Tasks 02-01 - Systems of Linear Equations

Section 02: Equations & Problem-Solving Strategies

A: Skills

Problem 1: Method Selection Practice (x)

Solve each 2×2 system using the indicated method:

  1. Use substitution: \[\begin{align} x - 2y &= 8\\ 3x + y &= 5 \end{align}\]

  2. Use elimination: \[\begin{align} 2x + 5y &= 16\\ 3x + 5y &= 21 \end{align}\]

  3. Your choice of method: \[\begin{align} 4x + 3y &= 25\\ 2x - y &= 5 \end{align}\]

  1. Substitution method:
    • From equation 1: x = 8 + 2y
    • Substitute into equation 2: 3(8 + 2y) + y = 5
    • 24 + 6y + y = 5
    • 7y = -19
    • y = -19/7
    • x = 8 + 2(-19/7) = 8 - 38/7 = 56/7 - 38/7 = 18/7
    • Solution: (18/7, -19/7)
    • Check: 18/7 - 2(-19/7) = 18/7 + 38/7 = 56/7 = 8
  2. Elimination method:
    • Subtract equation 1 from equation 2:
    • (3x + 5y) - (2x + 5y) = 21 - 16
    • x = 5
    • Substitute into equation 1: 2(5) + 5y = 16
    • 10 + 5y = 16
    • 5y = 6
    • y = 6/5
    • Solution: (5, 6/5)
    • Check: 3(5) + 5(6/5) = 15 + 6 = 21
  3. Best method: Elimination (multiply equation 2 by 3):
    • 4x + 3y = 25
    • 6x - 3y = 15
    • Add: 10x = 40
    • x = 4
    • From equation 2: 2(4) - y = 5
    • y = 3
    • Solution: (4, 3)
    • Check: 4(4) + 3(3) = 16 + 9 = 25

Problem 2: Special Cases Recognition (x)

Decide without fully solving whether each system has a unique solution, no solution, or infinitely many solutions. Then verify your answer by solving.

  1. \[\begin{align} 3x - 6y &= 12\\ x - 2y &= 4 \end{align}\]

  2. \[\begin{align} 2x + 4y &= 10\\ 3x + 6y &= 18 \end{align}\]

  3. \[\begin{align} x + 3y &= 7\\ 2x - y &= 0 \end{align}\]

  1. Analysis: Second equation × 3 gives 3x - 6y = 12, which equals the first equation
    • Prediction: Infinite solutions (same line)
    • Verification: The equations are equivalent. Any point satisfying x - 2y = 4 is a solution.
    • Example solutions: (4, 0), (6, 1), (8, 2)
  2. Analysis: First equation × 3 gives 6x + 12y = 30 Second equation × 2 gives 6x + 12y = 36
    • Prediction: No solution (parallel lines)
    • Verification: Same left side, different right side → inconsistent system
  3. Analysis: Different slopes (coefficients not proportional)
    • Prediction: Unique solution
    • Verification:
    • From equation 2: x = y/2
    • Substitute: y/2 + 3y = 7
    • 7y/2 = 7
    • y = 2, x = 1
    • Unique solution: (1, 2)

Problem 3: Systems with Three Unknowns (xx)

Solve each system: eliminate one variable to get a 2×2 system, solve it and back-substitute. Look at the structure first, since not every system needs all of these steps. Verify your solution in all three equations.

  1. \[\begin{cases} x - 3y + z = 9 \\ y + 2z = 3 \\ 5z = 10 \end{cases}\]

  2. \[\begin{cases} x + y + z = 4 \\ 2x - y + 3z = -2 \\ 3x + 2y - z = 13 \end{cases}\]

  3. \[\begin{cases} 2x + y = 0 \\ 3y - z = -10 \\ x + z = 5 \end{cases}\]

Number the equations (I), (II), (III) from top to bottom.

