrt{2}+4)(\sqrt{2}-4)\)
    • \(= (2-1)(2-16) = 1 \times (-14) = -14\)
  3. At \(x = 6\):
    • Numerator: \((216 - 8) \cdot \sqrt{8} = 208 \cdot 2\sqrt{2} = 416\sqrt{2}\)
    • Denominator: \(4 \cdot (36 + 12 + 4) = 4 \cdot 52 = 208\)
    • Result: \(\frac{416\sqrt{2}}{208} = 2\sqrt{2}\)
  4. From \(2^x = 3\): \(x = \log_2(3)\) From \(3^y = 4\): \(y = \log_3(4)\)
    • \(xy = \log_2(3) \cdot \log_3(4) = \log_2(4) = 2\)
    • Therefore: \(6^{xy} = 6^2 = 36\)
  5. \((1 + \sqrt{2})^3\)
    • Using binomial: \(1 + 3\sqrt{2} + 3(\sqrt{2})^2 + (\sqrt{2})^3\)
    • \(= 1 + 3\sqrt{2} + 6 + 2\sqrt{2}\)
    • \(= 7 + 5\sqrt{2}\)

Problem 2: Exponential and Logarithmic Equations (xx)

  1. \(2^x \cdot 3^y = 72\) and \(x = 3\). Determine \(y\) and verify your result.

  2. \(\log_a(b) = 2\) and \(\log_b(c) = 3\). Determine \(\log_a(c)\) and \(\log_c(a)\).

Hint for b): write each logarithm as a quotient of natural logarithms with the change of base formula, e.g. \(\log_a(b) = \frac{\ln(b)}{\ln(a)}\).

  1. A bacteria culture doubles every 5 hours. Starting with 2000 bacteria, the number after \(t\) hours is \(2000 \cdot 2^{t/5}\). Determine after how many hours there are exactly \(10^6\) bacteria.
  1. Given \(x = 3\) and \(2^x \cdot 3^y = 72\):
    • \(2^3 \cdot 3^y = 72\)
    • \(8 \cdot 3^y = 72\)
    • \(3^y = 9\)
    • \(y = 2\)
    • Verify: \(72 = 8 \times 9 = 2^3 \times 3^2\) ✓
  2. Given \(\log_a(b) = 2\) and \(\log_b(c) = 3\):
    • \(\log_a(c) = \log_a(b) \cdot \log_b(c) = 2 \times 3 = 6\)
    • Change of base: \(\log_c(a) = \frac{\ln a}{\ln c} = \frac{1}{\ln c / \ln a} = \frac{1}{\log_a(c)}\)
    • \(\log_c(a) = \frac{1}{\log_a(c)} = \frac{1}{6}\)
  3. Solve \(10^6 = 2000 \cdot 2^{t/5}\):
    • \(500 = 2^{t/5}\)
    • \(t/5 = \log_2(500) = \frac{\ln(500)}{\ln(2)} = 8.97\)
    • \(t = 44.8\) hours

Problem 3: Linear Equations and Inequalities (x)

Solve the following inequalities and give the solution sets in interval notation.

  1. \(5x + 3 \leq 18\)

  2. \(21 - 2x > 9\)

  3. \(3(x + 4) \geq 5x - 8\)

  4. \(\frac{x-1}{4} < \frac{x+3}{2}\)

Solve the following equations and verify each solution by substituting it into the original equation.

  1. \(5(2x - 3) - 3(x - 4) = 2(x + 6)\)

  2. \(\frac{3x + 2}{5} - \frac{x - 4}{2} = 3\)

  3. \(\frac{x}{2} - \frac{x - 3}{5} = \frac{x + 6}{4}\)

Hint for f) and g): first multiply both sides by the least common denominator (LCD).

