Tasks 02-02 - Quadratic & Biquadratic Equations

Section 02: Equations & Problem-Solving Strategies

A: Skills

Problem 1: Converting to Zero Form (x)

Convert each equation to the form “expression = 0” and then solve it:

  1. \(5x + 15 = 30\)

  2. \(3x - 7 = 2x + 8\)

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

  4. \(\frac{2x + 6}{3} = x - 1\)

  5. \(0.5x + 2.5 = 1.5x - 3.5\)

  1. \(5x + 15 = 30\)
    • Subtract 30: \(5x + 15 - 30 = 0\)
    • Simplify: \(5x - 15 = 0\)
    • Solve: \(5x = 15\)
    • \(x = 3\)
    • Check: \(5(3) + 15 = 30\) ✓
  2. \(3x - 7 = 2x + 8\)
    • Subtract 2x: \(3x - 2x - 7 = 8\)
    • Subtract 8: \(x - 15 = 0\)
    • Solve: \(x = 15\)
    • Check: \(3(15) - 7 = 38\) and \(2(15) + 8 = 38\) ✓
  3. \(4(x - 2) = 3x + 5\)
    • Expand: \(4x - 8 = 3x + 5\)
    • Move all terms: \(4x - 8 - 3x - 5 = 0\)
    • Simplify: \(x - 13 = 0\)
    • Solve: \(x = 13\)
    • Check: \(4(13 - 2) = 44\) and \(3(13) + 5 = 44\) ✓
  4. \(\frac{2x + 6}{3} = x - 1\)
    • Multiply by 3: \(2x + 6 = 3x - 3\)
    • Move all terms: \(2x + 6 - 3x + 3 = 0\)
    • Simplify: \(-x + 9 = 0\)
    • Solve: \(x = 9\)
    • Check: \(\frac{2(9) + 6}{3} = 8\) and \(9 - 1 = 8\) ✓
  5. \(0.5x + 2.5 = 1.5x - 3.5\)
    • Move all terms: \(0.5x + 2.5 - 1.5x + 3.5 = 0\)
    • Simplify: \(-x + 6 = 0\)
    • Solve: \(x = 6\)
    • Check: \(0.5(6) + 2.5 = 5.5\) and \(1.5(6) - 3.5 = 5.5\) ✓

Problem 2: Zero Product Property (x)

Solve each equation with the zero product property:

  1. \((x - 5)(x + 3) = 0\)

  2. \((2x - 8)(3x + 6) = 0\)

  3. \(x(x - 7) = 0\)

  4. \((4x + 12)(x - 1)(x + 2) = 0\)

  5. \(3x(2x - 10) = 0\)

  1. \((x - 5)(x + 3) = 0\)
    • Either \(x - 5 = 0\) or \(x + 3 = 0\)
    • Solutions: \(x = 5\) or \(x = -3\)
  2. \((2x - 8)(3x + 6) = 0\)
    • Either \(2x - 8 = 0\) or \(3x + 6 = 0\)
    • From first: \(2x = 8\), so \(x = 4\)
    • From second: \(3x = -6\), so \(x = -2\)
    • Solutions: \(x = 4\) or \(x = -2\)
  3. \(x(x - 7) = 0\)
    • Either \(x = 0\) or \(x - 7 = 0\)
    • Solutions: \(x = 0\) or \(x = 7\)
    • Note: Don’t forget that x = 0 is a solution!
  4. \((4x + 12)(x - 1)(x + 2) = 0\)
    • Any factor can equal zero:
    • \(4x + 12 = 0\) gives \(x = -3\)
    • \(x - 1 = 0\) gives \(x = 1\)
    • \(x + 2 = 0\) gives \(x = -2\)
    • Solutions: \(x = -3, -2,\) or \(1\)
  5. \(3x(2x - 10) = 0\)
    • Either \(3x = 0\) or \(2x - 10 = 0\)
    • From first: \(x = 0\)
    • From second: \(2x = 10\), so \(x = 5\)
    • Solutions: \(x = 0\) or \(x = 5\)

Problem 3: Discriminant Analysis (x)

Compute the discriminant of each quadratic equation and decide how many real solutions it has. Then solve every equation that has real solutions with the quadratic formula.

