Tasks 01-04 - Advanced Factorization & Radicals

Mastering Complex Algebraic Techniques

A: Skills

Problem 1: Advanced Factorization of Quadratics (x)

Factor the following quadratic expressions completely:

  1. \(x^2 + 9x + 20\)

  2. \(x^2 - 3x - 18\)

  3. \(2x^2 + 7x + 3\)

  4. \(3x^2 - 11x + 6\)

  5. \(6x^2 + 13x - 5\)

  6. \(4x^2 - 12x + 9\)

  1. \(x^2 + 9x + 20\)
    • Need factors of 20 that sum to 9: (4,5)
    • \((x + 4)(x + 5)\)
  2. \(x^2 - 3x - 18\)
    • Need factors of -18 that sum to -3: (3,-6)
    • \((x + 3)(x - 6)\)
  3. \(2x^2 + 7x + 3\)
    • AC = 6, need factors that sum to 7: (1,6)
    • \(2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1)\)
    • \((x + 3)(2x + 1)\)
  4. \(3x^2 - 11x + 6\)
    • AC = 18, need factors that sum to -11: (-2,-9)
    • \(3x^2 - 2x - 9x + 6 = x(3x - 2) - 3(3x - 2)\)
    • \((x - 3)(3x - 2)\)
  5. \(6x^2 + 13x - 5\)
    • AC = -30, need factors that sum to 13: (15,-2)
    • \(6x^2 + 15x - 2x - 5 = 3x(2x + 5) - 1(2x + 5)\)
    • \((3x - 1)(2x + 5)\)
  6. \(4x^2 - 12x + 9\)
    • Perfect square: \((2x - 3)^2\)

Problem 2: Factoring by Grouping & Cubes (xx)

Factor the following expressions completely:

  1. \(x^3 + 3x^2 - 4x - 12\)

  2. \(2x^3 - 5x^2 - 8x + 20\)

  3. \(x^3 - 125\)

  4. \(8x^3 + 27\)

  5. \(64x^3 - 1\)

  6. \(x^3 + 4x^2 - 9x - 36\)

  1. \(x^3 + 3x^2 - 4x - 12\)
    • Group: \((x^3 + 3x^2) + (-4x - 12)\)
    • \(= x^2(x + 3) - 4(x + 3)\)
    • \(= (x + 3)(x^2 - 4)\)
    • \(= (x + 3)(x + 2)(x - 2)\)
  2. \(2x^3 - 5x^2 - 8x + 20\)
    • Group: \((2x^3 - 5x^2) + (-8x + 20)\)
    • \(= x^2(2x - 5) - 4(2x - 5)\)
    • \(= (2x - 5)(x^2 - 4)\)
    • \(= (2x - 5)(x + 2)(x - 2)\)
  3. \(x^3 - 125\)
    • Difference of cubes: \(x^3 - 5^3\)
    • \(= (x - 5)(x^2 + 5x + 25)\)
  4. \(8x^3 + 27\)
    • Sum of cubes: \((2x)^3 + 3^3\)
    • \(= (2x + 3)(4x^2 - 6x + 9)\)
  5. \(64x^3 - 1\)
    • Difference of cubes: \((4x)^3 - 1^3\)
    • \(= (4x - 1)(16x^2 + 4x + 1)\)
  6. \(x^3 + 4x^2 - 9x - 36\)
    • Group: \((x^3 + 4x^2) + (-9x - 36)\)
    • \(= x^2(x + 4) - 9(x + 4)\)
    • \(= (x + 4)(x^2 - 9)\)
    • \(= (x + 4)(x + 3)(x - 3)\)

Problem 3: Simplifying Radicals (x)

Simplify the following radical expressions:

  1. \(\sqrt{72}\)

  2. \(\sqrt{108x^5y^3}\)

  3. \(\sqrt[3]{54a^7b^4}\)

  4. \(3\sqrt{50} + 2\sqrt{32} - \sqrt{200}\)

  5. \(\sqrt{45x^3} - x\sqrt{20x} + 2\sqrt{80x^3}\)

  6. \(\frac{\sqrt{48x^5}}{\sqrt{3x}}\)

  1. \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\)

  2. \(\sqrt{108x^5y^3}\)

    • \(= \sqrt{36 \times 3 \times x^4 \times x \times y^2 \times y}\)
    • \(= 6x^2y\sqrt{3xy}\)

