FSP Training 02

Preparation for Mock Exam 03

Instructions

This training page covers all of Section 02 and prepares you for Mock Exam 03. Set A trains setting up: you translate a text into mathematics and check your translation. Set B is a practice mock in the format of Mock Exam 03: 90 minutes, 50 points, and all 3 problems must be completed. Write down every step as you would in the exam. The points show what each part is worth.

In the real FSP, decimals are written with a comma (1,25 means 1.25).

Set A: Set Up & Check (40 minutes)

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, MU = monetary units)

  1. “A cinema sold 460 tickets on Saturday, regular tickets and reduced tickets. It sold 30 % more regular tickets than reduced tickets.”

    Check: Did the cinema sell 260 regular tickets?

  2. “An office orders paper and ink cartridges twice. The first order, 3 boxes of paper and 2 cartridges, costs 260 EUR. The second order, 6 boxes of paper and 4 cartridges, costs 520 EUR. The office manager wants to find the price of one box and the price of one cartridge from these two orders.”

    Check: Do the two orders show that a box costs 60 EUR and a cartridge 40 EUR?

  3. “In a market, the demand price is \(p_D(x) = 30 - 2x\) (the price at which buyers take \(x\) thousand units) and the supply price is \(p_S(x) = 6 + x\) (the price at which sellers offer \(x\) thousand units), both in EUR per unit. The market is in equilibrium (market equilibrium: buyers want to buy exactly the quantity that sellers offer, so at this quantity the demand price equals the supply price).”

    Check: Is the equilibrium price 8 EUR per unit?

  4. “The profit of a company is \(P = -x^2 + 15x - 44\), with \(x\) in QU and \(P\) in 1,000 EUR. The owner claims that a profit of exactly 13,000 EUR is possible.”

    Check: Is the owner right?

  5. “A company has the total costs \(C = x^2 + 2x + 36\), with \(x\) in QU and \(C\) in MU. It sells 1 QU for 14 MU. At a break-even point, the revenue equals the total costs.”

    Check: Does the company make a profit at 5 QU?

  6. “A company can sell \(x\) QU if it asks the price \(p(x) = 30 - 3x\) MU per QU (price-demand function). Its revenue is 72 MU.”

    Check: Is there only one output with this revenue?

  7. “A car-sharing company has 40 cars, 12 of them electric. It buys \(x\) new electric cars and no other cars. Afterwards, at least 40 % of its cars are to be electric.”

    Check: Are 4 new electric cars enough?

  8. “A supplier of wooden pallets offers \(x\) QU at the supply price \(p_S(x) = 4\sqrt{x + 1}\) EUR per pallet. We look for the quantity at which the supply price is 12 EUR.”

    Check: Does the supplier offer 35 QU at this price?

  9. “An online shop has 300 customers in region A and 1,200 customers in region B. In region A, the number of customers grows by 32 % per year, in region B by 10 % per year. After \(t\) years, region A has more customers than region B.”

    Check: Does region A have more customers than region B after 7 years?

  1. With \(r\) regular and \(d\) reduced tickets: \(r + d = 460\) and \(r = 1.3d\). “More than the reduced tickets” makes the reduced tickets the base.

    Substitute: \(1.3d + d = 2.3d = 460 \iff d = 200\), so \(r = 260\).

    Check: yes, \(1.3 \cdot 200 = 260\) and \(260 + 200 = 460\).

    Typical mistake: \(d = 1.3r\). This swaps the base. It gives 200 regular and 260 reduced tickets.

  2. With \(x\) the price of a box and \(y\) the price of a cartridge in EUR: (I) \(3x + 2y = 260\) and (II) \(6x + 4y = 520\).

    Eliminate: \(2 \cdot \text{(I)} - \text{(II)}\) gives \(0 = 0\). The second order is exactly twice the first, so both equations describe the same line: the system has infinitely many solutions.

    Check: no. \((x, y) = (60, 40)\) fits both orders, but so does \((40, 70)\): \(120 + 140 = 260\) and \(240 + 280 = 520\). The two orders cannot fix the prices. Only an order with a different mix of boxes and cartridges would.

    Typical mistake: assuming that two equations always fix two unknowns. If one equation is a multiple of the other, it adds no new information.

  3. Equilibrium: \(p_D(x) = p_S(x)\).

    \[30 - 2x = 6 + x \iff 24 = 3x \iff x = 8\]

    Price: \(p_D(8) = 30 - 16 = 14\) EUR and \(p_S(8) = 6 + 8 = 14\) EUR ✓

    Check: no. 8 is the quantity, i.e. 8,000 units. The equilibrium price is 14 EUR per unit.

    Typical mistake: reading \(x = 8\) as the price. Here the unknown is the quantity. Substitute it into \(p_D\) or \(p_S\) to get the price.