  1. \(3x^2 - 7x + 2 = 0\)

  2. \(x^2 + 6x + 4 = 0\)

  3. \(2x^2 - 3x + 5 = 0\)

  4. \(-x^2 + 6x - 9 = 0\)

  1. \(3x^2 - 7x + 2 = 0\)
    • Discriminant: \(\Delta = (-7)^2 - 4(3)(2) = 49 - 24 = 25 > 0\)
    • Prediction: Two distinct rational solutions
    • Solutions: \(x = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6}\)
    • \(x_1 = 2\), \(x_2 = \frac{1}{3}\)
    • Check: \(3(2)^2 - 7(2) + 2 = 12 - 14 + 2 = 0\) ✓
  2. \(x^2 + 6x + 4 = 0\)
    • Discriminant: \(\Delta = 6^2 - 4(1)(4) = 36 - 16 = 20 > 0\), not a perfect square
    • Prediction: Two distinct irrational solutions
    • Solutions: \(x = \frac{-6 \pm \sqrt{20}}{2} = \frac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5}\)
    • \(x_1 \approx -0.764\), \(x_2 \approx -5.236\)
  3. \(2x^2 - 3x + 5 = 0\)
    • Discriminant: \(\Delta = 9 - 40 = -31 < 0\)
    • Prediction: No real solutions
  4. \(-x^2 + 6x - 9 = 0\)
    • Discriminant: \(\Delta = 36 - 36 = 0\)
    • Prediction: One repeated real solution
    • Solution: \(x = \frac{-6}{-2} = 3\)
    • Note: This is \(-(x - 3)^2 = 0\)

Problem 4: Method Selection Practice (x)

Solve each equation with the indicated method:

  1. Factor: \(x^2 + 2x - 35 = 0\)

  2. Complete the square: \(x^2 + 8x + 3 = 0\)

  3. Quadratic formula: \(3x^2 + 5x - 7 = 0\)

  4. Your choice: \(9x^2 + 30x + 25 = 0\)

  1. Factoring:
    • Find two numbers that multiply to -35 and add to 2
    • Those are 7 and -5
    • \((x + 7)(x - 5) = 0\)
    • Solutions: \(x = -7\) or \(x = 5\)
  2. Completing the square:
    • \(x^2 + 8x + 3 = 0\)
    • \(x^2 + 8x = -3\)
    • \(x^2 + 8x + 16 = -3 + 16\)
    • \((x + 4)^2 = 13\)
    • \(x + 4 = \pm\sqrt{13}\)
    • \(x = -4 \pm \sqrt{13}\)
  3. Quadratic formula:
    • \(a = 3, b = 5, c = -7\)
    • \(x = \frac{-5 \pm \sqrt{25 + 84}}{6} = \frac{-5 \pm \sqrt{109}}{6}\)
    • \(x_1 \approx 0.907\), \(x_2 \approx -2.573\)
  4. Best method: Recognize perfect square
    • \(9x^2 + 30x + 25 = (3x + 5)^2 = 0\)
    • Solution: \(x = -\frac{5}{3}\) (repeated)

Problem 5: Biquadratic and Cubic Equations (xx)

Solve each equation:

  1. \(x^4 - 20x^2 + 64 = 0\)

  2. \(x^4 + 3x^2 - 28 = 0\)

  3. \(2x^4 - 9x^2 + 4 = 0\)

  4. \(x^3 + 5x^2 - 4x - 20 = 0\)

  5. \(x^3 + 4x^2 - 6x = 0\)

Hint: a) to c) are biquadratic, so substitute \(u = x^2\). For the cubic equations d) and e), factor first.