  3. \(\sqrt[3]{54a^7b^4}\)

    • \(= \sqrt[3]{27 \times 2 \times a^6 \times a \times b^3 \times b}\)
    • \(= 3a^2b\sqrt[3]{2ab}\)

  4. \(3\sqrt{50} + 2\sqrt{32} - \sqrt{200}\)

    • \(= 3 \times 5\sqrt{2} + 2 \times 4\sqrt{2} - 10\sqrt{2}\)
    • \(= 15\sqrt{2} + 8\sqrt{2} - 10\sqrt{2}\)
    • \(= 13\sqrt{2}\)

  5. \(\sqrt{45x^3} - x\sqrt{20x} + 2\sqrt{80x^3}\)

    • \(= 3x\sqrt{5x} - x \times 2\sqrt{5x} + 2 \times 4x\sqrt{5x}\)
    • \(= 3x\sqrt{5x} - 2x\sqrt{5x} + 8x\sqrt{5x}\)
    • \(= 9x\sqrt{5x}\)

  6. \(\frac{\sqrt{48x^5}}{\sqrt{3x}} = \sqrt{\frac{48x^5}{3x}} = \sqrt{16x^4} = 4x^2\)

Problem 4: Rationalizing Denominators (xx)

Rationalize the denominators of the following expressions and simplify the results:

  1. \(\frac{5}{\sqrt{3}}\)

  2. \(\frac{6}{\sqrt[3]{9}}\)

  3. \(\frac{3}{\sqrt{7} - 2}\)

  4. \(\frac{4}{3 + \sqrt{5}}\)

  5. \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)

  6. \(\frac{2\sqrt{3}}{2\sqrt{3} - 3\sqrt{2}}\)

  1. \(\frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}\)

  2. \(\frac{6}{\sqrt[3]{9}} = \frac{6}{\sqrt[3]{9}} \times \frac{\sqrt[3]{3}}{\sqrt[3]{3}} = \frac{6\sqrt[3]{3}}{3} = 2\sqrt[3]{3}\)

  3. \(\frac{3}{\sqrt{7} - 2}\)

    • Multiply by \(\frac{\sqrt{7} + 2}{\sqrt{7} + 2}\)
    • \(= \frac{3(\sqrt{7} + 2)}{7 - 4} = \frac{3\sqrt{7} + 6}{3} = \sqrt{7} + 2\)

  4. \(\frac{4}{3 + \sqrt{5}}\)

    • Multiply by \(\frac{3 - \sqrt{5}}{3 - \sqrt{5}}\)
    • \(= \frac{4(3 - \sqrt{5})}{9 - 5} = \frac{12 - 4\sqrt{5}}{4} = 3 - \sqrt{5}\)

  5. \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)

    • Multiply by \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}\)
    • Numerator: \((\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} + 2 = 8 - 4\sqrt{3}\)
    • Denominator: \(6 - 2 = 4\)
    • Result: \(\frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}\)

  6. \(\frac{2\sqrt{3}}{2\sqrt{3} - 3\sqrt{2}}\)

    • Multiply by \(\frac{2\sqrt{3} + 3\sqrt{2}}{2\sqrt{3} + 3\sqrt{2}}\)
    • Numerator: \(2\sqrt{3}(2\sqrt{3} + 3\sqrt{2}) = 12 + 6\sqrt{6}\)
    • Denominator: \((2\sqrt{3})^2 - (3\sqrt{2})^2 = 12 - 18 = -6\)
    • Result: \(\frac{12 + 6\sqrt{6}}{-6} = -2 - \sqrt{6}\)

Problem 5: Discriminant and Combined Techniques (xx)

Compute without a calculator:

  1. \(\sqrt[3]{-125} + \sqrt[4]{16}\)

Decide with the discriminant whether each expression can be factored over the integers. If it can, factor it.

