Tasks 08-03 - Cost Analysis & Pricing Decisions

Section 08: Financial Mathematics

A: Skills

Problem 1: Fixed Costs, Variable Costs and Costs per Unit (x)

On this sheet, \(C(x) = C_f + C_v(x)\) are the total costs at an output of \(x\) units, with the fixed costs \(C_f\) and the variable costs \(C_v(x)\). The average total cost is \(\overline{C}(x) = \frac{C(x)}{x}\). The FSP calls it the “unit costs” (the total costs per unit). The variable cost per unit is \(\overline{C}_v(x) = \frac{C(x) - C_f}{x}\). Earlier sheets wrote \(k(x)\) for \(\overline{C}(x)\) and \(k_v(x)\) for \(\overline{C}_v(x)\): they are the same functions.

Give costs and prices in EUR to the cent (two decimal places) and keep intermediate results unrounded. Give quantities as the lecture does: to two decimal places, or in whole units if you check the whole numbers on both sides. A lower limit price is the smallest price that still covers the costs, so it is a minimum: if you have to round it, round up to the cent. This rule holds for the whole sheet.

The total costs of three products are (\(C\) in EUR, \(x\) in units per week):

(1) \(C(x) = 1{,}480 + 23x\)

(2) \(C(x) = 760 + 31x + 0.025x^2\)

(3) \(C(x) = 0.002x^3 - 0.27x^2 + 44x + 2{,}100\)

  1. Give the fixed costs \(C_f\) and the variable costs \(C_v(x)\) for each product.

  2. Determine \(\overline{C}_v(x)\) and \(\overline{C}(x)\) for each product.

  3. Compute the variable cost per unit and the average total cost at an output of 120 units per week for each product.

  4. Show that \(\overline{C}(x) - \overline{C}_v(x) = \frac{C_f}{x}\) for every cost function and explain what happens to this difference when the output grows.

  1. The fixed costs are the costs at \(x = 0\), i.e. the term without \(x\). The rest are the variable costs.

    (1) \(C_f = 1{,}480\) EUR, \(C_v(x) = 23x\)

    (2) \(C_f = 760\) EUR, \(C_v(x) = 31x + 0.025x^2\)

    (3) \(C_f = 2{,}100\) EUR, \(C_v(x) = 0.002x^3 - 0.27x^2 + 44x\)

  2. Divide the variable costs and the total costs by \(x\):

    (1) \(\overline{C}_v(x) = \frac{23x}{x} = 23\) (constant), \(\quad \overline{C}(x) = \frac{1480}{x} + 23\)

    (2) \(\overline{C}_v(x) = 31 + 0.025x\), \(\quad \overline{C}(x) = \frac{760}{x} + 31 + 0.025x\)

    (3) \(\overline{C}_v(x) = 0.002x^2 - 0.27x + 44\), \(\quad \overline{C}(x) = 0.002x^2 - 0.27x + 44 + \frac{2100}{x}\)

  3. At \(x = 120\):

    (1) \(\overline{C}_v(120) = 23.00\) EUR, \(\quad \overline{C}(120) = \frac{1480}{120} + 23 \approx 12.33 + 23 = 35.33\) EUR

    (2) \(\overline{C}_v(120) = 31 + 3 = 34.00\) EUR, \(\quad \overline{C}(120) = \frac{760}{120} + 34 \approx 40.33\) EUR

    (3) \(\overline{C}_v(120) = 28.8 - 32.4 + 44 = 40.40\) EUR, \(\quad \overline{C}(120) = 40.4 + \frac{2100}{120} = 40.4 + 17.5 = 57.90\) EUR

  4. \(\overline{C}(x) - \overline{C}_v(x) = \frac{C(x)}{x} - \frac{C(x) - C_f}{x} = \frac{C(x) - C(x) + C_f}{x} = \frac{C_f}{x}\)

    This is the fixed cost per unit (the lecture calls it the average fixed cost). The fixed costs stay the same, but they are shared by more units, so the difference gets smaller when the output grows. The graph of \(\overline{C}\) approaches the graph of \(\overline{C}_v\), but it always stays above it.

Typical mistake: in a) (3), taking \(C_f = 0.002\) because it is the first number. The fixed costs are the term without \(x\), wherever it stands.