  4. Zero form: \(-x^2 + 15x - 44 = 13 \iff -x^2 + 15x - 57 = 0 \iff x^2 - 15x + 57 = 0\) (multiply by \(-1\)).

    \(\Delta = (-15)^2 - 4 \cdot 1 \cdot 57 = 225 - 228 = -3 < 0\)

    Check: no. The equation has no real solution, so no output gives a profit of exactly 13,000 EUR.

    Typical mistake: computing the discriminant of \(P = 0\). There, \(x^2 - 15x + 44\) has \(\Delta = 49 > 0\), which only shows that the company has two break-even points. Bring the target of 13 into the zero form first.

  5. Revenue \(R = 14x\). Break-even: \(14x = x^2 + 2x + 36 \iff x^2 - 12x + 36 = 0\).

    \(\Delta = (-12)^2 - 4 \cdot 1 \cdot 36 = 144 - 144 = 0\): exactly one solution, \(x = \frac{12}{2} = 6\). Indeed \(x^2 - 12x + 36 = (x - 6)^2\).

    Check: no. \(P(5) = 70 - (25 + 10 + 36) = -1\) MU, a loss. In fact \(P = 14x - (x^2 + 2x + 36) = -(x - 6)^2\), and a square is never negative. So the profit is never positive: the company breaks even at 6 QU and makes a loss at every other output.

    Typical mistake: expecting two break-even points with a profit zone between them. \(\Delta = 0\) means that there is only one break-even point and no profit zone at all.

  6. Revenue = quantity × price: \(R = x \cdot p(x) = x(30 - 3x)\).

    \[x(30 - 3x) = 72 \iff -3x^2 + 30x - 72 = 0\]

    Divide by \(-3\): \(x^2 - 10x + 24 = 0\).

    Check: no. \((x - 4)(x - 6) = 0\), so 4 QU at \(p(4) = 18\) MU per QU and 6 QU at \(p(6) = 12\) MU per QU both bring 72 MU.

    Typical mistake: \(30 - 3x = 72\), i.e. \(x = -14\). The price-demand function gives the price per QU, not the revenue.

  7. After the purchase, the company has \(40 + x\) cars, and \(12 + x\) of them are electric:

    \[\frac{12 + x}{40 + x} \geq 0.4 \quad \text{with } x \geq 0 \text{ (so } 40 + x \neq 0\text{)}\]

    Multiply by \(40 + x > 0\) (the sign stays): \(12 + x \geq 16 + 0.4x \iff 0.6x \geq 4 \iff x \geq 6.67\). Only whole cars can be bought, so round up: at least 7 new electric cars.

    Check: no. With 4 new cars, \(\frac{16}{44} \approx 36.4\,\% < 40\,\%\). With 7 new cars, \(\frac{19}{47} \approx 40.4\,\%\).

    Typical mistake: \(12 + x = 0.4 \cdot 40 = 16\), i.e. \(x = 4\). The new cars also enlarge the fleet, so the base grows from 40 to \(40 + x\).

  8. \(4\sqrt{x + 1} = 12\) with the domain \(x \geq -1\) (in the context \(x \geq 0\)). Isolate the root first:

    \[\sqrt{x + 1} = 3 \iff x + 1 = 9 \iff x = 8\]

    In the original equation: \(4\sqrt{9} = 12\) ✓. So the supplier offers 8 QU at 12 EUR per pallet.

    Check: no. \(p_S(35) = 4\sqrt{36} = 24\) EUR per pallet, twice the price.

    Typical mistake: squaring \(4\sqrt{x + 1}\) to \(4(x + 1)\). Then \(4(x + 1) = 144\) gives \(x = 35\). The factor 4 must be squared as well: \(\left(4\sqrt{x + 1}\right)^2 = 16(x + 1)\). Isolating the root first avoids the problem.

  9. \(300 \cdot 1.32^t > 1200 \cdot 1.1^t\). Divide by \(300 \cdot 1.1^t > 0\) and use \(\frac{1.32^t}{1.1^t} = \left(\frac{1.32}{1.1}\right)^t = 1.2^t\):

    \[1.2^t > 4 = 1.2^{\log_{1.2} 4}\]

    The base 1.2 is greater than 1, so compare the exponents: \(t > \log_{1.2} 4 = \frac{\ln 4}{\ln 1.2} \approx 7.60\).

    Check: no. After 7 years, region A has \(300 \cdot 1.32^7 \approx 2{,}095\) customers and region B has \(1200 \cdot 1.1^7 \approx 2{,}338\). Region A takes the lead only after about 7.60 years.