  1. \(3x^2 + 5x - 2\)

  2. \(2x^2 + 7x + 4\)

  3. \(5x^2 - 13x + 6\)

Simplify completely:

  1. \(\frac{x^2 - 9}{3x^2 + 8x - 3}\)

  2. \(\frac{x^3 - 27}{x^2 - 6x + 9}\)

  3. \(\frac{\frac{x^2}{4}}{\frac{x}{6}}\)

  1. \(\sqrt[3]{-125} + \sqrt[4]{16}\)
    • \(\sqrt[3]{-125} = -5\), because \((-5)^3 = -125\) (odd roots keep the sign)
    • \(\sqrt[4]{16} = 2\), because \(2^4 = 16\)
    • \(= -5 + 2 = -3\)
  2. \(3x^2 + 5x - 2\)
    • \(\Delta = 5^2 - 4 \cdot 3 \cdot (-2) = 25 + 24 = 49 = 7^2\), so it factors
    • \(ac = -6\), factors with sum 5: \((6, -1)\)
    • \(3x^2 + 6x - x - 2 = 3x(x + 2) - 1(x + 2)\)
    • \(= (3x - 1)(x + 2)\)
  3. \(2x^2 + 7x + 4\)
    • \(\Delta = 7^2 - 4 \cdot 2 \cdot 4 = 49 - 32 = 17\)
    • 17 is not a perfect square, so the expression cannot be factored over the integers.
  4. \(5x^2 - 13x + 6\)
    • \(\Delta = (-13)^2 - 4 \cdot 5 \cdot 6 = 169 - 120 = 49 = 7^2\), so it factors
    • \(ac = 30\), factors with sum \(-13\): \((-3, -10)\)
    • \(5x^2 - 3x - 10x + 6 = x(5x - 3) - 2(5x - 3)\)
    • \(= (x - 2)(5x - 3)\)
  5. \(\frac{x^2 - 9}{3x^2 + 8x - 3}\)
    • Numerator: \(x^2 - 9 = (x + 3)(x - 3)\)
    • Denominator (AC method, \(ac = -9\), factors \((9, -1)\)): \(3x^2 + 9x - x - 3 = 3x(x + 3) - 1(x + 3) = (3x - 1)(x + 3)\)
    • \(= \frac{(x + 3)(x - 3)}{(3x - 1)(x + 3)} = \frac{x - 3}{3x - 1}\)
  6. \(\frac{x^3 - 27}{x^2 - 6x + 9}\)
    • Numerator (difference of cubes): \(x^3 - 3^3 = (x - 3)(x^2 + 3x + 9)\)
    • Denominator (perfect square): \((x - 3)^2\)
    • \(= \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)^2} = \frac{x^2 + 3x + 9}{x - 3}\)
  7. \(\frac{\frac{x^2}{4}}{\frac{x}{6}}\)
    • Multiply by the reciprocal: \(\frac{x^2}{4} \cdot \frac{6}{x}\)
    • \(= \frac{6x^2}{4x} = \frac{3x}{2}\)

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 \((2x - 7)(12 - x)\) (in 100 EUR) of a bike repair shop at an output of \(x\) QU is zero.”

    Check: Is \(x = 7\) such an output?

  2. “A square display wall has an area of 50 m².”

    Check: Is the side length about 7.07 m?

  3. “A snack kiosk with the profit \(-x^2 + 17x - 42\) (in 100 EUR) reaches its break-even point at 3 QU.” The break-even point is the output at which the profit changes from negative (a loss) to positive.

    Check: Compute the profit at 2 QU and at 4 QU. Do the values confirm the statement?

  4. “For the profit \(-x^2 + 11x - 18\) (in 100 EUR), the profit limit lies at 9 QU.” The profit limit is the larger output with zero profit. Above it, the profit turns negative again.

    Check: Is the profit at 10 QU negative?

  5. “The volume of the cube-shaped container B is 8 times as large as the volume of the cube-shaped container A.”

    Check: Container A has a side length of 0.5 m. Is the side length of B 4 m?

  1. \((2x - 7)(12 - x) = 0 \iff 2x - 7 = 0\) or \(12 - x = 0\) (a product is zero if one of its factors is zero), so \(x = 3.5\) or \(x = 12\).

    Check: no, \(x = 7\) gives \((14 - 7)(12 - 7) = 7 \cdot 5 = 35 \neq 0\).

    Typical mistake: reading \(x = 7\) off the factor \(2x - 7\). Set the factor equal to zero and solve: \(2x = 7\), so \(x = 3.5\).

  2. With \(s\) the side length in m: \(s^2 = 50\) and \(s > 0\), so \(s = \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\) m.

    Check: yes, \(5\sqrt{2} \approx 7.07\), and \(7.07^2 \approx 49.98 \approx 50\).