  1. \(5x \leq 15 \implies x \leq 3\). Interval: \((-\infty; 3]\)

  2. \(-2x > -12 \implies x < 6\) (sign flips). Interval: \((-\infty; 6)\)

  3. \(3x + 12 \geq 5x - 8 \implies 20 \geq 2x \implies 10 \geq x\). Interval: \((-\infty; 10]\)

  4. Multiply by 4: \(x - 1 < 2(x + 3) \implies x - 1 < 2x + 6 \implies -7 < x\). Interval: \((-7; \infty)\)

  5. Expand: \(10x - 15 - 3x + 12 = 2x + 12 \implies 7x - 3 = 2x + 12 \implies 5x = 15 \implies x = 3\). Check: \(5 \cdot 3 - 3 \cdot (-1) = 18\) and \(2 \cdot 9 = 18\) ✓

  6. LCD 10: \(2(3x + 2) - 5(x - 4) = 30 \implies 6x + 4 - 5x + 20 = 30 \implies x + 24 = 30 \implies x = 6\). Check: \(\frac{20}{5} - \frac{2}{2} = 4 - 1 = 3\) ✓

  7. LCD 20: \(10x - 4(x - 3) = 5(x + 6)\). The minus sign in front of the fraction applies to the whole numerator: \(-4(x - 3) = -4x + 12\). \(10x - 4x + 12 = 5x + 30 \implies 6x + 12 = 5x + 30 \implies x = 18\). Check: \(\frac{18}{2} - \frac{15}{5} = 9 - 3 = 6\) and \(\frac{24}{4} = 6\) ✓

Problem 4: Motion and Meeting (xxx)

Two delivery services are coordinating a package handoff.

  • QuickShip leaves City A heading east toward City B at 65 km/h.
  • At the same time, FastTrack leaves City B heading west toward City A at 85 km/h.
  • The cities are 375 km apart.

(distance = speed × time)

  1. Determine how long it takes until the drivers meet.

  2. Determine how far from City A they meet.

  3. Determine where they would meet if QuickShip had left 30 minutes earlier.

  1. Meeting time:
    • Let t = time in hours until meeting
    • Distance by QuickShip: 65t
    • Distance by FastTrack: 85t
    • Total distance: 65t + 85t = 375
    • 150t = 375
    • t = 2.5 hours
  2. Distance from City A:
    • QuickShip travels: 65 × 2.5 = 162.5 km from City A
    • Verification: FastTrack travels 85 × 2.5 = 212.5 km
    • Total: 162.5 + 212.5 = 375 km
  3. With 30-minute head start:
    • QuickShip travels alone for 0.5 hours: 65 × 0.5 = 32.5 km
    • Remaining distance: 375 - 32.5 = 342.5 km
    • Let t = time after FastTrack starts
    • 65t + 85t = 342.5
    • 150t = 342.5
    • t = 2.283 hours
    • Distance from City A: 32.5 + 65(2.283) = 32.5 + 148.4 = 180.9 km

B: Set Up & Check

Problem 5: From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works. (QU = quantity units)

  1. “A shop buys a backpack for 64 EUR and sells it with a markup of 25 %, i.e. its selling price is 25 % above the purchase price.”

    Check: Is the margin (the profit as a percentage of the selling price) also 25 %?

  2. “The temperature \(T\) in a cold store may deviate from −18 °C by at most 2 °C.”

    Check: Is a temperature of −21 °C within the tolerance?

  3. “A company with the profit \(-(x - 4)(x - 13)\) (in 100 EUR) at an output of \(x\) QU makes a profit.”

    Check: Does the company make a profit at \(x = 15\)?

  4. “An investment fund grows by 12 % per year. Its doubling time (the time after which the value has doubled) is \(T\) years.”

    Check: The rule of 72 estimates \(\frac{72}{12} = 6\) years. Has the value doubled after 6 years?

  5. “A market stall sells lemonade at 3.20 EUR per cup. The ingredients of one cup cost 0.80 EUR, and the stall fee is 400 EUR per day. To break even, the stall has to sell at least a certain number of cups per day.”

    Check: Is 166 cups per day enough?

  1. Selling price: \(64 \cdot 1.25 = 80\) EUR, profit: \(80 - 64 = 16\) EUR.

    Check: no, the margin is \(\frac{16}{80} = 20\,\%\).

    Typical mistake: “margin = markup = 25 %”. The markup refers to the purchase price (64 EUR), the margin to the selling price (80 EUR). The same 16 EUR are a smaller percentage of the larger base.

  2. \(|T - (-18)| \leq 2 \iff |T + 18| \leq 2 \iff -2 \leq T + 18 \leq 2 \iff -20 \leq T \leq -16\)

    Check: no, \(|-21 + 18| = 3 > 2\).