  1. The system is already triangular, so start at the bottom:

    • \(5z = 10 \Rightarrow z = 2\)
    • \(y + 2 \cdot 2 = 3 \Rightarrow y = -1\)
    • \(x - 3 \cdot (-1) + 2 = 9 \Rightarrow x + 5 = 9 \Rightarrow x = 4\)
    • Solution: \((x, y, z) = (4, -1, 2)\)
    • Check: \(4 + 3 + 2 = 9\) ✓, \(-1 + 4 = 3\) ✓, \(5 \cdot 2 = 10\) ✓

  2. Eliminate \(z\) (it has the coefficients \(1\), \(3\) and \(-1\)):

    • \(\text{(I)} + \text{(III)}\): \(4x + 3y = 17\) … (IV)
    • \(\text{(II)} + 3 \cdot \text{(III)}\): \(2x - y + 3z + 9x + 6y - 3z = -2 + 39 \Rightarrow 11x + 5y = 37\) … (V)
    • \(5 \cdot \text{(IV)} - 3 \cdot \text{(V)}\): \(20x - 33x = 85 - 111 \Rightarrow -13x = -26 \Rightarrow x = 2\)
    • In (IV): \(8 + 3y = 17 \Rightarrow y = 3\)
    • In (I): \(2 + 3 + z = 4 \Rightarrow z = -1\)
    • Solution: \((x, y, z) = (2, 3, -1)\)
    • Check: \(2 + 3 - 1 = 4\) ✓, \(4 - 3 - 3 = -2\) ✓, \(6 + 6 + 1 = 13\) ✓

  3. Each equation contains only two unknowns, so substitution is quickest:

    • From (I): \(y = -2x\). From (III): \(z = 5 - x\)
    • In (II): \(3 \cdot (-2x) - (5 - x) = -10 \Rightarrow -6x - 5 + x = -10 \Rightarrow -5x = -5 \Rightarrow x = 1\)
    • \(y = -2\), \(z = 4\)
    • Solution: \((x, y, z) = (1, -2, 4)\)
    • Check: \(2 - 2 = 0\) ✓, \(-6 - 4 = -10\) ✓, \(1 + 4 = 5\) ✓

Problem 4: Fractions, Decimals and Brackets (xx)

Solve each system. First tidy up every equation: multiply out the brackets and multiply by a suitable number to clear fractions or decimals.

  1. \[\begin{cases} \frac{x}{2} + \frac{y}{3} = 4 \\ \frac{x}{4} - \frac{y}{6} = 0 \end{cases}\]

  2. \[\begin{cases} 0.4x + 0.25y = 5.5 \\ 0.2x - 0.5y = -1 \end{cases}\]

  3. \[\begin{cases} 2(x + y) - 3(x - y) = 13 \\ \frac{x + 1}{3} = y - 2 \end{cases}\]

Number the equations (I) and (II) from top to bottom.

  1. Multiply (I) by 6 and (II) by 12:

    • \(3x + 2y = 24\)
    • \(3x - 2y = 0\)
    • Add: \(6x = 24 \Rightarrow x = 4\)
    • In \(3x - 2y = 0\): \(12 = 2y \Rightarrow y = 6\)
    • Solution: \((4, 6)\)
    • Check: \(2 + 2 = 4\) ✓, \(1 - 1 = 0\) ✓

  2. Multiply (I) by 20 and (II) by 10:

    • \(8x + 5y = 110\)
    • \(2x - 5y = -10\)
    • Add: \(10x = 100 \Rightarrow x = 10\)
    • In \(2x - 5y = -10\): \(20 - 5y = -10 \Rightarrow y = 6\)
    • Solution: \((10, 6)\)
    • Check: \(4 + 1.5 = 5.5\) ✓, \(2 - 3 = -1\) ✓

  3. Multiply out the brackets in (I) and multiply (II) by 3:

    • (I): \(2x + 2y - 3x + 3y = 13 \Rightarrow -x + 5y = 13\)
    • (II): \(x + 1 = 3y - 6 \Rightarrow x = 3y - 7\)
    • Substitute: \(-(3y - 7) + 5y = 13 \Rightarrow 2y + 7 = 13 \Rightarrow y = 3\)
    • \(x = 9 - 7 = 2\)
    • Solution: \((2, 3)\)
    • Check: \(2 \cdot 5 - 3 \cdot (-1) = 13\) ✓, \(\frac{3}{3} = 1 = 3 - 2\) ✓

Problem 5: Parameters and Special Cases (xx)

Hint: eliminate one variable. If this leaves \(0 \cdot x = d\) with a number \(d \neq 0\), there is no solution. If it leaves \(0 = 0\), there are infinitely many solutions.