  1. \(x^4 - 20x^2 + 64 = 0\)
    • Let u = x²: u² - 20u + 64 = 0
    • Factor: (u - 4)(u - 16) = 0
    • u = 4 or u = 16
    • If u = 4: x² = 4, so x = ±2
    • If u = 16: x² = 16, so x = ±4
    • Solutions: x = -4, -2, 2, 4
  2. \(x^4 + 3x^2 - 28 = 0\)
    • Let u = x²: u² + 3u - 28 = 0
    • Factor: (u + 7)(u - 4) = 0
    • u = -7 or u = 4
    • If u = -7: x² = -7 (no real solutions)
    • If u = 4: x² = 4, so x = ±2
    • Real solutions: x = -2, 2
  3. \(2x^4 - 9x^2 + 4 = 0\)
    • Let u = x²: 2u² - 9u + 4 = 0
    • Using quadratic formula: u = (9 ± √(81-32))/4 = (9 ± 7)/4
    • u = 4 or u = 1/2
    • If u = 4: x = ±2
    • If u = 1/2: x = ±1/√2 = ±√2/2
    • Solutions: x = -2, -√2/2, √2/2, 2
  4. \(x^3 + 5x^2 - 4x - 20 = 0\)
    • Group in pairs: x²(x + 5) - 4(x + 5) = 0
    • Factor out (x + 5): (x + 5)(x² - 4) = 0
    • Difference of squares: (x + 5)(x - 2)(x + 2) = 0
    • Solutions: x = -5, -2, 2
  5. \(x^3 + 4x^2 - 6x = 0\)
    • Factor out x: x(x² + 4x - 6) = 0
    • So x = 0 or x² + 4x - 6 = 0
    • Δ = 16 + 24 = 40 is not a perfect square, so use the quadratic formula: x = (-4 ± √40)/2 = (-4 ± 2√10)/2 = -2 ± √10
    • Solutions: x = 0, x = -2 + √10 ≈ 1.162, x = -2 - √10 ≈ -5.162
    • Note: dividing the equation by x instead of factoring it out loses the solution x = 0.

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. (QU = quantity units)

  1. “The profit \(P = -3x^2 + 21x - 30\) of a company is zero, with \(x\) in QU and \(P\) in 1,000 EUR.”

    Check: Is \(x = 2\) such an output?

  2. “The profit limit of a company with the profit \(P = -2x^2 + 25x - 63\) lies at 9 QU.”

    Check: Does the company make a profit at 10 QU?

  3. “At the break-even point, the revenue \(R = 24x\) equals the total costs \(C = x^2 + 5x + 48\), with \(x\) in QU and \(R\), \(C\) in 100 EUR.”

    Check: Is \(x = 3\) a break-even point?

  4. “The equation \(2x^2 + 12x + k = 0\) has no real solution.”

    Check: Does \(k = 18\) fulfil this condition?

  5. “A drinks producer with the profit \(P = -x^2 + 10x - 19\) (\(P\) in 100 EUR, \(x\) in QU, 1 QU = 100 bottles) makes a profit only above its break-even point \(x = 5 - \sqrt{6}\). So at least … bottles must be sold.”

    Check: Are 255 bottles enough for a profit?

  1. Zero profit: \(-3x^2 + 21x - 30 = 0\). Divide every term by \(-3\): \(x^2 - 7x + 10 = 0\).

    Check: yes, \(2^2 - 7 \cdot 2 + 10 = 4 - 14 + 10 = 0\). Factored, \(x^2 - 7x + 10 = (x - 2)(x - 5)\), so the profit is also zero at 5 QU.

    Typical mistake: dividing only the first term by \(-3\), e.g. \(x^2 + 21x - 30 = 0\). Every term is divided, so every sign changes.

  2. The profit limit (upper break-even point) is the larger output at which the profit is zero, and above it the profit is negative: \(P(9) = 0\).

    Check: \(P(9) = -162 + 225 - 63 = 0\) ✓. At 10 QU, \(P(10) = -200 + 250 - 63 = -13 < 0\). So no, beyond the profit limit the company makes a loss.

    Typical mistake: reading “profit limit” as the highest possible profit. It is an output (a zero of \(P\)), not an amount of money.

  3. \(R = C\): \(24x = x^2 + 5x + 48 \iff x^2 - 19x + 48 = 0\).

    Check: yes, \(R = 24 \cdot 3 = 72\) and \(C = 9 + 15 + 48 = 72\). Factored, \(x^2 - 19x + 48 = (x - 3)(x - 16)\), so the profit limit (upper break-even point) lies at 16 QU.

    Typical mistake: solving \(C = 0\) or \(R = 0\). At the break-even point, revenue and costs are equal, so their difference \(P = R - C\) is zero.

  4. No real solution: \(\Delta = 12^2 - 4 \cdot 2 \cdot k < 0 \iff 144 - 8k < 0 \iff k > 18\).

    Check: no. For \(k = 18\), \(\Delta = 144 - 144 = 0\), so the equation has exactly one solution, \(x = -\frac{12}{2 \cdot 2} = -3\).

    Typical mistake: \(\Delta = b^2 - 4c\) without \(a\), which gives \(k > 36\). The discriminant is \(b^2 - 4ac\), here with \(a = 2\).

  5. A profit needs \(x > 5 - \sqrt{6} \approx 2.5505\) QU, i.e. more than 255.05 bottles. Only whole bottles are sold, so round up: at least 256 bottles.