    Typical mistake: subtracting the growth rates: “region A grows by 22 % more”, i.e. \(300 \cdot 1.22^t > 1200\), which gives \(t > 6.97\). Growth factors multiply, so compare them by dividing: \(\frac{1.32}{1.1} = 1.2\).

One Paragraph, All Conditions (xx)

A family business builds wooden garden benches. It models its monthly total costs by a quadratic function \(C(x) = ax^2 + bx + c\) of the output \(x\) in QU, with \(C\) in MU. Its fixed costs are 1,200 EUR per month. In a month with 20 benches, the total costs were 3,400 EUR, and in a month with 50 benches, they were 7,450 EUR. The business sells each bench for 190 EUR. It can build at most 60 benches per month. The owner also wants the average cost per bench (total costs divided by the number of benches) to be at most 165 EUR.

(1 QU = 10 benches, 1 MU = 100 EUR)

List every condition in this paragraph and the terms for the revenue and the profit, with \(x\) the monthly output in QU. Give the domain restriction of your average-cost condition. Do not solve anything.

Check: Which of your conditions rules out 20 benches per month?

  1. Fixed costs: \(c = C(0) = 12\) MU
  2. 20 benches \(= 2\) QU cost 3,400 EUR \(= 34\) MU: \(C(2) = 4a + 2b + 12 = 34\)
  3. 50 benches \(= 5\) QU cost 7,450 EUR \(= 74.5\) MU: \(C(5) = 25a + 5b + 12 = 74.5\)
  4. Price per QU: \(10 \cdot 190 = 1900\) EUR \(= 19\) MU, so the revenue is \(R = 19x\) MU
  5. Profit: \(P = R - C = 19x - (ax^2 + bx + 12)\) MU
  6. Capacity: \(0 \leq x \leq 6\)
  7. Average cost: \(\frac{100 \cdot C(x)}{10x} = \frac{10 \cdot C(x)}{x} \leq 165\) EUR per bench, i.e. \(\frac{C(x)}{x} \leq 16.5\). The fraction is defined only for \(x \neq 0\).

Conditions 1 to 3 form a system of linear equations for \(a\), \(b\) and \(c\).

Check: 20 benches cost \(\frac{3400}{20} = 170\) EUR per bench, more than 165 EUR. Condition 7 rules them out. The capacity (condition 6) would allow them.

Typical mistake: \(\frac{C(x)}{x} \leq 165\). \(C(x)\) is in MU and \(x\) in QU, so \(\frac{C(x)}{x}\) is in MU per QU: 1 MU per QU is 100 EUR per 10 benches, i.e. 10 EUR per bench.

FSP Words So Far

The FSP problems use these words from Sections 01 and 02. Learn what each one means and how to write it in mathematics.

Word Meaning Mathematics
quantity unit (QU) a fixed number of items, e.g. 1 QU = 100 items the output \(x\) is counted in QU
monetary unit (MU) a fixed amount of money, e.g. 1 MU = 100 EUR results in MU, convert to EUR at the end
fixed costs costs that arise even if nothing is produced the constant term of \(C(x)\)
variable costs costs that grow with the output output times costs per QU
total costs fixed costs + variable costs \(C(x)\)
revenue quantity times price \(R(x) = x \cdot \text{price}\)
profit revenue − total costs \(P(x) = R(x) - C(x)\)
break-even point the output at which the profit changes from negative (a loss) to positive \(P(x) = 0\), the smaller zero
profit limit (upper break-even point) the larger output with zero profit: above it, the profit turns negative again \(P(x) = 0\), the larger zero
profit zone all outputs with a positive profit \(P(x) > 0\): from the break-even point to the profit limit (or the capacity)
capacity the maximum possible output an upper bound: \(x \leq\) capacity
growth factor the factor by which a quantity is multiplied each period \(q\) in (starting value) \(\cdot\, q^t\)
growth rate the change per period in percent \(q = 1 + r\) with \(r\) as a decimal, a loss of 4 % gives \(q = 0.96\)
doubling time the time after which a quantity has doubled \(q^T = 2\)
half-life (halving time) the time after which a quantity has halved \(q^T = 0.5\)
“deviates from \(a\) by at most \(d\)” (tolerance) the distance from \(a\) is at most \(d\) \(\lvert x - a \rvert \leq d \iff a - d \leq x \leq a + d\)
“share of \(A\) among \(B\)” the part of \(B\) that also belongs to \(A\): the base is \(B\) \(\dfrac{\text{number in } A \cap B}{\text{number in } B}\)
“neither \(A\) nor \(B\)” not in \(A\) and not in \(B\), i.e. outside both circles \(\overline{A \cup B}\)
contingency table a table of counts for two properties: the inner fields are the regions of the Venn diagram, the margins are the totals inner fields \(A \cap B\), \(A \cap \overline{B}\), \(\overline{A} \cap B\), \(\overline{A} \cap \overline{B}\)
gross price, net price gross price = net price + VAT with 19 % VAT: gross \(= 1.19 \cdot\) net
system of linear equations several linear equations that must hold at the same time solve by substitution or elimination, check the solution in every equation
no solution, infinitely many solutions the conditions contradict each other, or one condition only repeats another elimination leaves \(0 = d\) with \(d \neq 0\), or \(0 = 0\)
demand, supply the quantity buyers want to buy and the quantity sellers offer at a given price as terms in the price \(P\), or as prices \(p_D(x)\), \(p_S(x)\) for the quantity \(x\)
market equilibrium the quantity demanded equals the quantity supplied demand = supply, or \(p_D(x) = p_S(x)\)
price-demand function gives the price at which \(x\) QU can be sold \(p(x)\), revenue \(R(x) = x \cdot p(x)\)
average cost total costs divided by the number of units \(\dfrac{C(x)}{x}\) with \(x \neq 0\)
zero form all terms on one side, 0 on the other side \(ax^2 + bx + c = 0\), then factor or use the quadratic formula
discriminant decides how many real solutions a quadratic equation has \(\Delta = b^2 - 4ac\): \(> 0\) two, \(= 0\) one, \(< 0\) none
biquadratic equation an equation with only \(x^4\), \(x^2\) and a constant substitute \(u = x^2\) in \(ax^4 + bx^2 + c\), solve for \(u\), then \(x = \pm\sqrt{u}\) for \(u \geq 0\)
domain restrictions (of an equation) the conditions the unknown must meet so that every term is defined denominators \(\neq 0\), under a root \(\geq 0\), inside a logarithm \(> 0\)
extraneous solution a candidate that solves the squared equation but not the original one check every candidate in the original equation