  1. Determine the value of \(k\) for which the system has no solution. Explain what the two lines look like for this value of \(k\).

\[\begin{cases} 3x - 2y = 5 \\ kx + 4y = 1 \end{cases}\]

  1. Determine the value of \(c\) for which the system has infinitely many solutions. Decide whether there is a value of \(c\) for which the system has exactly one solution.

\[\begin{cases} x - 4y = 3 \\ -3x + 12y = c \end{cases}\]

Number the equations (I) and (II) from top to bottom.

  1. Eliminate \(y\) with \(2 \cdot \text{(I)} + \text{(II)}\):

    • \(6x - 4y + kx + 4y = 10 + 1 \Rightarrow (6 + k)x = 11\)
    • For \(6 + k \neq 0\) this gives exactly one \(x\), and then one \(y\): a unique solution.
    • For \(k = -6\) the equation becomes \(0 \cdot x = 11\), which is false for every \(x\): no solution.
    • For \(k = -6\) the second equation is \(-6x + 4y = 1\), while \(-2\) times the first gives \(-6x + 4y = -10\). Same left sides, different right sides: the two lines are parallel.

  2. Eliminate \(x\) with \(3 \cdot \text{(I)} + \text{(II)}\):

    • \(3x - 12y - 3x + 12y = 9 + c \Rightarrow 0 = 9 + c\)
    • For \(c = -9\) this is \(0 = 0\), true for every point on the line \(x - 4y = 3\): infinitely many solutions. The second equation is then \(-3\) times the first, so both describe the same line.
    • For every other \(c\) it is a false statement: no solution (parallel lines).
    • There is no value of \(c\) with exactly one solution: the left sides are always multiples of each other, so the lines always have the same slope.

B: Set Up & Check

Problem 6: From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

  1. “A garden centre sells twice as many herb pots as flower pots.”

    Check: On Saturday, the garden centre sells 40 herb pots and 80 flower pots. Does this fit the statement?

  2. “A truck carries 60 crates, small ones of 20 kg and large ones of 45 kg. Together, the crates weigh 1,950 kg.”

    Check: Can the truck carry 30 small and 30 large crates?

  3. “A laptop and its docking station cost 1,150 EUR together. The laptop costs 850 EUR more than the docking station.”

    Check: Does the docking station cost 300 EUR?

  4. “A shop mixes 5 kg of a snack mix from cashews at 18 EUR per kg and raisins at 8 EUR per kg. The mix is to cost 12 EUR per kg.”

    Check: The shop uses 2 kg of cashews and 3 kg of raisins. Does this meet both conditions?

  5. “In a market, the demand (the quantity buyers want to buy at the price \(P\)) is \(Q = 150 - 5P\), and the supply (the quantity sellers offer at the price \(P\)) is \(Q = 10 + 2P\), with \(Q\) in units and \(P\) in EUR per unit. The market is in equilibrium: the quantity demanded equals the quantity supplied.”

    Check: Is the equilibrium price 20 EUR, and how many units are sold at this price?

  1. With \(h\) herb pots and \(f\) flower pots: \(h = 2f\).

    Check: no, \(2 \cdot 80 = 160 \neq 40\). Here the garden centre sells twice as many flower pots as herb pots.

    Typical mistake: \(2h = f\). Test the equation with small numbers: 10 flower pots mean 20 herb pots, and \(h = 2f\) gives exactly that.

  2. With \(s\) small and \(l\) large crates: \(s + l = 60\) (number of crates) and \(20s + 45l = 1950\) (weight in kg).

    Check: yes, \(30 + 30 = 60\) and \(20 \cdot 30 + 45 \cdot 30 = 600 + 1350 = 1950\).

    Typical mistake: \(20s + 45l = 60\). One equation counts crates, the other adds kilograms. Never mix the two in one equation.

  3. With \(l\) the price of the laptop and \(d\) the price of the docking station in EUR: \(l + d = 1150\) and \(l = d + 850\).

    Substitute: \(d + 850 + d = 1150 \Rightarrow 2d = 300 \Rightarrow d = 150\), so \(l = 1000\).

    Check: no, the docking station costs 150 EUR. With 300 EUR, the laptop would cost \(300 + 850 = 1150\) EUR, and both together 1,450 EUR.