    Typical mistake: \(|T - 18| \leq 2\), i.e. \(16 \leq T \leq 20\). The distance from \(-18\) is \(|T - (-18)| = |T + 18|\).

  3. \(-(x - 4)(x - 13) > 0 \iff (x - 4)(x - 13) < 0\) (multiplying by \(-1\) reverses the sign). A product is zero if one of its factors is zero, so the profit is zero at \(x = 4\) and \(x = 13\). The product is negative if the two factors have opposite signs: \(x - 4 > 0\) and \(x - 13 < 0\), i.e. \(4 < x < 13\). (The other case, \(x < 4\) and \(x > 13\), is impossible.)

    Check: no, at \(x = 15\) the profit is \(-(11)(2) = -22\), a loss of 2,200 EUR.

    Typical mistake: “\(x - 4 > 0\) and \(x - 13 > 0\)”, i.e. \(x > 13\). This ignores the minus sign in front of the product, and \(x = 15\) shows that the translation fails.

  4. With \(V_0\) the value today: \(V_0 \cdot 1.12^T = 2V_0 \iff 1.12^T = 2 \iff T = \frac{\ln(2)}{\ln(1.12)} \approx 6.12\) years.

    Check: no, \(1.12^6 \approx 1.974\), so the value has not quite doubled after 6 years. The rule of 72 gives a good estimate, not the exact value.

    Typical mistake: \(T = \frac{100}{12} \approx 8.3\) years. That would only hold if the fund gained 12 % of the original value every year. With compound growth, each year’s gain also grows by 12 % in the following years.

  5. With \(n\) the number of cups: revenue \(3.20n\) EUR, total costs \(0.80n + 400\) EUR. Break even or better:

    \[3.20n \geq 0.80n + 400 \iff 2.40n \geq 400 \iff n \geq 166.67\]

    Only whole cups are sold, so the stall has to sell at least 167 cups.

    Check: no, with 166 cups: \(2.40 \cdot 166 = 398.40 < 400\), a loss of 1.60 EUR.

    Typical mistake: \(3.20n \geq 400\), i.e. 125 cups. Every cup also costs 0.80 EUR for its ingredients, so these variable costs \(0.80n\) belong on the cost side.

Problem 6: One Paragraph, All Conditions (xx)

A farm shop sells 20 kg bags of bird seed. Each bag is a blend of sunflower seeds (3.50 EUR per kg) and wheat (0.90 EUR per kg). Each bag must contain at least 40 % sunflower seeds, and the seeds in one bag may cost at most 45 EUR.

List every condition in this paragraph as a mathematical statement, with \(s\) the amount of sunflower seeds per bag in kg. Do not solve anything.

Check: Does a bag with 11 kg of sunflower seeds meet all your conditions? If not, which condition does it violate?

  1. Wheat per bag: \(20 - s\) kg
  2. No negative amounts: \(s \geq 0\) and \(20 - s \geq 0\), so \(0 \leq s \leq 20\)
  3. Sunflower share: \(s \geq 0.40 \cdot 20 = 8\)
  4. Seed costs per bag: \(3.50s + 0.90(20 - s)\) EUR
  5. Budget: \(3.50s + 0.90(20 - s) \leq 45\)

Check: \(s = 11\) meets conditions 2 and 3, but the seeds cost \(3.50 \cdot 11 + 0.90 \cdot 9 = 38.50 + 8.10 = 46.60\) EUR \(> 45\) EUR. The bag violates the budget (condition 5).

Typical mistake: \(s \geq 40\). The 40 % refer to the 20 kg in the bag, so the share condition is \(s \geq 0.40 \cdot 20 = 8\) kg.

C: Applied

Problem 7: Break-Even of a New Smartwatch (xx)

A company launches a new smartwatch. Its fixed costs (rent, salaries, insurance) amount to 13,500 EUR per month. They arise even if no watch is produced. Each watch causes variable costs of 85 EUR for materials and assembly. The planned price is 249 EUR per watch. The assembly line can produce at most 300 watches per month. This is the capacity (the maximum possible output). Let \(x\) be the number of watches produced and sold per month.