    Check: no. 255 bottles are \(x = 2.55\) QU, below the break-even point: \(P = -2.55^2 + 25.5 - 19 = -0.0025\), a small loss.

    Typical mistake: rounding 255.05 to the nearest whole number, 255. “At least” asks for the first whole number above the break-even point, so round up.

Problem 7: One Paragraph, All Conditions (xx)

A workshop builds wooden longboards. Its fixed costs, which arise even when no board is built, are 2,700 EUR per month. Because larger orders need overtime, the variable costs (the costs that grow with the output) for \(x\) QU are \((0.5x^2 + 6x)\) MU. The workshop sells every board for 150 EUR. It can build at most 160 boards per month, and a contract with a surf school requires at least 40 boards per month. The owner wants to know at which outputs the profit is exactly zero.

(1 QU = 10 boards, MU = monetary units, 1 MU = 100 EUR)

List every condition in this paragraph as a mathematical statement, with \(x\) the monthly output in QU. Write the zero-profit condition in zero form. Do not solve anything.

Check: How many outputs with zero profit are there? Decide with the discriminant of your zero form, without solving.

  1. Fixed costs: \(27\) MU
  2. Variable costs: \(0.5x^2 + 6x\) MU
  3. Total costs: \(C = 0.5x^2 + 6x + 27\) MU
  4. Price per QU: \(10 \cdot 150 = 1500\) EUR \(= 15\) MU, so the revenue is \(R = 15x\) MU
  5. Profit: \(P = R - C = -0.5x^2 + 9x - 27\) MU
  6. Zero profit: \(-0.5x^2 + 9x - 27 = 0 \iff x^2 - 18x + 54 = 0\) (multiply by \(-2\))
  7. Capacity (the maximum possible output): \(x \leq 16\)
  8. Contract: \(x \geq 4\)

Together: \(x \in [4; 16]\).

Check: \(\Delta = (-18)^2 - 4 \cdot 1 \cdot 54 = 324 - 216 = 108 > 0\), so there are two outputs with zero profit. 108 is not a perfect square, so they are irrational and need the quadratic formula.

Typical mistake: \(R = 150x\). The price is given per board, but \(x\) counts QU of 10 boards and the result must be in MU: 1 QU brings 1,500 EUR, i.e. 15 MU.

C: Applied

Problem 8: Break-Even of a Fresh Pasta Manufactory (xx)

A manufactory produces fresh pasta for restaurants. Its fixed costs are 3,000 EUR per week. Larger quantities need overtime and an extra shift, so the variable costs grow faster than the output: for \(x\) QU they amount to \((x^2 + 3x)\) MU. The manufactory sells its pasta for 16 EUR per kg. (1 QU = 100 kg, 1 MU = 100 EUR)

  1. Give terms for the total costs \(C\) and the revenue \(R\) in MU, and show that the weekly profit is \(P = -x^2 + 13x - 30\).

  2. Convert the break-even condition \(P = 0\) to zero form with a positive coefficient of \(x^2\). Decide with the discriminant whether the left side can be factored, and determine the break-even point and the profit limit in kg.

(for verification: 300 kg and 1,000 kg)

  1. At the moment, the manufactory produces and sells 800 kg per week. Compute its weekly profit in EUR.

  2. The owner wants a weekly profit of at least 1,100 EUR. Determine all outputs that reach this target, in whole kg, and decide whether the current output of 800 kg does.

Note: bring the condition to zero form first. An expression can only change its sign at a zero.

  1. Decide with the discriminant whether a weekly profit of 1,500 EUR is possible.

a) Fixed costs: 3,000 EUR \(= 30\) MU. Price per QU: \(100 \cdot 16 = 1600\) EUR \(= 16\) MU.

\[C = x^2 + 3x + 30, \qquad R = 16x\]

\[P = R - C = 16x - (x^2 + 3x + 30) = -x^2 + 13x - 30 \quad \checkmark\]

b) Multiply \(P = 0\) by \(-1\):

\[x^2 - 13x + 30 = 0\]

\(\Delta = (-13)^2 - 4 \cdot 1 \cdot 30 = 169 - 120 = 49 = 7^2\) is a perfect square, so the left side can be factored. Factors of 30 with sum \(-13\): \(-3\) and \(-10\).

\[(x - 3)(x - 10) = 0 \iff x = 3 \text{ or } x = 10\]