Problem 2: Short-Term Lower Limit Price (xx)

The short-term lower limit price is the minimum of the variable cost per unit \(\overline{C}_v(x)\). Below this price, a company does not even cover its variable costs.

  1. Compute the short-term lower limit price for each cost function (\(C\) in EUR, \(x\) in units). For (3) and (4), also give the output at which it is reached and prove that it is a minimum.

(1) \(C(x) = 2{,}250 + 17.5x\)

(2) \(C(x) = 940 + 26x + 0.04x^2\)

(3) \(C(x) = 0.004x^3 - 0.64x^2 + 45x + 2{,}450\)

(4) \(C(x) = 2x^3 - 22x^2 + 85x + 120\) with \(C\) in MU and \(x\) in QU (1 QU = 100 units, 1 MU = 100 EUR). Give the result in EUR per unit.

Hint: In (2), check the sign of \(\overline{C}_v'(x)\). If it is positive for every \(x > 0\), the lowest value is the value that \(\overline{C}_v\) approaches for very small \(x\).

  1. Check your result for (4) with the shortcut \(C'(x) \cdot x = C(x) - C_f\) from the lecture.

  1. (1) \(\overline{C}_v(x) = \frac{17.5x}{x} = 17.5\) is constant. Every unit costs 17.50 EUR in variable costs, whatever the output. The short-term lower limit price is 17.50 EUR.

    (2) \(\overline{C}_v(x) = 26 + 0.04x\) and \(\overline{C}_v'(x) = 0.04 > 0\). The variable cost per unit grows with every unit, so it has no minimum at a positive output. For small outputs it approaches 26. The short-term lower limit price is the coefficient of \(x\): 26.00 EUR.

    (3) Step 1: \(\overline{C}_v(x) = \frac{0.004x^3 - 0.64x^2 + 45x}{x} = 0.004x^2 - 0.64x + 45\)

    Step 2 and 3: \(\overline{C}_v'(x) = 0.008x - 0.64 = 0 \iff x = 80\)

    Sufficient condition: \(\overline{C}_v''(x) = 0.008 > 0\), so \(\overline{C}_v\) has a minimum at \(x = 80\).

    Step 4: \(\overline{C}_v(80) = 0.004 \cdot 6400 - 0.64 \cdot 80 + 45 = 25.6 - 51.2 + 45 = 19.4\)

    The short-term lower limit price is 19.40 EUR, reached at 80 units.

    (4) \(\overline{C}_v(x) = 2x^2 - 22x + 85\), \(\quad \overline{C}_v'(x) = 4x - 22 = 0 \iff x = 5.5\)

    Sufficient condition: \(\overline{C}_v''(x) = 4 > 0\), so this is a minimum.

    \(\overline{C}_v(5.5) = 2 \cdot 30.25 - 22 \cdot 5.5 + 85 = 60.5 - 121 + 85 = 24.5\) MU per QU

    At 5.5 QU, i.e. 550 units, the variable cost per unit is lowest. 1 MU per QU is 100 EUR per 100 units, i.e. 1 EUR per unit: the short-term lower limit price is 24.50 EUR per unit.

  2. \(C'(x) = 6x^2 - 44x + 85\), so \(C'(5.5) = 181.5 - 242 + 85 = 24.5\) and

    \[C'(5.5) \cdot 5.5 = 24.5 \cdot 5.5 = 134.75\]

    \[C(5.5) - C_f = 2 \cdot 166.375 - 22 \cdot 30.25 + 85 \cdot 5.5 = 332.75 - 665.5 + 467.5 = 134.75 \quad \checkmark\]

    Both sides agree, so \(x = 5.5\) is correct. Also, \(C'(5.5) = \overline{C}_v(5.5) = 24.5\): at the minimum of the variable cost per unit, the marginal costs equal the variable cost per unit.

Typical mistake: in (2), setting \(\overline{C}_v'(x) = 0.04 = 0\) and concluding that there is no short-term lower limit price. \(\overline{C}_v\) is increasing, so the lowest value is the one it approaches at the smallest outputs, 26 EUR.

Problem 3: Long-Term Lower Limit Price for Quadratic Costs (xx)

The long-term lower limit price is the minimum of the average total cost \(\overline{C}(x)\). Below this price, a company cannot cover all its costs in the long run.