  1. Give terms for the total costs \(C(x)\) and the revenue \(R(x)\), and show that the monthly profit is \(P(x) = 164x - 13500\).

  2. Determine the break-even point and interpret it: how many watches does the company have to sell at least to avoid a loss?

(for verification: \(x \approx 82.32\))

  1. The investors expect a monthly profit of at least 25,000 EUR. Determine all monthly sales volumes that meet this target within the capacity.

(for verification: at least 235 watches)

  1. A competitor sells a similar watch for 199 EUR. Determine the lowest price at which the company still breaks even when it sells 250 watches per month. Decide whether the company can match the competitor’s price at this volume and still meet the investors’ target from c).

a) Total costs are fixed costs plus variable costs. Revenue is price times quantity:

\[C(x) = 13500 + 85x, \qquad R(x) = 249x\]

\[P(x) = R(x) - C(x) = 249x - (13500 + 85x) = 164x - 13500 \quad \checkmark\]

b) At the break-even point, revenue equals total costs, i.e. \(P(x) = 0\):

\[249x = 13500 + 85x \iff 164x = 13500 \iff x = \frac{13500}{164} \approx 82.32\]

Only whole watches are sold, so round up: the company has to sell at least 83 watches per month. With 82 watches it makes a small loss: \(164 \cdot 82 - 13500 = -52\) EUR.

c)

\[164x - 13500 \geq 25000 \iff 164x \geq 38500 \iff x \geq 234.76\]

Round up to 235 watches. Together with the capacity of 300 watches:

\[x \in \{235, 236, \dots, 300\}\]

Check: \(164 \cdot 235 - 13500 = 25040\) EUR \(\checkmark\), while 234 watches give only \(164 \cdot 234 - 13500 = 24876\) EUR.

d) With price \(p\) and 250 watches, break-even means revenue = total costs:

\[250p = 13500 + 85 \cdot 250 = 34750 \iff p = 139 \text{ EUR}\]

At 199 EUR the company still makes a profit: \(250 \cdot (199 - 85) - 13500 = 28500 - 13500 = 15000\) EUR. It can match the competitor’s price, but 15,000 EUR is below the investors’ target of 25,000 EUR.

Problem 8: Pricing Plans of a Freelance Designer (xx)

A freelance designer offers a client two pricing plans for a project:

  • Plan A: a flat fee of 1,200 EUR.
  • Plan B: an initial fee of 500 EUR plus 35 EUR per hour.

Let \(h\) be the number of hours the project takes.

  1. Give terms for the costs \(A(h)\) and \(B(h)\) of both plans and determine the number of hours for which both plans cost the same.

(for verification: 20 hours)

  1. Determine for which numbers of hours plan B is cheaper for the client. Give the answer as an interval.

  2. The designer estimates 25 hours, and the actual time may deviate from this estimate by at most 20 %. Formulate this condition as an absolute value inequality and solve it. Decide which plan the client should choose.

  3. The designer adds Plan C: 75 EUR per hour, without an initial fee. Determine for which numbers of hours each of the three plans is the cheapest.

(for verification: plan C is the cheapest for fewer than 12.5 hours)

a) \(A(h) = 1200\) and \(B(h) = 500 + 35h\) (in EUR).

\[500 + 35h = 1200 \iff 35h = 700 \iff h = 20\]

Both plans cost the same for a project of 20 hours.

b)

\[500 + 35h < 1200 \iff 35h < 700 \iff h < 20\]

A project takes a positive number of hours, so plan B is cheaper for \(h \in (0; 20)\).

c) 20 % of 25 hours: \(0.20 \cdot 25 = 5\) hours.

\[|h - 25| \leq 5 \iff -5 \leq h - 25 \leq 5 \iff 20 \leq h \leq 30\]

So \(h \in [20; 30]\). By b), plan B is cheaper only for fewer than 20 hours. At \(h = 20\) both plans cost 1,200 EUR. For every \(h\) in \((20; 30]\) plan A is cheaper, e.g. \(B(30) = 500 + 35 \cdot 30 = 1550\) EUR. The client should choose plan A: it is never more expensive than plan B in the whole range.

d) \(C(h) = 75h\).