In set-builder notation, \(\{x \in \mathbb{R} \mid \dots\}\) and \(\{x \in \mathbb{R} : \dots\}\) mean the same.

Set B: Practice Mock (90 minutes, 50 points)

Problem 1: Camping Tents [20 pts. total]

An outdoor equipment company produces camping tents. It models its monthly total costs by a quadratic function \(C(x) = ax^2 + bx + c\). In March, it produced 200 tents at total costs of 15,800 EUR, and in April 600 tents at total costs of 23,000 EUR. In May, it produced 1,000 tents at an average cost (total costs divided by the number of tents) of 31.80 EUR per tent. Market research shows a linear price-demand function \(p(x) = mx + n\), which gives the price at which \(x\) QU can be sold: at a price of 48 EUR per tent, the company sells 400 tents, and 1,200 tents bring a revenue of 43,200 EUR.

(QU = quantity units, MU = monetary units. 1 QU = 100 tents, 1 MU = 100 EUR)

  1. Determine the cost function \(C(x)\) in MU, with \(x\) in QU, by setting up and solving a system of three linear equations.

(for verification: \(C(x) = 0.5x^2 + 14x + 128\))

[6 pts.]

  1. Determine the price-demand function \(p(x)\) in MU per QU by setting up and solving a system of two linear equations. Give the revenue function \(R(x)\).

(for verification: \(p(x) = 54 - 1.5x\))

[3 pts.]

  1. Show that the profit function is given by \(P(x) = -2x^2 + 40x - 128\).

[2 pts.]

  1. Show by completing the square that \(P(x) = -2(x - 10)^2 + 72\). Use this form to determine the (lower) break-even point and the profit limit (upper break-even point).

[4 pts.]

  1. Use the form from d) and the fact that a square is never negative to determine the maximum profit and the output at which it occurs. Calculate the price per tent the company must ask to sell exactly this output.

[3 pts.]

  1. A new warehouse lease raises the fixed costs from 128 MU to \(F\) MU, and all other terms stay the same. Determine with the help of the discriminant the largest value of \(F\) for which the company can still avoid a loss. Interpret the result for this value of \(F\).

Hint: the company can still avoid a loss exactly if the equation \(P(x) = 0\) still has a solution, i.e. if its discriminant is not negative.

[2 pts.]

a) Convert to QU and MU: 200 tents \(= 2\) QU cost 15,800 EUR \(= 158\) MU, and 600 tents \(= 6\) QU cost 23,000 EUR \(= 230\) MU. 1,000 tents \(= 10\) QU cost \(1000 \cdot 31.80 = 31{,}800\) EUR \(= 318\) MU in total.