    Typical mistake: \(s = 50 \div 4 = 12.5\) m. Dividing by 4 turns a perimeter into a side. For an area, you need the square root.

  3. The profit is zero at \(x = 3\), negative just below 3 QU and positive just above 3 QU. At \(x = 3\): \(-9 + 51 - 42 = 0\).

    Check: at 2 QU: \(-4 + 34 - 42 = -12 < 0\). At 4 QU: \(-16 + 68 - 42 = 10 > 0\). Yes, the profit changes from negative to positive at 3 QU.

    Typical mistake: calling 14 QU the break-even point. There the profit turns negative, so it is the profit limit (upper break-even point).

  4. The profit is zero at \(x = 9\), and negative for outputs above 9 QU. At \(x = 9\): \(-81 + 99 - 18 = 0\).

    Check: yes, at 10 QU: \(-100 + 110 - 18 = -8 < 0\).

    Typical mistake: “the profit limit is the highest possible profit”. It is the output at which the profit ends, not the output with the largest profit.

  5. With \(a\) and \(b\) the side lengths of A and B: \(b^3 = 8a^3 \Rightarrow b = \sqrt[3]{8a^3} = 2a\).

    Check: no. For \(a = 0.5\) m: \(b = 1\) m, and indeed \(1^3 = 1 = 8 \cdot 0.5^3 = 8 \cdot 0.125\) (in m³). A side of 4 m would give \(64\) m³, i.e. 512 times the volume of A.

    Typical mistake: \(b = 8a\). The volume grows with the cube of the side, so 8 times the volume needs only \(\sqrt[3]{8} = 2\) times the side.

Problem 7: One Paragraph, All Conditions (xx)

A candle manufacturer has fixed costs of 1,800 EUR per week. These costs arise even if no candle is produced. Producing 100 candles (1 QU) causes variable costs (they grow with the output) of another 140 EUR. The manufacturer sells the candles at 440 EUR per QU. Its moulding machine makes at most 1,500 candles per week. This is the capacity (the maximum possible output).

List every condition in this paragraph and the terms for the total costs, the revenue and the profit, with \(x\) the weekly output in QU. Do not solve anything.

Check: What is the profit if no candle is produced, and which amount from the paragraph does it equal?

  1. Fixed costs: \(1800\) EUR
  2. Variable costs: \(140x\) EUR
  3. Total costs: \(C = 140x + 1800\) EUR
  4. Revenue: \(R = 440x\) EUR
  5. Profit: \(P = R - C = 440x - (140x + 1800) = 300x - 1800\) EUR
  6. Capacity: 1,500 candles \(= 15\) QU, so \(x \leq 15\)
  7. No negative output: \(x \geq 0\)

Together: \(x \in [0; 15]\).

Check: at \(x = 0\) the profit is \(-1800\) EUR. Without output there is no revenue, so the loss equals the fixed costs.

Typical mistake: \(x \leq 1500\). Here \(x\) counts QU, and 1,500 candles are 15 QU.

C: Applied

Problem 8: Profit of a Yoga Mat Producer (xxx)

A producer of yoga mats models its monthly profit (in 100 EUR) at an output of \(x\) QU (1 QU = 100 mats), with \(0 \leq x \leq 9\), by the expression \(-x^3 + 2x^2 + 49x - 98\).

  1. Compute the profit for \(x = 0\) and interpret the result.

  2. Factor the profit completely by grouping. Hint: group the first two and the last two terms and factor \(-x^2\) out of the first group.

(for verification: \((x - 2)(7 - x)(7 + x)\))

  1. Determine all outputs at which the profit is zero. Give the break-even point (the output at which the profit changes from negative to positive) and the profit limit (the output at which it turns negative again).

Note: a product is zero if one of its factors is zero.

  1. Compute the profit for \(x = 1\), \(x = 4\) and \(x = 8\). Determine the profit zone (all outputs with a positive profit) as an interval.

Note: the profit can only change its sign at an output where it is zero. So one test value between two neighbouring zeros shows the sign on that whole stretch.

a) \(-0^3 + 2 \cdot 0^2 + 49 \cdot 0 - 98 = -98\), i.e. a loss of 9,800 EUR per month.