  1. Compute the long-term lower limit price and the output at which it is reached for each cost function (\(C\) in EUR, \(x\) in units). Prove that it is a minimum.

(1) \(C(x) = 1{,}815 + 22x + 0.15x^2\)

(2) \(C(x) = 2{,}000 + 35x + 0.08x^2\). Here the output is not a whole number: give the best whole output as well.

  1. In (1), the two parts \(\frac{C_f}{x}\) and \(0.15x\) of \(\overline{C}(x)\) are equal at the minimum. Show that this holds for every cost function \(C(x) = C_f + bx + ax^2\) with \(a > 0\).

  1. (1) \(\overline{C}(x) = \frac{1815}{x} + 22 + 0.15x\)

    \[\overline{C}'(x) = -\frac{1815}{x^2} + 0.15 = 0 \iff x^2 = \frac{1815}{0.15} = 12{,}100 \iff x = 110 \quad (x > 0)\]

    Sufficient condition: \(\overline{C}''(x) = \frac{3630}{x^3} > 0\) for every \(x > 0\), so this is a minimum.

    \[\overline{C}(110) = \frac{1815}{110} + 22 + 0.15 \cdot 110 = 16.5 + 22 + 16.5 = 55\]

    The long-term lower limit price is 55.00 EUR, reached at 110 units.

    (2) \(\overline{C}(x) = \frac{2000}{x} + 35 + 0.08x\)

    \[\overline{C}'(x) = -\frac{2000}{x^2} + 0.08 = 0 \iff x^2 = 25{,}000 \iff x = \sqrt{25{,}000} \approx 158.11\]

    Sufficient condition: \(\overline{C}''(x) = \frac{4000}{x^3} > 0\) for \(x > 0\), so this is a minimum.

    \[\overline{C}(158.11) \approx 12.649 + 35 + 12.649 \approx 60.2982\]

    The long-term lower limit price is a minimum price, so round up: 60.30 EUR. Whole outputs: \(\overline{C}(158) \approx 60.29823\) and \(\overline{C}(159) \approx 60.29862\), so 158 units is the best whole output.

  2. \(\overline{C}(x) = \frac{C_f}{x} + b + ax\) and \(\overline{C}'(x) = -\frac{C_f}{x^2} + a = 0 \iff x^2 = \frac{C_f}{a}\). Divide both sides of \(ax^2 = C_f\) by \(x\):

    \[ax = \frac{C_f}{x}\]

    At the minimum, the fixed cost per unit equals the part \(ax\). So the long-term lower limit price is \(b + 2ax\), e.g. \(22 + 2 \cdot 16.5 = 55\) in (1).

Typical mistake: minimising \(C(x)\) instead of \(\overline{C}(x)\). \(C'(x) = 22 + 0.3x\) is never 0 for \(x > 0\): the total costs grow with every unit, only the costs per unit have a minimum.

Problem 4: Long-Term Lower Limit Price for Cubic Costs (xx)

A small company sews dog beds. Its weekly total costs are

\[C(x) = 0.005x^3 - 0.7x^2 + 38x + 3{,}000\]

in EUR for an output of \(x\) dog beds per week.

  1. Determine \(\overline{C}(x)\) and show that \(\overline{C}'(x) = 0\) leads to the equation \(x^3 - 70x^2 - 300{,}000 = 0\).

  2. Solve the equation with the polynomial solver (MENU → A → 2, degree 3) or by testing whole numbers. Prove that the solution gives a minimum of \(\overline{C}\) and compute the long-term lower limit price.

Hint: The solution is a whole number between 50 and 150.

  1. The company currently produces 120 dog beds per week. Compute the average total cost at this output and state how much the company would save per dog bed at the output from b).