  • C cheaper than A: \(75h < 1200 \iff h < 16\)
  • C cheaper than B: \(75h < 500 + 35h \iff 40h < 500 \iff h < 12.5\)

Plan C is the cheapest if both conditions hold, i.e. for \(h < 12.5\). Together with b):

  • Plan C is the cheapest for \(0 < h < 12.5\)
  • Plan B is the cheapest for \(12.5 < h < 20\) (cheaper than C since \(h > 12.5\), cheaper than A since \(h < 20\))
  • Plan A is the cheapest for \(h > 20\) (there \(C(h) = 75h > 1500 > 1200\))

At \(h = 12.5\) plans B and C cost the same (937.50 EUR), at \(h = 20\) plans A and B (1,200 EUR).

Problem 9: Reserve of a Sports Club (xxx)

A sports club invests its reserve of 36,000 EUR for one year. It puts \(s\) EUR into an investment fund with an expected return of 7 % and the rest into a savings bond with a fixed return of 2.5 %. (return = the interest or gain after one year, as a percentage of the amount invested)

  1. Give a term for the total return after one year and show that it equals \(900 + 0.045s\) EUR.

  2. Determine the minimum amount in the fund for a total return of at least 1,800 EUR.

(for verification: 20,000 EUR)

  1. The club’s rules allow at most 60 % of the reserve in the fund. Give all amounts \(s\) that meet the target from b) and this rule as an interval.

  2. In a bad year the fund loses 10 % instead of gaining 7 %, while the bond still pays 2.5 %. Determine the largest amount \(s\) for which the club makes no loss even in a bad year. Decide whether an amount \(s\) exists that meets the return target, the 60 % rule and the no-loss condition at the same time.

(for verification: at most 7,200 EUR)

a) Fund: \(0.07s\) EUR, bond: \(0.025(36000 - s)\) EUR.

\[0.07s + 0.025(36000 - s) = 0.07s + 900 - 0.025s = 900 + 0.045s \quad \checkmark\]

b)

\[900 + 0.045s \geq 1800 \iff 0.045s \geq 900 \iff s \geq 20000\]

At least 20,000 EUR go into the fund (and at most 16,000 EUR into the bond).

Check: \(0.07 \cdot 20000 + 0.025 \cdot 16000 = 1400 + 400 = 1800\) EUR \(\checkmark\)

c) 60 % of 36,000 EUR: \(0.60 \cdot 36000 = 21600\) EUR, so \(s \leq 21600\). Together with b):

\[s \in [20000; 21600]\]

d) In a bad year the fund returns \(-0.10s\) EUR and the bond \(0.025(36000 - s)\) EUR:

\[-0.10s + 900 - 0.025s \geq 0 \iff -0.125s \geq -900 \iff s \leq 7200\]

Dividing by the negative number \(-0.125\) reverses the inequality sign. So the club makes no loss in a bad year only if at most 7,200 EUR go into the fund.

Decision: the return target needs at least 20,000 EUR in the fund, and \([20000; 21600] \cap [0; 7200] = \emptyset\). No amount meets all three conditions: the club has to lower its return target or accept a possible loss.

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: Tinted Wall Paint [26 pts. total]

A paint manufacturer mixes a tinted premium wall paint from a white base paint and a colour concentrate. The base paint contains 4 % pigment (the colouring substance) and costs 3.00 EUR per litre. The concentrate contains 44 % pigment and costs 13.00 EUR per litre. The finished paint must contain at least 18 % pigment. Filling and cans add 1.50 EUR per litre of paint. The fixed costs, which arise even when no paint is produced, amount to 7,000 EUR per week. To sell \(x\) QU per week, the manufacturer can charge \((35 - 2x)\) EUR per litre: the more paint it offers, the lower the price. The mixing plant can produce at most 900 litres per week.

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

Part a)

Let \(c\) be the amount of concentrate in litres per 100 litres of paint. Formulate the pigment condition as an inequality and solve it. Determine the cheapest allowed mix and its material costs per litre of paint. [5 pts.]

(for verification: at least 35 litres of concentrate, 6.50 EUR per litre)

Variables: \(c\) litres of concentrate and \(100 - c\) litres of base paint, with \(0 \leq c \leq 100\).