    Typical mistake: \(d = 1150 - 850 = 300\). The 850 EUR are the difference between the two prices, not between the total and one price.

  4. With \(c\) kg of cashews and \(r\) kg of raisins: \(c + r = 5\) (amount) and \(18c + 8r = 12 \cdot 5 = 60\) (value in EUR).

    Check: yes, \(2 + 3 = 5\) and \(18 \cdot 2 + 8 \cdot 3 = 36 + 24 = 60\).

    Typical mistake: \(18c + 8r = 12\). The left side is the value of the whole mix in EUR, so the right side must be a value as well: 5 kg at 12 EUR per kg, i.e. 60 EUR.

  5. Equilibrium: \(150 - 5P = 10 + 2P\).

    \[140 = 7P \iff P = 20\]

    Check: yes. At \(P = 20\), the demand is \(150 - 100 = 50\) units and the supply is \(10 + 40 = 50\) units, so 50 units are sold.

    Typical mistake: stopping at \(P = 20\). The equilibrium is a pair of price and quantity. Substitute the price into both equations: they must give the same quantity.

Problem 7: One Paragraph, All Conditions (xx)

A small workshop builds wooden birdhouses and insect hotels. A birdhouse needs 2 hours of sawing and 1 hour of painting, an insect hotel needs 3 hours of sawing and 2 hours of painting. Next week, the workshop has 60 hours of sawing time and 35 hours of painting time, and it wants to use all of them. The owner also wants to build 5 more birdhouses than insect hotels. A birdhouse sells for 45 EUR and an insect hotel for 70 EUR, and the owner hopes for a revenue of at least 1,300 EUR.

List every condition in this paragraph as a mathematical statement, with \(b\) the number of birdhouses and \(h\) the number of insect hotels. Do not solve anything.

Check: A plan with 12 birdhouses and 12 insect hotels uses exactly 60 hours of sawing time. Which of your conditions does it violate?

  1. Sawing time: \(2b + 3h = 60\)
  2. Painting time: \(b + 2h = 35\)
  3. More birdhouses: \(b = h + 5\)
  4. Revenue: \(45b + 70h \geq 1300\)
  5. Whole numbers, none negative: \(b, h \in \{0, 1, 2, \dots\}\)

Conditions 1 to 3 form a system of three equations with only two unknowns.

Check: with \(b = h = 12\), condition 1 holds (\(24 + 36 = 60\)) and so does condition 4 (\(540 + 840 = 1380 \geq 1300\)). But the painting time is \(12 + 24 = 36 \neq 35\) hours (condition 2), and \(12 \neq 12 + 5\) (condition 3).

Typical mistake: \(2b + h = 60\), i.e. reading the times of one product as one equation. Each equation belongs to one resource: all sawing hours go into the sawing equation, all painting hours into the painting equation.

C: Applied

Problem 8: Christmas Tree Markets (xx)

Christmas trees are sold in a city and in the rural area around it. The two markets are separate. With \(Q\) in 100 trees and \(P\) in EUR per tree:

  • City: demand \(Q = 90 - 2P\), supply \(Q = 3P - 10\)
  • Rural area: demand \(Q = 42 - P\), supply \(Q = 2P - 12\)
  1. Determine the equilibrium price and the equilibrium quantity in the city.

(for verification: \(P = 20\) EUR)

  1. Determine the equilibrium in the rural area. Decide in which market a tree is cheaper and by how many percent, measured against the higher price.

  2. The city charges the sellers a fee of 5 EUR per tree for their stands. Of every price \(P\), a seller keeps only \(P - 5\) EUR, so the supply in the city becomes \(Q = 3(P - 5) - 10\). Determine the new equilibrium, the price the buyers pay and the amount the sellers keep per tree.

(for verification: \(P = 23\) EUR)

  1. Compute the city’s income from the fee. Determine which share of the 5 EUR fee the buyers carry through the higher price and which share the sellers carry.

a) In equilibrium, demand equals supply:

\[90 - 2P = 3P - 10 \iff 100 = 5P \iff P = 20\]

\(Q = 90 - 2 \cdot 20 = 50\). Check with the supply: \(3 \cdot 20 - 10 = 50\) ✓

Equilibrium in the city: 20 EUR per tree, 5,000 trees.

b)

\[42 - P = 2P - 12 \iff 54 = 3P \iff P = 18\]

\(Q = 42 - 18 = 24\). Check: \(2 \cdot 18 - 12 = 24\) ✓

Equilibrium in the rural area: 18 EUR per tree, 2,400 trees.