Break-even point: 3 QU = 300 kg per week. Profit limit (upper break-even point): 10 QU = 1,000 kg per week.

c) 800 kg = 8 QU:

\[P = -64 + 104 - 30 = 10 \text{ MU} = 1{,}000 \text{ EUR per week}\]

d) 1,100 EUR \(= 11\) MU. Zero form:

\[-x^2 + 13x - 30 \geq 11 \iff -x^2 + 13x - 41 \geq 0 \iff x^2 - 13x + 41 \leq 0\]

(Multiplying by \(-1\) flips the inequality sign.) \(\Delta = 169 - 164 = 5\) is not a perfect square, so use the quadratic formula:

\[x = \frac{13 \pm \sqrt{5}}{2}, \qquad x_1 \approx 5.382, \quad x_2 \approx 7.618\]

Test value between the zeros, \(x = 6\): \(36 - 78 + 41 = -1 \leq 0\) ✓. Outside, e.g. at \(x = 8\): \(64 - 104 + 41 = 1 > 0\). So the target is reached for \(x \in [5.382; 7.618]\) QU, i.e. from 538.2 kg to 761.8 kg.

In whole kg, round the lower bound up and the upper bound down: from 539 kg to 761 kg per week.

The current 800 kg lie above this range (the profit is only 1,000 EUR). To reach the target, the manufactory would have to produce less: the overtime costs of the last kilograms are higher than the revenue they bring.

e) 1,500 EUR \(= 15\) MU:

\[-x^2 + 13x - 30 \geq 15 \iff x^2 - 13x + 45 \leq 0\]

\(\Delta = 169 - 180 = -11 < 0\), so \(x^2 - 13x + 45\) has no zero. At \(x = 0\) it equals \(45 > 0\), and it can only change its sign at a zero, so it is positive for every output. A weekly profit of 1,500 EUR is not possible.

Problem 9: Saving for an X-Ray Unit (xx)

A dental practice saves for a new X-ray unit. It deposited 25,000 EUR in a savings account and another 10,000 EUR exactly one year later. The bank pays a fixed interest rate \(r\) per year, compounded yearly. Two years after the first deposit, the balance was 37,440 EUR.

  1. Explain why the growth factor \(q = 1 + r\) fulfils \(25000q^2 + 10000q = 37440\). Determine \(q\) with the quadratic formula, decide which of the two solutions makes sense and give the interest rate.

Hint: divide the zero form by 5,000 first.

(for verification: \(r = 4\,\%\))

  1. A second bank had offered 3.9 % per year with continuous compounding: there, an amount of \(K\) EUR grows to \(K \cdot e^{0.039t}\) EUR in \(t\) years. Compute the balance the two deposits would have reached there two years after the first deposit, in EUR with whole cents, and decide which offer was better. Explain the result with the growth factor per year.

Hint: Your calculator computes \(e^x\) with SHIFT + ln.

  1. The X-ray unit costs 42,000 EUR. The practice leaves the 37,440 EUR in the first account and makes no further deposits. Determine after how many more full years the balance covers the price.

a) The first deposit earns interest for two years, so it grows to \(25000 \cdot q \cdot q = 25000q^2\). The second deposit earns interest for one year and grows to \(10000q\). Together they make up the balance: \(25000q^2 + 10000q = 37440\).

Zero form, then divide by 5,000:

\[25000q^2 + 10000q - 37440 = 0 \iff 5q^2 + 2q - 7.488 = 0\]

\[\Delta = 2^2 - 4 \cdot 5 \cdot (-7.488) = 4 + 149.76 = 153.76 = 12.4^2\]

\[q = \frac{-2 \pm 12.4}{10}, \qquad q_1 = 1.04, \quad q_2 = -1.44\]

A growth factor is positive, since interest cannot turn a positive balance negative. So \(q = 1.04\) and the interest rate is \(r = 4\,\%\).

Check: \(25000 \cdot 1.04^2 + 10000 \cdot 1.04 = 27040 + 10400 = 37440\) ✓

b) The first deposit grows for 2 years, the second for 1 year:

\[25000 \cdot e^{0.039 \cdot 2} + 10000 \cdot e^{0.039} \approx 27028.07 + 10397.70 = 37425.77 \text{ EUR}\]

The first bank was better, by about 14.23 EUR. Reason: at the second bank the growth factor per year is \(e^{0.039} \approx 1.0398\), i.e. about 3.98 % per year, just below the 4 % of the first bank.

c) After \(t\) more years the balance is \(37440 \cdot 1.04^t\) EUR. Set it equal to the price:

\[37440 \cdot 1.04^t = 42000 \iff 1.04^t = \frac{42000}{37440} \iff t = \frac{\ln(42000/37440)}{\ln 1.04} \approx 2.93\]

The balance grows every year, so from then on it covers the price. Only full years count, so round up: after 3 more years, i.e. five years after the first deposit.