Without output there is no revenue, so this loss equals the fixed costs, which arise even when nothing is produced.

b) Group the first two and the last two terms:

\[-x^3 + 2x^2 + 49x - 98 = (-x^3 + 2x^2) + (49x - 98) = -x^2(x - 2) + 49(x - 2)\]

Both groups contain the common factor \((x - 2)\):

\[= (x - 2)(49 - x^2)\]

\(49 - x^2 = 7^2 - x^2\) is a difference of squares:

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

c) A product is zero if one of its factors is zero:

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

\(x = -7\) is rejected, since the output cannot be negative.

Break-even point: \(x = 2\) QU (200 mats). Profit limit: \(x = 7\) QU (700 mats).

d) Use the factored form:

  • \(x = 1\): \((1 - 2)(7 - 1)(7 + 1) = (-1) \cdot 6 \cdot 8 = -48\), a loss
  • \(x = 4\): \((4 - 2)(7 - 4)(7 + 4) = 2 \cdot 3 \cdot 11 = 66\), a profit
  • \(x = 8\): \((8 - 2)(7 - 8)(7 + 8) = 6 \cdot (-1) \cdot 15 = -90\), a loss

The sign can only change at the zeros 2 and 7, so the producer makes a profit exactly between them.

Profit zone: \((2; 7)\). The interval is open, because at 2 QU and 7 QU the profit is exactly zero.

Problem 9: Gift Boxes of a Chocolate Maker (xx)

A chocolate maker packs pralines in boxes with square lids. All lengths are in cm.

  1. The lid of the large box has an area of 150 cm². Determine its side length exactly, with the radical simplified, and rounded to 0.1 cm.

Hint: Without a calculator, use \(\sqrt{6} \approx 2.449\) and \(\sqrt{5} \approx 2.236\).

(for verification: \(5\sqrt{6}\) cm)

  1. The lid of the small box has an area of 96 cm². A gold ribbon runs once around all four sides of each lid. Compute the ribbon length for one large and one small box, exactly and rounded to 0.1 cm.

  2. A sample box has a lid with an area of 28 cm². For a gift edition, its side length is to grow to 14 cm. Determine the factor by which the side length grows, exactly and without a radical in the denominator. Explain by which factor the lid area grows.

  3. The label on the large lid is a rectangle 6 cm wide. Its length is \(\frac{2}{\sqrt{5} - 1}\) times its width, a ratio many designers like. Rationalize the denominator of this ratio, compute the length of the label exactly and rounded to 0.1 cm, and decide whether the label fits on the large lid.

a) With \(s\) the side length: \(s^2 = 150\), so

\[s = \sqrt{150} = \sqrt{25 \cdot 6} = 5\sqrt{6} \approx 12.2 \text{ cm}\]

b) Small lid: \(\sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6} \approx 9.8\) cm.

Ribbon: \(4 \cdot 5\sqrt{6} + 4 \cdot 4\sqrt{6} = 20\sqrt{6} + 16\sqrt{6} = 36\sqrt{6} \approx 88.2\) cm.

Both terms contain \(\sqrt{6}\), so they can be combined.

c) Old side length: \(\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}\) cm.

\[\frac{14}{2\sqrt{7}} = \frac{7}{\sqrt{7}} = \frac{7}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{7\sqrt{7}}{7} = \sqrt{7} \approx 2.65\]

The side length grows by the factor \(\sqrt{7}\). The area is the side length squared, so it grows by the factor \((\sqrt{7})^2 = 7\). Check: \(14^2 = 196 = 7 \cdot 28\).

d) Multiply by the conjugate \(\sqrt{5} + 1\):

\[\frac{2}{\sqrt{5} - 1} \cdot \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{2(\sqrt{5} + 1)}{5 - 1} = \frac{\sqrt{5} + 1}{2} \approx 1.618\]

Length: \(6 \cdot \frac{\sqrt{5} + 1}{2} = 3\sqrt{5} + 3 \approx 9.7\) cm.

The label (\(6 \times 9.7\) cm) fits on the large lid, whose side is about 12.2 cm.

Problem 10: Shipping Crates of a Fish Wholesaler (xx)

A fish wholesaler ships fresh fish in cube-shaped insulated crates. A crate has an outer side length of \(x\) dm (1 dm = 10 cm). Its walls, bottom and lid are 1 dm thick, so the inner space is a cube with side length \(x - 2\) dm. The insulation fills the space between the outer and the inner cube.