A tree is cheaper in the rural area, by \(20 - 18 = 2\) EUR, i.e. by \(\frac{2}{20} = 10\,\%\) of the city price.

c) New supply: \(Q = 3P - 15 - 10 = 3P - 25\).

\[90 - 2P = 3P - 25 \iff 115 = 5P \iff P = 23\]

\(Q = 90 - 46 = 44\). Check: \(3 \cdot 23 - 25 = 44\) ✓

The buyers pay 23 EUR per tree, the sellers keep \(23 - 5 = 18\) EUR. 4,400 trees are sold.

d) Income: \(5 \cdot 4400 = 22{,}000\) EUR.

The buyers pay \(23 - 20 = 3\) EUR more than before, which is \(\frac{3}{5} = 60\,\%\) of the fee. The sellers keep \(20 - 18 = 2\) EUR less, which is \(\frac{2}{5} = 40\,\%\) of the fee.

Problem 9: Two Plants of a Pet-Food Manufacturer (xx)

A pet-food manufacturer has to deliver 1,200 bags of dog food to a retail chain. It can produce in two plants. The capacity of a plant is the maximum possible output.

  • Plant North: production 9.00 EUR per bag, shipping to the retail chain 3.00 EUR per bag, capacity 600 bags
  • Plant South: production 6.50 EUR per bag, shipping to the retail chain 3.50 EUR per bag, capacity 800 bags

The retail chain pays 15 EUR per bag. Let \(x\) be the number of bags from Plant North and \(y\) the number from Plant South.

  1. Compute the costs per bag (production and shipping) for each plant.

(for verification: 12 EUR and 10 EUR)

  1. The planning department expects total costs of exactly 13,000 EUR for the order. Set up a system of two equations and determine how many bags each plant produces. Check both capacities.

(for verification: \(x = 500\))

  1. Express the total costs as a term in \(x\) only. Determine all values of \(x\) that respect both capacities, and decide which split of the order is the cheapest. Compute the profit of the order for this split.

  2. The shipping company raises its price for Plant South by 2.50 EUR per bag. Decide which split is now the cheapest and compute the new profit.

a) North: \(9 + 3 = 12\) EUR per bag. South: \(6.50 + 3.50 = 10\) EUR per bag.

b)

\[\begin{aligned} \text{(I)} \quad x + y &= 1200 \\ \text{(II)} \quad 12x + 10y &= 13000 \end{aligned}\]

From (I): \(y = 1200 - x\). In (II):

\[12x + 10(1200 - x) = 13000 \iff 2x + 12000 = 13000 \iff x = 500\]

\(y = 700\). Check: \(12 \cdot 500 + 10 \cdot 700 = 6000 + 7000 = 13000\) ✓

Capacities: \(500 \leq 600\) and \(700 \leq 800\) ✓. North produces 500 bags and South 700 bags.

c) With \(y = 1200 - x\):

\[C = 12x + 10(1200 - x) = 2x + 12000\]

Capacities: \(x \leq 600\) and \(1200 - x \leq 800 \iff x \geq 400\). So \(x \in [400; 600]\).

\(C = 2x + 12000\) grows with \(x\) (every bag from North costs 2 EUR more), so the smallest \(x\) is the cheapest: \(x = 400\), \(y = 800\).

Costs: \(2 \cdot 400 + 12000 = 12{,}800\) EUR. Revenue: \(15 \cdot 1200 = 18{,}000\) EUR. Profit: \(18000 - 12800 = 5{,}200\) EUR.

d) South now costs \(10 + 2.50 = 12.50\) EUR per bag:

\[C = 12x + 12.5(1200 - x) = 15000 - 0.5x\]

Now \(C\) falls with \(x\), so the largest \(x\) is the cheapest: \(x = 600\), \(y = 600\).

Costs: \(15000 - 300 = 14{,}700\) EUR. Profit: \(18000 - 14700 = 3{,}300\) EUR.

The cheapest split switches from “as much as possible in South” to “as much as possible in North”.