Pigment condition: the pigment in the mix must be at least 18 % of 100 litres:

\[0.04(100 - c) + 0.44c \geq 18\]

\[4 - 0.04c + 0.44c \geq 18 \iff 0.40c \geq 14 \iff c \geq 35\]

Material costs per 100 litres of paint:

\[3.00(100 - c) + 13.00c = 300 + 10c \text{ EUR}\]

The costs grow with \(c\): every litre of concentrate instead of base paint costs 10 EUR more. So the cheapest allowed mix uses the smallest allowed \(c\): 35 litres of concentrate and 65 litres of base paint.

\[300 + 10 \cdot 35 = 650 \text{ EUR per 100 litres} = 6.50 \text{ EUR per litre} \quad \checkmark\]

Check: \(0.04 \cdot 65 + 0.44 \cdot 35 = 2.6 + 15.4 = 18\) litres of pigment, i.e. exactly 18 % \(\checkmark\)

Grading: 1 Pt. variables with \(100 - c\), 1 Pt. pigment inequality, 1 Pt. \(c \geq 35\), 1 Pt. cost term and reason for the smallest \(c\), 1 Pt. 6.50 EUR per litre.

Part b)

Give terms for the total costs \(C\) and the revenue \(R\) per week in MU if the manufacturer produces and sells \(x\) QU of this cheapest mix. [3 pts.]

(for verification: \(C = 8x + 70\), \(\; R = 35x - 2x^2\))

Convert to MU. Variable costs per litre: \(6.50 + 1.50 = 8.00\) EUR, per QU \(100 \cdot 8.00 = 800\) EUR \(= 8\) MU. Fixed costs: 7,000 EUR \(= 70\) MU. Price per QU: \(100 \cdot (35 - 2x)\) EUR \(= (35 - 2x)\) MU.

Total costs are variable costs plus fixed costs:

\[C = 8x + 70 \quad \checkmark\]

Revenue is quantity times price:

\[R = x(35 - 2x) = 35x - 2x^2 \quad \checkmark\]

Grading: 1 Pt. conversion of variable costs, fixed costs and price to MU, 1 Pt. \(C\), 1 Pt. \(R\) multiplied out.

Part c)

Show that the weekly profit is given by \(P = -2x^2 + 27x - 70\). [2 pts.]

\[P = R - C = 35x - 2x^2 - (8x + 70)\]

Remove the bracket. The minus sign changes every sign inside it:

\[P = 35x - 2x^2 - 8x - 70 = -2x^2 + 27x - 70 \quad \checkmark\]

Grading: 1 Pt. ansatz \(P = R - C\), 1 Pt. sign change in the bracket and collecting terms.

Part d)

Note: the break-even point is the output at which the profit changes from negative to positive. The profit limit is the output at which it turns negative again.

Factor \(P\) completely. Determine the break-even point and the profit limit. [5 pts.]

(for verification: \(P = -(2x - 7)(x - 10)\))

Factor out \(-1\):

\[P = -(2x^2 - 27x + 70)\]

AC method. \(ac = 2 \cdot 70 = 140\). Two numbers with product 140 and sum \(-27\): \(-7\) and \(-20\).

\[2x^2 - 7x - 20x + 70 = x(2x - 7) - 10(2x - 7) = (x - 10)(2x - 7)\]

\[P = -(2x - 7)(x - 10) \quad \checkmark\]

Zero profit. A product is zero if one of its factors is zero:

\[2x - 7 = 0 \Rightarrow x = 3.5, \qquad x - 10 = 0 \Rightarrow x = 10\]

Check: \(-2 \cdot 3.5^2 + 27 \cdot 3.5 - 70 = -24.5 + 94.5 - 70 = 0\) \(\checkmark\)

Break-even point: \(x = 3.5\) QU (350 litres per week). Profit limit: \(x = 10\) QU (1,000 litres per week). It lies above the capacity of 9 QU.

Grading: 1 Pt. factoring out \(-1\), 1 Pt. \(ac = 140\) and the pair \((-7, -20)\), 1 Pt. factored form, 1 Pt. both zeros, 1 Pt. naming break-even point and profit limit.

Part e)

Decide in which output range the manufacturer makes a profit, taking the capacity into account. State this profit zone (all outputs with a positive profit) as an interval and in set-builder notation. [3 pts.]