  1. \(\overline{C}(x) = \frac{C(x)}{x} = 0.005x^2 - 0.7x + 38 + \frac{3000}{x}\)

    \[\overline{C}'(x) = 0.01x - 0.7 - \frac{3000}{x^2} = 0 \quad \Big| \cdot x^2\]

    \[0.01x^3 - 0.7x^2 - 3000 = 0 \quad \Big| \cdot 100 \quad \iff \quad x^3 - 70x^2 - 300{,}000 = 0 \quad \checkmark\]

  2. Polynomial solver with \(a = 1\), \(b = -70\), \(c = 0\), \(d = -300{,}000\) gives \(x = 100\). By testing: \(100^3 - 70 \cdot 100^2 - 300{,}000 = 1{,}000{,}000 - 700{,}000 - 300{,}000 = 0\) \(\checkmark\)

    Sufficient condition: \(\overline{C}''(x) = 0.01 + \frac{6000}{x^3} > 0\) for every \(x > 0\), so \(\overline{C}\) has a minimum at \(x = 100\).

    \[\overline{C}(100) = 0.005 \cdot 10{,}000 - 0.7 \cdot 100 + 38 + \frac{3000}{100} = 50 - 70 + 38 + 30 = 48\]

    The long-term lower limit price is 48.00 EUR per dog bed, reached at 100 dog beds per week.

  3. \(\overline{C}(120) = 0.005 \cdot 14{,}400 - 0.7 \cdot 120 + 38 + \frac{3000}{120} = 72 - 84 + 38 + 25 = 51\)

    At 120 dog beds, a dog bed costs 51.00 EUR on average. At 100 dog beds it would cost 48.00 EUR, i.e. 3.00 EUR less per dog bed.

Typical mistake: in a), forgetting the factor \(-1\) from the derivative of \(\frac{3000}{x} = 3000x^{-1}\). Its derivative is \(-\frac{3000}{x^2}\), so the constant of the cubic equation is negative.

Problem 5: Three Prices, Three Decisions (xx)

The dog bed company from Problem 4 has the long-term lower limit price 48.00 EUR.

  1. Compute the short-term lower limit price of the company.

(for verification: 13.50 EUR at 70 dog beds per week)

  1. The market price for a dog bed could be 11.90 EUR, 28.00 EUR or 52.50 EUR. Decide for each price in which of the three pricing zones it lies and what the company should do.

  2. At a price of 28.00 EUR, the company sells 90 dog beds per week. Compute the contribution margin per dog bed and the total contribution margin. Compute the weekly profit or loss and compare it with the loss if the company stops production.

  1. \(\overline{C}_v(x) = 0.005x^2 - 0.7x + 38\) and \(\overline{C}_v'(x) = 0.01x - 0.7 = 0 \iff x = 70\). Since \(\overline{C}_v''(x) = 0.01 > 0\), this is a minimum:

    \[\overline{C}_v(70) = 0.005 \cdot 4900 - 0.7 \cdot 70 + 38 = 24.5 - 49 + 38 = 13.5\]

    The short-term lower limit price is 13.50 EUR.

  2. The zones are bounded by 13.50 EUR and 48.00 EUR:

    • 11.90 EUR \(<\) 13.50 EUR: the price does not even cover the variable costs. Stop production: every dog bed sold makes the loss larger.
    • 13.50 EUR \(\le\) 28.00 EUR \(<\) 48.00 EUR: the price covers the variable costs and part of the fixed costs. Continue in the short term, but this is not sustainable.
    • 52.50 EUR \(\ge\) 48.00 EUR: the price covers all costs. Production is sustainable: at 100 dog beds the company earns \(52.50 - 48.00 = 4.50\) EUR per dog bed.

  3. \(\overline{C}_v(90) = 0.005 \cdot 8100 - 0.7 \cdot 90 + 38 = 40.5 - 63 + 38 = 15.5\)

    Contribution margin per dog bed: \(28.00 - 15.50 = 12.50\) EUR. Total: \(90 \cdot 12.50 = 1{,}125\) EUR per week.

    Profit: total contribution minus fixed costs, \(1{,}125 - 3{,}000 = -1{,}875\) EUR. Check: \(R(90) - C(90) = 2{,}520 - 4{,}395 = -1{,}875\) EUR \(\checkmark\)

    If the company stops, it still has to pay the fixed costs and loses 3,000 EUR per week. Producing loses 1,125 EUR less, because each dog bed contributes 12.50 EUR to the fixed costs.

Typical mistake: in b), calling 28.00 EUR a “stop” price because the company makes a loss. The loss is smaller than without production. Stop only below the short-term lower limit price.

B: Set Up & Check

Problem 6: From Words to Functions (xx)

Translate each question into the function you need and compute the result. Then answer the check question: it tells you whether your translation works.