Problem 10: Heating Costs in an Office Building (xxx)

The four tenants of an office building share the annual heating costs of 60,000 EUR: an architecture office, a law firm, a tax advisor and a design studio. Their contract lists five conditions:

  1. The architecture office and the law firm together pay exactly 40 % of the total.
  2. The design studio pays 1.5 times as much as the tax advisor.
  3. The tax advisor pays 2,000 EUR more than the law firm.
  4. The architecture office pays twice as much as the law firm.
  5. Together, the four tenants pay the total of 60,000 EUR.
  1. Translate the five conditions into equations, with \(A\), \(L\), \(T\) and \(D\) the amounts in EUR paid by the architecture office, the law firm, the tax advisor and the design studio.

(for verification: condition 1 gives \(A + L = 24000\))

  1. Determine \(A\), \(L\), \(T\) and \(D\) from conditions 1 to 4.

(for verification: \(L = 8000\))

  1. Check condition 5, the total of 60,000 EUR, and decide whether the contract can be fulfilled. Explain what this means for the system of all five equations, and give the amount that conditions 1 to 4 leave unassigned.

  2. The tenants drop condition 1 and keep conditions 2 to 5. Determine the new amounts and the share of the total that the architecture office and the law firm now pay together.

(for verification: \(L = 10000\))

a)

\[\begin{aligned} \text{(1)} \quad A + L &= 0.4 \cdot 60000 = 24000 \\ \text{(2)} \quad D &= 1.5\,T \\ \text{(3)} \quad T &= L + 2000 \\ \text{(4)} \quad A &= 2L \\ \text{(5)} \quad A + L + T + D &= 60000 \end{aligned}\]

b) Substitute (4) into (1):

\[2L + L = 24000 \iff L = 8000\]

Then \(A = 2 \cdot 8000 = 16{,}000\), \(T = 8000 + 2000 = 10{,}000\) and \(D = 1.5 \cdot 10000 = 15{,}000\) (all in EUR).

c) \(A + L + T + D = 16000 + 8000 + 10000 + 15000 = 49{,}000 \neq 60{,}000\).

Conditions 1 to 4 fix all four amounts, and these amounts violate condition 5. So the system of all five equations has no solution: the conditions contradict each other, and 11,000 EUR of the heating costs would not be assigned to anyone. The contract cannot be fulfilled as written.

d) Express everything in \(L\): \(A = 2L\), \(T = L + 2000\), \(D = 1.5(L + 2000) = 1.5L + 3000\). In (5):

\[2L + L + (L + 2000) + (1.5L + 3000) = 60000 \iff 5.5L + 5000 = 60000 \iff L = 10000\]

So \(L = 10{,}000\), \(A = 20{,}000\), \(T = 12{,}000\) and \(D = 18{,}000\) (all in EUR). Check: \(20000 + 10000 + 12000 + 18000 = 60000\) ✓

Architecture office and law firm together: \(\frac{30000}{60000} = 50\,\%\) of the total instead of 40 %.

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: Board Game Publisher [21 pts. total]

A small publisher produces a strategy board game. Its fixed costs, which arise even when no game is produced, amount to 4,500 EUR per month. For larger print runs, the printing company charges extra for overtime, so the total costs rise faster than the output. The publisher therefore models its total costs by a quadratic function \(C(x) = ax^2 + bx + c\). In a month with 300 games, the total costs were 11,100 EUR, and in a month with 500 games they were 17,500 EUR. The publisher sells each game to retailers for 39 EUR. The printing company can print at most 1,200 games per month, and the publisher plans to produce and sell 700 games per month.

(QU = quantity units, MU = monetary units. 1 QU = 100 games, 1 MU = 100 EUR. Total costs = fixed costs + variable costs, where the variable costs grow with the output. Revenue = quantity × price. Profit = revenue − total costs)

Part a)

Determine the cost function \(C(x)\) in MU, with \(x\) in QU, by setting up and solving a system of linear equations. [5 pts.]

Hint: the fixed costs are the total costs at \(x = 0\). Each month gives one equation: insert its output in QU for \(x\) and its total costs in MU for \(C(x)\).

(for verification: \(C(x) = 2x^2 + 16x + 45\))

Convert to QU and MU. Fixed costs: 4,500 EUR \(= 45\) MU, so \(c = C(0) = 45\). 300 games \(= 3\) QU cost 11,100 EUR \(= 111\) MU. 500 games \(= 5\) QU cost 17,500 EUR \(= 175\) MU.