Check: \(37440 \cdot 1.04^2 \approx 40{,}495.10\) EUR is not enough, \(37440 \cdot 1.04^3 \approx 42{,}114.91\) EUR is.

Problem 10: Sales Team of a Heat-Pump Installer (xx)

A heat-pump installer builds a sales team for a new region. A single sales adviser would sign 26 contracts per month. Every further adviser lowers the average number of contracts per adviser by 2, because the advisers compete for the same customers. Let \(x\) be the number of advisers and \(y\) the average number of contracts per adviser and month.

  1. Express \(y\) in terms of \(x\).

  2. To keep its installation teams busy, the company needs exactly 90 contracts per month. Set up the condition for the total number of contracts, substitute \(y\) and convert the equation to zero form.

(for verification: \(x^2 - 14x + 45 = 0\))

  1. Solve the equation and give both possible team sizes with the contracts per adviser. Every adviser earns the same salary. Decide which team the company should hire.

  2. The company would like more contracts. Decide with the discriminant how many team sizes give exactly 98 contracts per month and how many give exactly 100. Give the team in the case with exactly one solution.

a) Start at 26 contracts for one adviser and subtract 2 for each of the \(x - 1\) further advisers:

\[y = 26 - 2(x - 1) = 28 - 2x\]

Check: \(x = 1\) gives \(y = 26\) ✓

b) Total contracts: \(x \cdot y = 90\). Substitute \(y\):

\[x(28 - 2x) = 90 \iff 28x - 2x^2 - 90 = 0 \;\big|:(-2) \iff x^2 - 14x + 45 = 0\]

c) \(\Delta = 196 - 180 = 16 = 4^2\), so factor. Factors of 45 with sum \(-14\): \(-5\) and \(-9\).

\[(x - 5)(x - 9) = 0 \iff x = 5 \text{ or } x = 9\]

  • \(x = 5\): \(y = 28 - 10 = 18\) contracts per adviser
  • \(x = 9\): \(y = 28 - 18 = 10\) contracts per adviser

Check: \(5 \cdot 18 = 9 \cdot 10 = 90\) ✓

Both teams sign 90 contracts, but 5 advisers cost fewer salaries than 9. The company should hire 5 advisers.

d) For a target of \(T\) contracts: \(x(28 - 2x) = T \iff x^2 - 14x + \frac{T}{2} = 0\).

  • \(T = 98\): \(x^2 - 14x + 49 = 0\), \(\Delta = 196 - 196 = 0\). Exactly one team size: \(x = \frac{14}{2} = 7\) advisers with \(y = 28 - 14 = 14\) contracts each.
  • \(T = 100\): \(x^2 - 14x + 50 = 0\), \(\Delta = 196 - 200 = -4 < 0\). No team size signs exactly 100 contracts per month.

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: Pizza Oven Manufacturer [20 pts. total]

A young company builds portable pizza ovens for gardens and balconies. Its fixed costs, which arise even when no oven is built, amount to 56,000 EUR per month. Material and assembly cost 160 EUR per oven. Market research shows that the company can sell \(x\) QU per month if it asks the price \(p(x) = 400 - 20x\) EUR per oven.

(QU = quantity units, MU = monetary units. 1 QU = 100 ovens, 1 MU = 1,000 EUR. Total costs = fixed costs + variable costs, where the variable costs grow with the output. Revenue = quantity × price. Profit = revenue − total costs. The price-demand function \(p(x)\) gives the price at which \(x\) QU can be sold.)

Part a)

Give terms for the total costs \(C(x)\) and the revenue \(R(x)\) in MU, with \(x\) in QU. Show that the profit function is given by \(P(x) = -2x^2 + 24x - 56\). [3 pts.]