  1. A copy centre has the weekly costs \(C(x) = 1{,}350 + 4.2x + 0.003x^2\) in EUR for \(x\) posters. The manager asks for the unit costs and for the variable costs per unit at an output of 300 posters per week.

    Check: A classmate gets 6.00 EUR for the unit costs. Explain which function the classmate used.

  2. A company has the costs \(C(x) = 0.5x^3 - 8x^2 + 45x + 200\) (\(C\) in MU, \(x\) in QU). The bank asks for the long term price floor (the long-term lower limit price). Write the function you minimise and the equation for its minimum, and solve it by testing whole numbers from 8 to 12.

    Check: A classmate computes the minimum of \(C'(x)\) and gets about 2.33 MU per QU. Explain why this cannot be the price floor.

  3. A company has the costs \(C(x) = 0.002x^3 - 0.36x^2 + 31x + 1{,}900\) in EUR for \(x\) units. Compute the short-term lower limit price.

    Check: A classmate says: “The short-term lower limit price is 31 EUR, the variable cost per unit of the very first unit.” Explain why this is wrong here, but would be right for \(C(x) = 1{,}900 + 31x + 0.002x^2\).

  4. A company has the monthly costs \(C(x) = 6{,}400 + 14.6x\) in EUR for \(x\) units. It sells 1,000 units per month and has spare capacity, i.e. it could produce more without new machines or staff. A customer offers a special order of 250 units at 13.90 EUR per unit. Decide whether the company should accept and compute the contribution margin per unit and in total.

    Check: A colleague says: “We have spare capacity and pay the fixed costs anyway, so every special order brings in money.” Explain what is wrong with this argument.

  5. A company has the weekly costs \(C(x) = 2{,}650 + 19x + 0.05x^2\) in EUR and sells 120 units per week at 34 EUR per unit. Compute the contribution margin per unit and in total, and the weekly profit.

    Check: A classmate says: “The contribution margin is positive, so the company makes a profit.” Explain why this is wrong.

  1. “Unit costs” means the average total cost \(\overline{C}(x) = \frac{C(x)}{x}\), “variable costs per unit” means \(\overline{C}_v(x) = \frac{C(x) - C_f}{x}\):

    \[\overline{C}(300) = \frac{1350}{300} + 4.2 + 0.003 \cdot 300 = 4.5 + 4.2 + 0.9 = 9.60 \text{ EUR}, \qquad \overline{C}_v(300) = 4.2 + 0.9 = 5.10 \text{ EUR}\]

    Check: 6.00 EUR is the marginal cost \(C'(300) = 4.2 + 0.006 \cdot 300 = 6.0\), i.e. the cost of one more poster. The unit costs are the costs of all posters divided by their number.

    Typical mistake: dividing only the variable costs by \(x\) when the question asks for the unit costs. The unit costs include the fixed costs.

  2. The long term price floor is the minimum of the unit costs \(\overline{C}(x) = 0.5x^2 - 8x + 45 + \frac{200}{x}\):

    \[\overline{C}'(x) = x - 8 - \frac{200}{x^2} = 0 \quad \Big| \cdot x^2 \quad \iff \quad x^3 - 8x^2 - 200 = 0\]

    Test: \(x = 8\) gives \(-200\), \(x = 9\) gives \(-119\), \(x = 10\) gives \(1000 - 800 - 200 = 0\) \(\checkmark\). Since \(\overline{C}''(x) = 1 + \frac{400}{x^3} > 0\), this is a minimum, and \(\overline{C}(10) = 50 - 80 + 45 + 20 = 35\). The long term price floor is 35 MU per QU at 10 QU.

    Check: \(C'(x)\) is the marginal cost. Its minimum (at \(x = \frac{16}{3} \approx 5.33\)) is the inflection point of \(C\), where the total costs grow most slowly. It says nothing about the costs per unit, and it ignores the fixed costs of 200 MU.

    Typical mistake: minimising \(C(x)\) or \(C'(x)\) instead of \(\overline{C}(x)\). A price is compared with costs per unit, so the function to minimise is always a cost per unit.