Note: a product of two factors is negative if the factors have opposite signs.

Capacity: 900 litres \(= 9\) QU, so \(0 \leq x \leq 9\).

Sign of \(P = -(2x - 7)(x - 10)\): for \(3.5 < x < 10\) the factor \(2x - 7\) is positive and \(x - 10\) is negative, so their product is negative and \(P\) is positive. For \(x < 3.5\) both factors are negative, so their product is positive and \(P\) is negative (e.g. \(P(1) = -2 + 27 - 70 = -45\)).

Profit zone:

\[(3.5; 9] = \{x \in \mathbb{R} \mid 3.5 < x \leq 9\}\]

At \(x = 3.5\) the profit is exactly zero, so 3.5 is excluded. The capacity of 9 QU is included: \(P(9) = -162 + 243 - 70 = 11\) MU, i.e. 1,100 EUR.

Grading: 1 Pt. sign argument or test values, 1 Pt. capacity limit 9 QU, 1 Pt. half-open interval in both notations.

Part f)

The sales manager compares the price per litre at 500 and at 750 litres per week and says: “At 750 litres our price is 20 % lower than at 500 litres, so at 500 litres it is 20 % higher than at 750 litres.” Compute both percentage changes and explain the manager’s mistake. [3 pts.]

Prices: at 5 QU: \(35 - 2 \cdot 5 = 25\) EUR per litre, at 7.5 QU: \(35 - 2 \cdot 7.5 = 20\) EUR per litre.

From 500 to 750 litres (old value 25 EUR):

\[\frac{20 - 25}{25} = -0.20 = -20\,\%\]

The first half of the statement is correct.

From 750 to 500 litres (old value 20 EUR):

\[\frac{25 - 20}{20} = 0.25 = +25\,\%\]

Explanation: a percentage change always refers to the old value, the base. The difference is 5 EUR both times, but 5 EUR are 20 % of 25 EUR and 25 % of 20 EUR. With growth factors (the factor by which a value is multiplied): a fall by 20 % is the factor \(0.8\), and it is undone by \(\frac{1}{0.8} = 1.25\), not by \(1.2\).

Grading: 1 Pt. both prices and −20 %, 1 Pt. +25 % with base 20 EUR, 1 Pt. explanation with the base.

Part g)

Because new staff must be trained first, the manufacturer starts with an output of 200 litres per week and raises its weekly output by 6 % every month. Give the weekly output \(O(n)\) in QU after \(n\) monthly raises. Determine after how many raises the manufacturer first makes a profit, and after how many raises the planned output would exceed the capacity. [5 pts.]

Hint: Taking the logarithm of both sides keeps the inequality sign, because the larger of two positive numbers has the larger logarithm.

(for verification: \(O(n) = 2 \cdot 1.06^n\))

Model: start 200 litres \(= 2\) QU, growth factor \(1.06\) per raise:

\[O(n) = 2 \cdot 1.06^n\]

Profit: the output must exceed the break-even point of 3.5 QU from d):

\[2 \cdot 1.06^n > 3.5 \iff 1.06^n > 1.75\]

Take the natural logarithm of both sides. The logarithm is increasing, so the sign stays. \(\ln(1.06) > 0\), so dividing by it keeps the sign too:

\[n \cdot \ln(1.06) > \ln(1.75) \iff n > \frac{\ln(1.75)}{\ln(1.06)} \approx 9.60\]

The manufacturer first makes a profit after 10 raises (10 months). Check: \(O(9) \approx 3.38 < 3.5\) and \(O(10) \approx 3.58 > 3.5\) \(\checkmark\)

Capacity:

\[2 \cdot 1.06^n > 9 \iff 1.06^n > 4.5 \iff n > \frac{\ln(4.5)}{\ln(1.06)} \approx 25.81\]

The 26th raise would exceed the capacity: \(O(25) \approx 8.58\) QU, but \(O(26) \approx 9.10\) QU \(> 9\) QU. The plan works for 25 raises. After that the output has to stay at 9 QU, or the plant must be expanded.

Grading: 1 Pt. \(O(n)\), 1 Pt. inequality with the break-even point, 1 Pt. logarithm and 10 raises, 1 Pt. capacity inequality, 1 Pt. 26th raise with interpretation.