Insert the three data points into \(C(x) = ax^2 + bx + c\):

\[\begin{aligned} \text{(I)} \quad C(2) &= 4a + 2b + c = 158 \\ \text{(II)} \quad C(6) &= 36a + 6b + c = 230 \\ \text{(III)} \quad C(10) &= 100a + 10b + c = 318 \end{aligned}\]

Eliminate \(c\) by subtracting (I) from (II) and (II) from (III):

\[\text{(II)} - \text{(I)}: \quad 32a + 4b = 72 \quad \big| : 4 \implies \text{(IV)} \quad 8a + b = 18\]

\[\text{(III)} - \text{(II)}: \quad 64a + 4b = 88 \quad \big| : 4 \implies \text{(V)} \quad 16a + b = 22\]

Eliminate \(b\):

\[\text{(V)} - \text{(IV)}: \quad 8a = 4 \implies a = 0.5\]

Back-substitution: in (IV): \(4 + b = 18 \implies b = 14\). In (I): \(2 + 28 + c = 158 \implies c = 128\).

Check in (III): \(100 \cdot 0.5 + 10 \cdot 14 + 128 = 50 + 140 + 128 = 318 \; \checkmark\)

\[C(x) = 0.5x^2 + 14x + 128 \quad \checkmark\]

The constant term \(c = 128\) MU is the fixed cost, 12,800 EUR per month.

Grading: 1 Pt. conversion including the total costs from the average cost, 1 Pt. system of three equations, 2 Pts. reduction to a 2×2 system, 1 Pt. solving and back-substitution, 1 Pt. check and cost function.

b) 400 tents \(= 4\) QU. A price of 48 EUR per tent is \(100 \cdot 48 = 4{,}800\) EUR \(= 48\) MU per QU, so \(p(4) = 48\). 1,200 tents \(= 12\) QU bring 43,200 EUR \(= 432\) MU. Revenue is quantity times price, so \(p(12) = \frac{432}{12} = 36\) MU per QU.

\[\begin{aligned} \text{(I)} \quad 4m + n &= 48 \\ \text{(II)} \quad 12m + n &= 36 \end{aligned}\]

\[\text{(II)} - \text{(I)}: \quad 8m = -12 \implies m = -1.5, \qquad \text{in (I)}: \; -6 + n = 48 \implies n = 54\]

\[p(x) = 54 - 1.5x \quad \checkmark\]

Revenue is quantity times price:

\[R(x) = x \cdot p(x) = x(54 - 1.5x) = -1.5x^2 + 54x\]

Grading: 1 Pt. both data points in QU and MU (the second price from the revenue), 1 Pt. system and \(m\), \(n\), 1 Pt. \(R(x)\).

c)

\[P(x) = R(x) - C(x) = (-1.5x^2 + 54x) - (0.5x^2 + 14x + 128)\]

The minus sign changes every sign in the second bracket:

\[P(x) = -1.5x^2 + 54x - 0.5x^2 - 14x - 128 = -2x^2 + 40x - 128 \quad \checkmark\]

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

d) Completing the square: factor out \(-2\) from the terms with \(x\):

\[P(x) = -2\left(x^2 - 20x\right) - 128\]

Add and subtract \(\left(\frac{20}{2}\right)^2 = 100\) inside the bracket:

\[P(x) = -2\left(x^2 - 20x + 100 - 100\right) - 128 = -2(x - 10)^2 + 200 - 128 = -2(x - 10)^2 + 72 \quad \checkmark\]

Break-even points: the profit is zero, \(P(x) = 0\):

\[-2(x - 10)^2 + 72 = 0 \iff (x - 10)^2 = 36 \iff x - 10 = \pm 6\]

\[x_1 = 4, \qquad x_2 = 16\]

Check: \(P(4) = -32 + 160 - 128 = 0 \; \checkmark\)

Break-even point at \(x = 4\) QU (400 tents), profit limit at \(x = 16\) QU (1,600 tents).

Grading: 2 Pts. completing the square, 1 Pt. \((x - 10)^2 = 36\) with both signs, 1 Pt. both points named in tents.

e) Maximum: a square is never negative, so \((x - 10)^2 \geq 0\). Multiplying by \(-2 < 0\) reverses the inequality: \(-2(x - 10)^2 \leq 0\). Hence

\[P(x) = -2(x - 10)^2 + 72 \leq 72 \quad \text{for all } x,\]

with equality exactly when \((x - 10)^2 = 0\), i.e. at \(x = 10\):

\[x_{\max} = 10 \text{ QU} = 1{,}000 \text{ tents}, \qquad P_{\max} = 72 \text{ MU} = 7{,}200 \text{ EUR per month}\]

Check: \(P(10) = -200 + 400 - 128 = 72 \; \checkmark\)

Price: \(p(10) = 54 - 1.5 \cdot 10 = 39\) MU per QU \(= 3{,}900\) EUR per 100 tents, i.e. 39 EUR per tent.