Convert to MU. Fixed costs: 56,000 EUR \(= 56\) MU. Variable costs per QU: \(100 \cdot 160 = 16{,}000\) EUR \(= 16\) MU.

\[C(x) = 16x + 56\]

Revenue. \(x\) QU are \(100x\) ovens, each sold at \(p(x)\) EUR:

\[100x \cdot (400 - 20x) = 40000x - 2000x^2 \text{ EUR} \implies R(x) = 40x - 2x^2\]

Profit:

\[P(x) = R(x) - C(x) = 40x - 2x^2 - (16x + 56) = -2x^2 + 24x - 56 \quad \checkmark\]

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

Part b)

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: \(x_1 = 6 - 2\sqrt{2} \approx 3.17\))

Zero form. At a break-even point the profit is zero, \(P(x) = 0\). Divide by \(-2\):

\[-2x^2 + 24x - 56 = 0 \;\big|:(-2) \iff x^2 - 12x + 28 = 0\]

Discriminant: \(\Delta = (-12)^2 - 4 \cdot 1 \cdot 28 = 144 - 112 = 32 > 0\). So there are two zeros. 32 is not a perfect square, so factoring over the integers fails and the quadratic formula is needed:

\[x_{1,2} = \frac{12 \pm \sqrt{32}}{2} = \frac{12 \pm 4\sqrt{2}}{2} = 6 \pm 2\sqrt{2}\]

\[x_1 = 6 - 2\sqrt{2} \approx 3.17, \qquad x_2 = 6 + 2\sqrt{2} \approx 8.83\]

Check \(x_1\) with the perfect square formula \((a - b)^2 = a^2 - 2ab + b^2\): \((6 - 2\sqrt{2})^2 - 12(6 - 2\sqrt{2}) + 28 = 44 - 24\sqrt{2} - 72 + 24\sqrt{2} + 28 = 0\) ✓

Break-even point: \(x_1 = 6 - 2\sqrt{2} \approx 3.17\) QU. Profit limit: \(x_2 = 6 + 2\sqrt{2} \approx 8.83\) QU.

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

Part c)

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

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

Test values below, between and above the zeros:

  • \(x = 1\): \(P(1) = -2 + 24 - 56 = -34 < 0\), a loss
  • \(x = 5\): \(P(5) = -50 + 120 - 56 = 14 > 0\), a profit
  • \(x = 10\): \(P(10) = -200 + 240 - 56 = -16 < 0\), a loss

Profit zone: \((6 - 2\sqrt{2}; 6 + 2\sqrt{2}) \approx (3.17; 8.83)\). The interval is open, since the profit is zero at both ends.

Whole ovens. \(x_1 \approx 3.1716\) QU \(= 317.16\) ovens and \(x_2 \approx 8.8284\) QU \(= 882.84\) ovens. Round the lower bound up and the upper bound down: the company makes a profit with 318 to 882 ovens per month.

The rounded values from b) are not enough here: 317 ovens (\(x = 3.17\)) give \(P = -0.0178\) MU, a loss of 17.80 EUR, and 883 ovens (\(x = 8.83\)) give the same loss.

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

Part d)

Compute the number of ovens sold per month and the monthly profit in EUR if the company asks 340 EUR per oven, and if it asks 300 EUR per oven. Explain the difference with the result of part c). [4 pts.]

Price 340 EUR: \(400 - 20x = 340 \iff 20x = 60 \iff x = 3\) QU, i.e. 300 ovens.

\[P(3) = -18 + 72 - 56 = -2 \text{ MU} = -2{,}000 \text{ EUR}\]

The company makes a loss of 2,000 EUR per month.

Price 300 EUR: \(400 - 20x = 300 \iff 20x = 100 \iff x = 5\) QU, i.e. 500 ovens.

\[P(5) = -50 + 120 - 56 = 14 \text{ MU} = 14{,}000 \text{ EUR}\]

Check in EUR: revenue \(500 \cdot 300 = 150{,}000\) EUR, total costs \(56{,}000 + 500 \cdot 160 = 136{,}000\) EUR, profit 14,000 EUR ✓

Explanation: at 340 EUR the company sells only 300 ovens, fewer than the 318 ovens it needs for a profit (part c). At 300 EUR it sells 500 ovens, which lie inside the profit zone.

Grading: 1 Pt. both quantities, 1 Pt. profit at 340 EUR in EUR, 1 Pt. profit at 300 EUR in EUR, 1 Pt. explanation with the profit zone.