Insert the data points into \(C(x) = ax^2 + bx + 45\):

\[\begin{aligned} \text{(I)} \quad C(3) &= 9a + 3b + 45 = 111 \\ \text{(II)} \quad C(5) &= 25a + 5b + 45 = 175 \end{aligned}\]

Subtract 45 on both sides and divide:

\[\text{(I)}: \; 9a + 3b = 66 \;\big|: 3 \implies 3a + b = 22, \qquad \text{(II)}: \; 25a + 5b = 130 \;\big|: 5 \implies 5a + b = 26\]

Eliminate \(b\) by subtracting the first of these equations from the second:

\[2a = 4 \implies a = 2\]

Back-substitution: \(3 \cdot 2 + b = 22 \implies b = 16\).

Check with the data: \(C(3) = 18 + 48 + 45 = 111\) ✓, \(\; C(5) = 50 + 80 + 45 = 175\) ✓

\[C(x) = 2x^2 + 16x + 45 \quad \checkmark\]

Grading: 1 Pt. conversion to QU and MU, 1 Pt. \(c = 45\) from the fixed costs, 1 Pt. both equations, 1 Pt. solving by elimination, 1 Pt. check and cost function.

Part b)

In a month with 800 games, the total costs were 30,100 EUR. Verify computationally that this value also fits the cost function. [2 pts.]

800 games \(= 8\) QU:

\[C(8) = 2 \cdot 64 + 16 \cdot 8 + 45 = 128 + 128 + 45 = 301 \text{ MU} = 30{,}100 \text{ EUR} \quad \checkmark\]

The third month gives a third equation, \(64a + 8b + c = 301\), and the solution of part a) fulfils it as well. The model matches all three months.

Grading: 1 Pt. \(x = 8\) and \(C(8)\), 1 Pt. comparison in EUR.

Part c)

Give a term for the revenue \(R(x)\) in MU. Show that the profit function is given by \(P(x) = -2x^2 + 23x - 45\). [3 pts.]

Price per QU: \(100 \cdot 39 = 3{,}900\) EUR \(= 39\) MU. Revenue is quantity times price:

\[R(x) = 39x\]

Profit:

\[P(x) = R(x) - C(x) = 39x - (2x^2 + 16x + 45)\]

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

\[P(x) = 39x - 2x^2 - 16x - 45 = -2x^2 + 23x - 45 \quad \checkmark\]

Grading: 1 Pt. \(R(x)\) with the price per QU in MU, 1 Pt. ansatz \(P = R - C\), 1 Pt. sign change and collecting terms.

Part d)

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

Factor \(P(x)\) completely with the AC method. Determine the (lower) break-even point and the profit limit (upper break-even point). [5 pts.]

Hint: first factor out \(-1\).

(for verification: \(P(x) = -(2x - 5)(x - 9)\))

Factor out \(-1\):

\[P(x) = -(2x^2 - 23x + 45)\]

Check: \(\Delta = (-23)^2 - 4 \cdot 2 \cdot 45 = 529 - 360 = 169 = 13^2\) is a perfect square, so the bracket factors.

AC method. \(ac = 2 \cdot 45 = 90\). Factors of 90 with sum \(-23\): \((-5, -18)\).

\[2x^2 - 5x - 18x + 45 = x(2x - 5) - 9(2x - 5) = (x - 9)(2x - 5)\]

\[P(x) = -(2x - 5)(x - 9) \quad \checkmark\]

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

\[2x - 5 = 0 \Rightarrow x = 2.5, \qquad x - 9 = 0 \Rightarrow x = 9\]

Check: \(P(9) = -162 + 207 - 45 = 0\) ✓

Break-even point: \(x = 2.5\) QU (250 games). Profit limit: \(x = 9\) QU (900 games). Both lie below the maximum print run of 12 QU.

Grading: 1 Pt. factoring out \(-1\), 1 Pt. \(ac = 90\) and the pair \((-5, -18)\), 1 Pt. grouping to the factored form, 1 Pt. both zeros, 1 Pt. naming break-even point and profit limit.

Part e)

Decide in which output range the publisher makes a profit and state this profit zone as an interval. Compute the monthly profit in EUR at the planned output of 700 games. [3 pts.]