Grading: 1 Pt. argument \(-2(x - 10)^2 \leq 0\), 1 Pt. maximum profit in EUR and output in tents, 1 Pt. price per tent.

f) New profit function: \(P_F(x) = -2x^2 + 40x - F\). Break-even: \(P_F(x) = 0 \iff 2x^2 - 40x + F = 0\).

A break-even point exists only if the discriminant is not negative (hint):

\[\Delta = (-40)^2 - 4 \cdot 2 \cdot F = 1600 - 8F \geq 0 \iff F \leq 200\]

\[F_{\max} = 200 \text{ MU} = 20{,}000 \text{ EUR per month}\]

Interpretation: for \(F = 200\) we get \(\Delta = 0\), so there is exactly one break-even point, \(x = \frac{40}{2 \cdot 2} = 10\) QU. There, \(P_{200}(x) = -2x^2 + 40x - 200 = -2(x - 10)^2\) is zero, and at every other output it is negative. The company can only just cover its costs, with 1,000 tents. This matches e): the maximum profit of 72 MU is used up by the additional \(200 - 128 = 72\) MU of fixed costs. For \(F > 200\) we get \(\Delta < 0\), and the company makes a loss at every output.

Grading: 1 Pt. discriminant condition and \(F = 200\), 1 Pt. interpretation of \(\Delta = 0\).

Problem 2: Olive Oil Importer [17 pts. total]

An importer of olive oil supplies restaurants and fine-food shops.

  1. One delivery tour leads to a customer 180 km away: 150 km on the motorway, then 30 km on country roads. On country roads, the delivery van drives 40 km/h slower than on the motorway, and the whole trip takes 2 hours. Let \(v\) be the speed on the motorway in km/h (time = distance ÷ speed). Give the domain restrictions and the condition that the situation adds. Show that the situation leads to the equation

\[v^2 - 130v + 3000 = 0\]

[4 pts.]

  1. Solve the equation. Explain why the equation has two solutions, but the situation only one. Give both speeds and the driving time on each part of the route.

[3 pts.]

  1. In the regional market for this olive oil, the demand price is \(p_D(x) = 4\sqrt{10 - x}\) and the supply price is \(p_S(x) = x + 2\) (\(x\) in 1,000 litres, \(p\) in EUR per litre). Determine the market equilibrium (quantity and price). Explain why one of the solutions of the squared equation does not describe an equilibrium.

[5 pts.]

  1. Four years ago, the importer bought a bottling machine for 25,600 EUR, and two years ago a labelling machine for 19,200 EUR. Both machines lose the same percentage of their value every year: each year, their values are multiplied by the same factor \(q\). Today, the two machines are worth 18,900 EUR together. Show that \(q\) satisfies

\[256q^4 + 192q^2 - 189 = 0\]

and determine by how many percent the machines lose value every year.

[5 pts.]

a) Driving times (time = distance ÷ speed): \(\frac{150}{v}\) hours on the motorway and \(\frac{30}{v - 40}\) hours on country roads. Together they take 2 hours:

\[\frac{150}{v} + \frac{30}{v - 40} = 2\]

Domain restrictions: the denominators must not be zero, so \(v \neq 0\) and \(v \neq 40\). The situation adds: both speeds are positive, so \(v - 40 > 0\), i.e. \(v > 40\).

Multiply by the common denominator \(v(v - 40)\):

\[150(v - 40) + 30v = 2v(v - 40)\]

\[150v - 6000 + 30v = 2v^2 - 80v\]

\[0 = 2v^2 - 260v + 6000 \quad \big| : 2\]

\[v^2 - 130v + 3000 = 0 \quad \checkmark\]

Grading: 1 Pt. both driving times and the equation, 1 Pt. domain restrictions and \(v > 40\), 1 Pt. multiplying by the common denominator, 1 Pt. simplifying.

b) \(\Delta = (-130)^2 - 4 \cdot 1 \cdot 3000 = 16900 - 12000 = 4900 = 70^2\):

\[v_{1,2} = \frac{130 \pm 70}{2}, \qquad v_1 = 100, \quad v_2 = 30\]

Both values satisfy the domain restrictions \(v \neq 0\) and \(v \neq 40\), so both solve the equation: \(\frac{150}{30} + \frac{30}{-10} = 5 - 3 = 2\). But \(v = 30\) violates \(v > 40\): on country roads, the van would drive \(30 - 40 = -10\) km/h. So \(v = 100\).

Motorway: 100 km/h for \(\frac{150}{100} = 1.5\) hours. Country roads: 60 km/h for \(\frac{30}{60} = 0.5\) hours.