  1. Show with the difference of cubes that the insulation volume (in dm³) is \(2(3x^2 - 6x + 4)\). Hint: use \(a = x\) and \(b = x - 2\), and treat \((x - 2)\) as one term.

  2. Compute the insulation volume of a crate with \(x = 10\) and determine its share of the outer volume in percent.

(for verification: 488 dm³)

  1. The wholesaler buys \(2.5 \cdot 10^4\) crates with \(x = 10\) per year. Compute the total volume of insulation material in m³ and give it in scientific notation (1 m³ = 1,000 dm³).

  2. A customer needs crates with an inner volume of exactly 729 dm³. Determine the outer side length and verify the insulation volume from a) for this crate by subtracting the two volumes.

a) The insulation volume is outer volume minus inner volume: \(x^3 - (x - 2)^3\). With \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), \(a = x\) and \(b = x - 2\):

  • \(a - b = x - (x - 2) = 2\)
  • \(a^2 + ab + b^2 = x^2 + x(x - 2) + (x - 2)^2 = x^2 + x^2 - 2x + x^2 - 4x + 4 = 3x^2 - 6x + 4\)

\[x^3 - (x - 2)^3 = 2(3x^2 - 6x + 4) \quad \checkmark\]

b) \(2(3 \cdot 100 - 60 + 4) = 2 \cdot 244 = 488\) dm³ (check: \(10^3 - 8^3 = 1000 - 512 = 488\)).

Share: \(\frac{488}{1000} = 0.488 = 48.8\,\%\) of the outer volume is insulation.

c)

\[488 \cdot 2.5 \cdot 10^4 = 1220 \cdot 10^4 = 1.22 \cdot 10^7 \text{ dm}^3\]

Divide by \(10^3\) to convert to m³:

\[1.22 \cdot 10^7 \div 10^3 = 1.22 \cdot 10^4 \text{ m}^3\]

d) Inner side length: \(\sqrt[3]{729} = 9\) dm, because \(9^3 = 729\). So \(x - 2 = 9\), i.e. \(x = 11\) dm.

Formula: \(2(3 \cdot 121 - 66 + 4) = 2 \cdot 301 = 602\) dm³. Subtraction: \(11^3 - 9^3 = 1331 - 729 = 602\) dm³ \(\checkmark\)

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: Ceramic Workshop [20 pts. total]

A ceramic workshop makes handmade mugs. Its fixed costs, which arise even when no mug is made, amount to 2,400 EUR per month. Each mug causes variable costs of 6 EUR for clay, glaze and energy. The more mugs the workshop offers, the lower the price it can charge: to sell \(x\) QU per month, it can charge \((25 - 2x)\) EUR per mug. The kiln can fire at most 1,000 mugs per month. The workshop plans to make and sell 700 mugs per month.

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

Part a)

Give terms for the total costs \(C\) and the revenue \(R\) per month in MU if the workshop makes and sells \(x\) QU. [4 pts.]

(for verification: \(C = 6x + 24\), \(\; R = 25x - 2x^2\))

Convert to MU per QU. Fixed costs: 2,400 EUR \(= 24\) MU. Variable costs per QU: \(100 \cdot 6 = 600\) EUR \(= 6\) MU. Price per QU: \(100 \cdot (25 - 2x)\) EUR \(= (25 - 2x)\) MU.

Total costs are variable costs plus fixed costs:

\[C = 6x + 24 \quad \checkmark\]

Revenue is quantity times price:

\[R = x(25 - 2x) = 25x - 2x^2 \quad \checkmark\]

Grading: 1 Pt. fixed costs in MU, 1 Pt. variable costs and price per QU in MU, 1 Pt. \(C\), 1 Pt. \(R\) multiplied out.

Part b)

Show that the monthly profit is given by \(P = -2x^2 + 19x - 24\). [2 pts.]

\[P = R - C = 25x - 2x^2 - (6x + 24)\]

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

\[P = 25x - 2x^2 - 6x - 24 = -2x^2 + 19x - 24 \quad \checkmark\]

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

Part c)

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 with the AC method. Determine the break-even point and the profit limit (upper break-even point). [5 pts.]

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

(for verification: \(P = -(2x - 3)(x - 8)\))

Factor out \(-1\):

\[P = -(2x^2 - 19x + 24)\]

Check: \(\Delta = (-19)^2 - 4 \cdot 2 \cdot 24 = 361 - 192 = 169 = 13^2\) is a perfect square, so the bracket factors.