Note: the profit can only change its sign at an output where it is zero.

Test values between and beyond the zeros 2.5 and 9 (at most 12 QU), with the factored form:

  • \(x = 1\): \(-(2 - 5)(1 - 9) = -(-3)(-8) = -24 < 0\), a loss
  • \(x = 5\): \(-(10 - 5)(5 - 9) = -5 \cdot (-4) = 20 > 0\), a profit
  • \(x = 10\): \(-(20 - 5)(10 - 9) = -15 < 0\), a loss

Profit zone: \((2.5; 9)\). The interval is open: at 2.5 QU and 9 QU the profit is exactly zero.

Planned output: 700 games \(= 7\) QU lies in the profit zone.

\[P(7) = -(14 - 5)(7 - 9) = -9 \cdot (-2) = 18 \text{ MU} = 1{,}800 \text{ EUR per month}\]

Grading: 1 Pt. test values or sign argument, 1 Pt. open interval, 1 Pt. profit at 7 QU in EUR.

Part f)

A retail chain offers to buy the whole planned output of 700 games per month, but only at a lower price per game. Determine the lowest price per game, in EUR with whole cents, at which the publisher still avoids a loss. Give the largest possible price cut in percent of the current price, rounded to one decimal place. [3 pts.]

Total costs at 7 QU:

\[C(7) = 2 \cdot 49 + 16 \cdot 7 + 45 = 98 + 112 + 45 = 255 \text{ MU} = 25{,}500 \text{ EUR}\]

No loss at a price of \(p\) EUR per game means that the revenue covers the total costs:

\[700p \geq 25500 \iff p \geq 36.428\ldots\]

Round up to whole cents: \(p = 36.43\) EUR. With 36.42 EUR, the revenue \(700 \cdot 36.42 = 25{,}494\) EUR does not cover the costs.

Price cut:

\[\frac{39 - 36.43}{39} = \frac{2.57}{39} \approx 0.0659 = 6.6\,\%\]

Check: at 39 EUR, the profit of 1,800 EUR per month is \(\frac{1800}{700} \approx 2.57\) EUR per game. This is exactly the room for the price cut.

Grading: 1 Pt. total costs at 700 games, 1 Pt. inequality and rounding up with reason, 1 Pt. percentage with the current price as base.

Calculator Practice

Problem 11: Calculator: Systems of Linear Equations (x)

Use the simultaneous equation solver:

  1. \[\begin{cases} 2x + 3y = 13 \\ x - y = 1 \end{cases}\]

  2. \[\begin{cases} 5x + 2y = 24 \\ 3x - 4y = 4 \end{cases}\]

  3. \[\begin{cases} x + 2y + z = 3 \\ 3x - y + 2z = 13 \\ 2x + y - z = 0 \end{cases}\]

  1. System 1:
    • Enter the coefficients row by row: [2, 3, 13; 1, -1, 1]
    • Solution: \(x = 3.2 = \frac{16}{5}\), \(y = 2.2 = \frac{11}{5}\)
  2. System 2:
    • Enter the coefficients row by row: [5, 2, 24; 3, -4, 4]
    • Solution: \(x = 4\), \(y = 2\)
    • Verification: \(5(4) + 2(2) = 24\) ✓, \(3(4) - 4(2) = 4\) ✓
  3. System 3:
    • Enter the coefficients row by row: [1, 2, 1, 3; 3, -1, 2, 13; 2, 1, -1, 0]
    • Solution: \(x = 2\), \(y = -1\), \(z = 3\)

Problem 12: Calculator: Market Equilibrium (xx)

Given supply and demand functions:

  • Demand: \(Q_d = 200 - 4P\)
  • Supply: \(Q_s = 50 + P\)

Set up and solve the system to find equilibrium price and quantity.

Setting up the system: - At equilibrium: \(Q_d = Q_s\), let’s call this Q - System: \(Q + 4P = 200\) and \(Q - P = 50\)

Using the 2-unknown solver: - Enter the coefficients row by row: [1, 4, 200; 1, -1, 50] - Solution: P = 30, Q = 80

Verification: - Demand: \(Q_d = 200 - 4(30) = 200 - 120 = 80\) ✓ - Supply: \(Q_s = 50 + 30 = 80\) ✓