Check: \(1.5 + 0.5 = 2 \; \checkmark\)

Grading: 1 Pt. solving the quadratic equation, 1 Pt. rejecting \(v = 30\) with reason, 1 Pt. both speeds and driving times.

c) Equilibrium: demand price equals supply price:

\[4\sqrt{10 - x} = x + 2\]

Domain: \(10 - x \geq 0\). Since a root is never negative, also \(x + 2 \geq 0\). Together: \(-2 \leq x \leq 10\).

Square both sides (perfect square formula on the right):

\[16(10 - x) = (x + 2)^2\]

\[160 - 16x = x^2 + 4x + 4\]

\[x^2 + 20x - 156 = 0\]

Quadratic formula:

\[x_{1,2} = \frac{-20 \pm \sqrt{400 + 624}}{2} = \frac{-20 \pm \sqrt{1024}}{2} = \frac{-20 \pm 32}{2} \implies x_1 = 6, \quad x_2 = -26\]

Check in the original equation (squaring can create extraneous solutions):

  • \(x = 6\): \(4\sqrt{4} = 8\) and \(6 + 2 = 8\) ✓
  • \(x = -26\): \(4\sqrt{36} = 24\), but \(-26 + 2 = -24\) ✗

Equilibrium: \(x = 6\), i.e. 6,000 litres, at a price of 8 EUR per litre.

Explanation: \(x = -26\) solves the squared equation only because squaring turns \(-24\) into \(576 = 24^2\) as well. It lies outside the domain \(x \geq -2\): a negative quantity with a negative supply price has no meaning.

Grading: 1 Pt. equation and domain, 1 Pt. squaring correctly, 1 Pt. solving the quadratic equation, 1 Pt. check and equilibrium, 1 Pt. explanation of the extraneous solution.

d) Values today: the bottling machine has lost value for 4 years, the labelling machine for 2 years:

\[25600q^4 + 19200q^2 = 18900\]

Zero form, then divide by 100:

\[25600q^4 + 19200q^2 - 18900 = 0 \quad \big| : 100 \implies 256q^4 + 192q^2 - 189 = 0 \quad \checkmark\]

Substitution \(u = q^2\) turns the biquadratic equation into a quadratic one:

\[256u^2 + 192u - 189 = 0\]

\[\Delta = 192^2 - 4 \cdot 256 \cdot (-189) = 36864 + 193536 = 230400 = 480^2\]

\[u_{1,2} = \frac{-192 \pm 480}{512}, \qquad u_1 = \frac{288}{512} = 0.5625, \quad u_2 = \frac{-672}{512} = -1.3125\]

Back-substitution \(q^2 = u\): \(q^2 = -1.3125\) has no real solution, since a square is never negative. \(q^2 = 0.5625\) gives \(q = \pm 0.75\). A negative factor would turn the values of the machines negative, so \(q = 0.75 = 1 - 0.25\).

The machines lose 25 % of their value every year.

Check: \(25600 \cdot 0.75^4 = 8100\) EUR and \(19200 \cdot 0.75^2 = 10800\) EUR, together 18,900 EUR ✓

Grading: 1 Pt. both values with \(q^4\) and \(q^2\), 1 Pt. zero form, 1 Pt. substitution and quadratic formula, 1 Pt. rejecting \(u < 0\) and \(q < 0\), 1 Pt. \(q = 0.75\) and 25 %.

Problem 3: Robot Lawn Mowers [13 pts. total]

A manufacturer of garden tools replaces its old robot lawn mower with a new model. \(t\) years after the launch of the new model, it sells \(N(t)\) thousand new mowers and \(O(t)\) thousand old mowers per year.

  1. At the launch, the new model sells 4,000 mowers per year, and two years later 9,000 mowers per year. Determine \(N(t) = a \cdot b^t\). The old model sells 36,000 mowers per year at the launch, and its sales fall by one third every year. Show that \(O(t) = 36 \cdot 1.5^{-t}\).

(for verification: \(N(t) = 4 \cdot 1.5^t\))

[3 pts.]

  1. Determine from which time on the new model sells more than the old one, exactly as a logarithm and rounded to two decimal places.

[3 pts.]

  1. Both models are made on the same production line. Determine the times at which the total sales \(S(t) = N(t) + O(t)\) are exactly 30,000 mowers per year, exactly and rounded to two decimal places.

Hint: substitute \(u = 1.5^t\). Then \(1.5^{-t} = \frac{1}{u}\).

[4 pts.]

  1. The production line only pays off as long as the total sales are at least 20,000 mowers per year. Decide with the discriminant whether the total sales ever fall to exactly 20,000 mowers per year. Explain why they always stay above this level.

Note: the total sales can only fall below 20,000 mowers per year at a time where they are exactly 20,000.