AC method. \(ac = 2 \cdot 24 = 48\). Factors of 48 with sum \(-19\): \((-3, -16)\).

\[2x^2 - 3x - 16x + 24 = x(2x - 3) - 8(2x - 3) = (x - 8)(2x - 3)\]

\[P = -(2x - 3)(x - 8) \quad \checkmark\]

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

\[2x - 3 = 0 \Rightarrow x = 1.5, \qquad x - 8 = 0 \Rightarrow x = 8\]

Check: \(-2 \cdot 64 + 19 \cdot 8 - 24 = -128 + 152 - 24 = 0\).

Break-even point: \(x = 1.5\) QU (150 mugs). Profit limit: \(x = 8\) QU (800 mugs). Both lie below the capacity of 10 QU.

Grading: 1 Pt. factoring out \(-1\), 1 Pt. \(ac = 48\) and the pair \((-3, -16)\), 1 Pt. grouping to the factored form, 1 Pt. both zeros, 1 Pt. naming break-even point and profit limit.

Part d)

Decide in which output range the workshop makes a profit. State this profit zone as an interval and in set-builder notation, and decide whether the planned output lies in it. [3 pts.]

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

Test values between and beyond the zeros 1.5 and 8 (capacity 10 QU):

  • \(x = 1\): \(-2 + 19 - 24 = -7 < 0\), a loss
  • \(x = 5\): \(-50 + 95 - 24 = 21 > 0\), a profit
  • \(x = 9\): \(-162 + 171 - 24 = -15 < 0\), a loss

Profit zone:

\[(1.5; 8) = \{x \in \mathbb{R} \mid 1.5 < x < 8\}\]

The interval is open: at 1.5 QU and 8 QU the profit is exactly zero.

Planned output: 700 mugs \(= 7\) QU, and \(1.5 < 7 < 8\), so the workshop makes a profit.

Grading: 1 Pt. test values or sign argument, 1 Pt. open interval in both notations, 1 Pt. decision for 7 QU.

Part e)

The clay supplier raises its prices, so the variable costs per mug rise by 50 %. Compute by how many percent the total monthly costs at the planned output of 700 mugs rise, and explain why this increase is smaller than 50 %. [3 pts.]

Old total costs at 7 QU: \(6 \cdot 7 + 24 = 66\) MU \(= 6{,}600\) EUR

New variable costs: \(6 \cdot 1.5 = 9\) EUR per mug, i.e. \(9\) MU per QU. New total costs at 7 QU: \(9 \cdot 7 + 24 = 87\) MU \(= 8{,}700\) EUR

Change:

\[\frac{87 - 66}{66} = \frac{21}{66} \approx 0.318 = 31.8\,\%\]

Explanation: only the variable costs rise by 50 %: from 42 MU to 63 MU, i.e. by 21 MU. The fixed costs of 24 MU stay the same. The increase of 21 MU is measured against the whole 66 MU, so the total costs rise by less than 50 %.

Grading: 1 Pt. old and new total costs, 1 Pt. percentage change, 1 Pt. explanation referring to the fixed costs.

Part f)

Show that after the price rise of the clay supplier the monthly profit is \(P_{\text{new}} = -2(x - 2)(x - 6)\). Decide whether the planned output of 700 mugs is still profitable and give the monthly profit or loss at this output in EUR. [3 pts.]

New profit with \(C_{\text{new}} = 9x + 24\):

\[P_{\text{new}} = 25x - 2x^2 - (9x + 24) = -2x^2 + 16x - 24\]

Given form multiplied out:

\[-2(x - 2)(x - 6) = -2(x^2 - 8x + 12) = -2x^2 + 16x - 24 \quad \checkmark\]

At 700 mugs (\(x = 7\)):

\[P_{\text{new}} = -2(7 - 2)(7 - 6) = -2 \cdot 5 \cdot 1 = -10 \text{ MU}\]

Decision: no, the planned output is no longer profitable. The workshop makes a loss of 1,000 EUR per month (before the price rise: a profit of \(-98 + 133 - 24 = 11\) MU \(= 1{,}100\) EUR).

Grading: 1 Pt. \(P_{\text{new}}\) from \(R - C_{\text{new}}\), 1 Pt. multiplying out the given form, 1 Pt. decision with the loss in EUR.