Part e)

The company decides on the price of 300 EUR from part d). Of its buyers, 36 % also order a weather cover, 130 order a pizza peel, and 40 order both. Let \(A\) be the set of buyers who order a cover and \(B\) the set of buyers who order a peel. The company earns an extra profit of 25 EUR on each cover and 12 EUR on each peel. Determine the number of buyers who order neither accessory. State the businesslike meaning of the set \(A \cap \overline{B}\) and give its size. Compute the monthly profit in EUR including the accessories, and the percentage by which the accessories raise the profit from d), rounded to one decimal place. [5 pts.]

Buyers: 500 per month (part d). \(A\) contains \(0.36 \cdot 500 = 180\) buyers, \(B\) contains 130 and \(A \cap B\) contains 40.

At least one accessory (\(A \cup B\)): split the Venn diagram into its regions. Only a cover: \(180 - 40 = 140\), only a peel: \(130 - 40 = 90\), both: 40. Together \(140 + 90 + 40 = 270\) buyers.

Neither accessory (\(\overline{A \cup B}\)): \(500 - 270 = 230\) buyers.

Meaning: \(A \cap \overline{B}\) is the set of buyers who order a cover but no peel. Size: \(180 - 40 = 140\) buyers.

Profit with accessories:

\[180 \cdot 25 + 130 \cdot 12 = 4500 + 1560 = 6{,}060 \text{ EUR}\]

\[14{,}000 + 6{,}060 = 20{,}060 \text{ EUR per month}\]

Increase, with the profit from d) as the base:

\[\frac{6060}{14000} \approx 0.4329 = 43.3\,\%\]

Grading: 1 Pt. \(A\) with 180 buyers, 1 Pt. \(A \cup B\) and 230 buyers with neither accessory, 1 Pt. meaning of \(A \cap \overline{B}\) and 140 buyers, 1 Pt. extra and total profit in EUR, 1 Pt. percentage with the profit from d) as the base.

Calculator Practice

Problem 11: Calculator: Quadratic Equations (x)

Use the polynomial equation solver to find the solutions of:

  1. \(x^2 - 13x + 40 = 0\)

  2. \(2x^2 + 7x - 4 = 0\)

  3. \(x^2 + 12x + 36 = 0\)

  4. \(x^2 + 4 = 0\) (What does the calculator show, and what does it mean?)

  1. \(x^2 - 13x + 40 = 0\)
    • Coefficients: a = 1, b = -13, c = 40
    • Solutions: \(x_1 = 8\), \(x_2 = 5\)
  2. \(2x^2 + 7x - 4 = 0\)
    • Coefficients: a = 2, b = 7, c = -4
    • Solutions: \(x_1 = 0.5\), \(x_2 = -4\)
  3. \(x^2 + 12x + 36 = 0\)
    • Coefficients: a = 1, b = 12, c = 36
    • Solution: \(x = -6\) (double root)
  4. \(x^2 + 4 = 0\)
    • Coefficients: a = 1, b = 0, c = 4
    • The calculator finds no real solutions \(x_1\), \(x_2\)
    • Meaning: the equation has no real solution, since \(\Delta = 0^2 - 4 \cdot 1 \cdot 4 = -16 < 0\)

Problem 12: Calculator: Break-Even Analysis (xx)

A company’s profit function is \(P(x) = -2x^2 + 44x - 210\) where \(x\) is in hundreds of units and \(P(x)\) in 100 EUR.

  1. Find the break-even points using the polynomial solver.

  2. Give the profit zone as an interval. Check it with the CALC key: evaluate \(P(x)\) at one output inside the zone and at one output outside it.

  3. Interpret the business meaning of both results.

  1. Break-even points where \(P(x) = 0\):
    • Coefficients: a = -2, b = 44, c = -210
    • Solutions: \(x = 7\) and \(x = 15\)
    • Break-even at 700 units and 1500 units
    • After the two solutions, the calculator also displays the vertex of the parabola. It is not needed here.
  2. Profit zone:
    • Profit zone: \((7; 15)\), open because the profit is zero at both ends
    • Enter \(-2x^2 + 44x - 210\), press CALC, enter the value and press =
    • Inside: \(P(10) = -200 + 440 - 210 = 30 > 0\) ✓
    • Outside: \(P(6) = -72 + 264 - 210 = -18 < 0\) ✓
  3. Business interpretation:
    • Break-even point at 700 units, profit limit at 1500 units
    • Between 700 and 1500 units the company makes a profit
    • Below 700 units and above 1500 units it makes a loss