[3 pts.]

a) Ansatz \(N(t) = a \cdot b^t\). At the launch: \(N(0) = a = 4\). Two years later:

\[4b^2 = 9 \implies b^2 = 2.25 \implies b = 1.5\]

A growth factor is positive, so \(b = -1.5\) is rejected: \(N(t) = 4 \cdot 1.5^t\).

Old model: falling by one third means the factor \(1 - \frac{1}{3} = \frac{2}{3}\) per year, so \(O(t) = 36 \cdot \left(\frac{2}{3}\right)^t\). Since \(\frac{2}{3} = \frac{1}{1.5}\) and \(\left(\frac{1}{1.5}\right)^t = 1.5^{-t}\):

\[O(t) = 36 \cdot 1.5^{-t} \quad \checkmark\]

Grading: 1 Pt. \(a = 4\) and \(b^2 = 2.25\), 1 Pt. \(b = 1.5\) with the negative root rejected, 1 Pt. factor \(\frac{2}{3}\) and rewriting as \(1.5^{-t}\).

b)

\[4 \cdot 1.5^t > 36 \cdot 1.5^{-t}\]

Multiply by \(1.5^t > 0\) (the sign stays) and use \(1.5^t \cdot 1.5^t = \left(1.5^t\right)^2\):

\[4 \cdot \left(1.5^t\right)^2 > 36 \iff \left(1.5^t\right)^2 > 9 \iff 1.5^t > 3\]

Only the positive root 3 matters, since \(1.5^t > 0\). Write \(3 = 1.5^{\log_{1.5} 3}\). The base 1.5 is greater than 1, so compare the exponents:

\[t > \log_{1.5} 3 = \frac{\ln 3}{\ln 1.5} \approx 2.71\]

From about 2.71 years after the launch on, i.e. in the third year, the new model sells more than the old one.

Check: \(t = 2\): \(N = 9 < O = 16\). \(\; t = 3\): \(N = 13.5 > O \approx 10.67 \; \checkmark\)

Grading: 1 Pt. inequality and multiplying by \(1.5^t > 0\), 1 Pt. \(1.5^t > 3\), 1 Pt. logarithm exactly and rounded.

c)

\[4 \cdot 1.5^t + 36 \cdot 1.5^{-t} = 30\]

Substitution \(u = 1.5^t\) with \(u > 0\), so \(1.5^{-t} = \frac{1}{u}\):

\[4u + \frac{36}{u} = 30\]

Multiply by \(u\) (allowed, since \(u > 0\)) and bring the equation to zero form:

\[4u^2 + 36 = 30u \iff 4u^2 - 30u + 36 = 0 \quad \big| : 2 \iff 2u^2 - 15u + 18 = 0\]

\[\Delta = (-15)^2 - 4 \cdot 2 \cdot 18 = 225 - 144 = 81, \qquad u_{1,2} = \frac{15 \pm 9}{4} \implies u_1 = 6, \quad u_2 = 1.5\]

Both values are positive, so both are allowed. Back-substitution:

\[1.5^t = 1.5 \implies t = 1, \qquad 1.5^t = 6 \implies t = \frac{\ln 6}{\ln 1.5} = \log_{1.5} 6 \approx 4.42\]

Check: \(S(1) = 6 + 24 = 30 \; \checkmark\). At \(t \approx 4.42\): \(N = 4 \cdot 6 = 24\) and \(O = \frac{36}{6} = 6\), together \(30 \; \checkmark\)

The total sales are exactly 30,000 mowers per year after 1 year and after about 4.42 years.

Grading: 1 Pt. equation with the substitution, 1 Pt. multiplying by \(u > 0\) and zero form, 1 Pt. solving for \(u\), 1 Pt. back-substitution with both times.

d) \(S(t) = 20\) gives, with \(u = 1.5^t > 0\):

\[4u + \frac{36}{u} = 20 \iff 4u^2 - 20u + 36 = 0 \quad \big| : 4 \iff u^2 - 5u + 9 = 0\]

\[\Delta = (-5)^2 - 4 \cdot 1 \cdot 9 = 25 - 36 = -11 < 0\]

There is no real \(u\), so there is no time \(t\) with \(S(t) = 20\): the total sales never fall to exactly 20,000 mowers per year.

At the launch, \(S(0) = 4 + 36 = 40 > 20\). To get below 20,000 mowers per year, the total sales would first have to be exactly 20,000, and this never happens. So they always stay above 20,000 mowers per year, and the production line pays off.

Grading: 1 Pt. zero form in \(u\), 1 Pt. discriminant, 1 Pt. decision with \(S(0) = 40\) and reason.

Exam Preparation Checklist

Before the exam, ensure you can:

Remember: Show all work clearly, label units, and always interpret your results in the business context!