  3. The short-term lower limit price is the minimum of \(\overline{C}_v(x) = 0.002x^2 - 0.36x + 31\):

    \[\overline{C}_v'(x) = 0.004x - 0.36 = 0 \iff x = 90, \qquad \overline{C}_v''(x) = 0.004 > 0\]

    \[\overline{C}_v(90) = 16.2 - 32.4 + 31 = 14.8\]

    The short-term lower limit price is 14.80 EUR at 90 units.

    Check: here \(\overline{C}_v\) first falls and then grows, so its lowest value 14.80 EUR is below 31 EUR. For \(C(x) = 1{,}900 + 31x + 0.002x^2\), the variable cost per unit \(31 + 0.002x\) grows from the start, so its lowest value is the one it approaches for small outputs, 31 EUR.

    Typical mistake: taking the coefficient of \(x\) as the short-term lower limit price for every cost function. This shortcut only works when \(\overline{C}_v\) is increasing, e.g. for quadratic costs with \(a > 0\).

  4. Variable cost per unit: \(\overline{C}_v(x) = 14.6\), so the short-term lower limit price is 14.60 EUR. The offer, 13.90 EUR, is below it: reject. Contribution margin: \(13.90 - 14.60 = -0.70\) EUR per unit, in total \(250 \cdot (-0.70) = -175\) EUR. The order would lower the monthly profit by 175 EUR.

    Check: the fixed costs are indeed paid anyway, but every unit of the order still causes variable costs of 14.60 EUR. At 13.90 EUR, each unit brings 0.70 EUR less than it costs, so the order makes the result worse. The revenue of \(250 \cdot 13.90 = 3{,}475\) EUR does not cover the extra variable costs of \(250 \cdot 14.60 = 3{,}650\) EUR.

    Typical mistake: looking only at the revenue of the order. A special order with spare capacity must cover its variable costs, so compare its price with the short-term lower limit price.

  5. \(\overline{C}_v(120) = 19 + 0.05 \cdot 120 = 25\), so the contribution margin is \(34 - 25 = 9.00\) EUR per unit and \(120 \cdot 9 = 1{,}080\) EUR in total.

    Profit: \(1{,}080 - 2{,}650 = -1{,}570\) EUR per week. Check: \(R(120) - C(120) = 4{,}080 - 5{,}650 = -1{,}570\) EUR \(\checkmark\)

    Check: the contribution margin only says how much each unit contributes towards the fixed costs. Here the total contribution, 1,080 EUR, covers only part of the 2,650 EUR fixed costs, so the company makes a loss.

    Typical mistake: forgetting the fixed costs. Profit = total contribution margin \(-\) fixed costs.

Problem 7: One Paragraph, All Questions (xx)

A company builds conveyor toasters (large toasters for hotel breakfast buffets) and sells them to catering wholesalers. Its monthly total costs are \(C(x) = 0.5x^3 - 11x^2 + 84x + 360\) MU for an output of \(x\) QU (1 QU = 10 toasters, 1 MU = 1,000 EUR). The wholesalers pay 60 MU per QU. The management wants to know the fixed costs, the output with the lowest unit costs and the long term price floor, the minimum variable cost per unit and the short-term lower limit price. It also asks for the break-even point and the profit limit (the upper break-even point) at the wholesalers’ price, for the output with the highest profit, and whether a café chain’s special order of 2 QU at 30 MU per QU is worth accepting while the company has spare capacity.

List every given quantity with its symbol and unit. Then write every question of the management as a function to minimise or maximise, an equation or an inequality. Do not compute anything.

Check: An assistant writes the long term price floor as the minimum of \(\overline{C}_v(x) = 0.5x^2 - 11x + 84\). Explain what is wrong.

Given:

  1. Total costs \(C(x) = 0.5x^3 - 11x^2 + 84x + 360\) in MU, output \(x\) in QU
  2. Units: 1 QU = 10 toasters, 1 MU = 1,000 EUR, so 1 MU per QU = 100 EUR per toaster
  3. Price \(p = 60\) MU per QU, so revenue \(R(x) = 60x\)
  4. Special order: 2 QU at 30 MU per QU, spare capacity

Questions:

  1. Fixed costs: \(C_f = C(0)\)
  2. Lowest unit costs and long term price floor: minimise \(\overline{C}(x) = \frac{C(x)}{x}\), i.e. solve \(\overline{C}'(x) = 0\) (multiplied by \(x^2\), a cubic equation), check \(\overline{C}'' > 0\), then the price floor is the minimum value of \(\overline{C}\)
  3. Minimum variable cost per unit and short-term lower limit price: minimise \(\overline{C}_v(x) = \frac{C(x) - C_f}{x}\), i.e. solve \(\overline{C}_v'(x) = 0\)
  4. Break-even point and profit limit: \(P(x) = R(x) - C(x) = 0\), the lower zero is the break-even point, the upper zero the profit limit
  5. Highest profit: \(P'(x) = 0\) with \(P''(x) < 0\)
  6. Special order: first condition \(30 \geq\) the short-term lower limit price from 7. To be sure, compare the extra revenue \(2 \cdot 30\) MU with the additional costs \(C(x_0 + 2) - C(x_0)\) of the 2 QU at the current output \(x_0\)

Check: \(\overline{C}_v\) leaves out the fixed costs. Its minimum is the short-term lower limit price. The long term price floor must cover all costs, so it is the minimum of \(\overline{C}(x) = 0.5x^2 - 11x + 84 + \frac{360}{x}\).

Typical mistake: comparing the special order with the wholesalers’ price or with the long term price floor. With spare capacity, the fixed costs are paid anyway, so the order only has to cover the additional costs it causes.

C: Applied

Problem 8: Climbing Holds from a Small Workshop (xx)

A small workshop casts climbing holds (the coloured plastic grips on the walls of climbing halls) and sells them to climbing halls. Its weekly total costs are

\[C(x) = 0.002x^3 - 0.48x^2 + 42x + 2{,}700\]

in EUR for an output of \(x\) holds per week.

  1. Give the fixed costs and \(\overline{C}_v(x)\). Determine the short-term lower limit price and the output at which it is reached.

  2. Show that the lowest average total cost leads to the equation \(x^3 - 120x^2 - 675{,}000 = 0\). Solve it, prove that you have found a minimum and compute the long-term lower limit price.

(for verification: 33.00 EUR at 150 holds per week)

  1. Write the three pricing zones of the workshop with their decisions. The owner says: “The difference between the two limits, 19.80 EUR, is exactly the fixed cost per hold at the best output.” Check this claim and explain the result.

  2. The landlord raises the rent of the workshop by 900 EUR per week. Explain without computing which of the two limits changes. Then compute the new long-term lower limit price with the polynomial solver.

a) \(C_f = C(0) = 2{,}700\) EUR per week and

\[\overline{C}_v(x) = \frac{C(x) - 2700}{x} = 0.002x^2 - 0.48x + 42\]

\[\overline{C}_v'(x) = 0.004x - 0.48 = 0 \iff x = 120, \qquad \overline{C}_v''(x) = 0.004 > 0 \;\Rightarrow\; \text{minimum}\]

\[\overline{C}_v(120) = 0.002 \cdot 14{,}400 - 0.48 \cdot 120 + 42 = 28.8 - 57.6 + 42 = 13.2\]

The short-term lower limit price is 13.20 EUR per hold, reached at 120 holds per week.

b) \(\overline{C}(x) = 0.002x^2 - 0.48x + 42 + \frac{2700}{x}\) and

\[\overline{C}'(x) = 0.004x - 0.48 - \frac{2700}{x^2} = 0 \quad \Big| \cdot x^2 \quad \iff \quad 0.004x^3 - 0.48x^2 - 2700 = 0 \quad \Big| \cdot 250\]

\[x^3 - 120x^2 - 675{,}000 = 0 \quad \checkmark\]

The polynomial solver (MENU → A → 2, degree 3) gives \(x = 150\). Test: \(3{,}375{,}000 - 2{,}700{,}000 - 675{,}000 = 0\) \(\checkmark\)

Sufficient condition: \(\overline{C}''(x) = 0.004 + \frac{5400}{x^3} > 0\) for every \(x > 0\), so this is a minimum.

\[\overline{C}(150) = 0.002 \cdot 22{,}500 - 0.48 \cdot 150 + 42 + \frac{2700}{150} = 45 - 72 + 42 + 18 = 33\]

The long-term lower limit price is 33.00 EUR per hold, reached at 150 holds per week.

c) Pricing zones:

Price per hold Decision
below 13.20 EUR stop production, the price does not cover the variable costs
13.20 EUR to below 33.00 EUR produce in the short term, the price covers the variable costs and part of the fixed costs
33.00 EUR or more production is sustainable, all costs are covered

The claim is wrong. The fixed cost per hold at the best output is \(\frac{2700}{150} = 18.00\) EUR, not 19.80 EUR. The two minima lie at different outputs: \(\overline{C}_v\) is lowest at 120 holds, \(\overline{C}\) at 150 holds. At 150 holds, the variable cost per unit is \(\overline{C}_v(150) = 45 - 72 + 42 = 15\) EUR, not 13.20 EUR. Only at the same output would the difference be the fixed cost per hold: \(\overline{C}(150) - \overline{C}_v(150) = 33 - 15 = 18\).

d) The rent belongs to the fixed costs. \(\overline{C}_v\) does not contain the fixed costs, so the short-term lower limit price stays 13.20 EUR. Only the long-term lower limit price rises.

New fixed costs: \(2{,}700 + 900 = 3{,}600\) EUR. As in b):

\[0.004x^3 - 0.48x^2 - 3600 = 0 \iff x^3 - 120x^2 - 900{,}000 = 0\]

The polynomial solver gives \(x \approx 156.67\) holds and

\[\overline{C}(156.67) = 0.002 \cdot 156.67^2 - 0.48 \cdot 156.67 + 42 + \frac{3600}{156.67} \approx 38.8676\]

(with the unrounded solution). The new long-term lower limit price is 38.87 EUR per hold (rounded up). In whole holds: \(\overline{C}(156) \approx 38.8689\) and \(\overline{C}(157) \approx 38.8679\), so 157 holds per week is the best whole output.

Typical mistake: in c), subtracting the two minimum values and calling the result the fixed cost per unit. Fixed cost per unit and variable cost per unit can only be compared at the same output.

Problem 9: Trekking Poles and a Special Order (xxx)

A company produces trekking poles (walking poles for hiking) and sells them in pairs to outdoor shops. Its monthly total costs are

\[C(x) = 0.25x^3 - 6x^2 + 60x + 512\]

MU for an output of \(x\) QU (1 QU = 100 pairs, 1 MU = 100 EUR). The company can produce up to 20 QU per month.

Note: The long-term lower limit price of the company is 60 MU per QU, reached at 16 QU.

  1. Compute the minimum variable cost per unit and the short-term lower limit price.

(for verification: 24 EUR per pair at 1,200 pairs per month)

  1. The outdoor shops currently buy 10 QU per month at 52 EUR per pair. Compute the contribution margin per pair and in total, and the monthly profit or loss in EUR. Decide whether the company should keep producing in the short term.

  2. A travel company that organises hiking tours offers a special order: 3 QU of poles with its logo at 30 EUR per pair. Compare the offer with both lower limit prices. Then compute the additional costs \(C(13) - C(10)\) of the order and decide whether the order raises the profit.

  3. After accepting, the travel company wants another 6 QU at the same price. Compute the additional costs \(C(19) - C(13)\) and decide again. Explain why the comparison with the short-term lower limit price alone is not enough here.

Hint: Compute the additional costs per QU of the second order and compare them with the offered 30 MU per QU.

a) \(\overline{C}_v(x) = \frac{C(x) - 512}{x} = 0.25x^2 - 6x + 60\)

\[\overline{C}_v'(x) = 0.5x - 6 = 0 \iff x = 12, \qquad \overline{C}_v''(x) = 0.5 > 0 \;\Rightarrow\; \text{minimum}\]

\[\overline{C}_v(12) = 0.25 \cdot 144 - 6 \cdot 12 + 60 = 36 - 72 + 60 = 24\]

The minimum variable cost per unit is 24 MU per QU at 12 QU. 1 MU per QU is 100 EUR per 100 pairs, i.e. 1 EUR per pair: the short-term lower limit price is 24 EUR per pair, reached at 1,200 pairs per month.

b) \(\overline{C}_v(10) = 25 - 60 + 60 = 25\) MU per QU, so the contribution margin is \(52 - 25 = 27\) MU per QU, i.e. 27 EUR per pair. In total: \(10 \cdot 27 = 270\) MU \(= 27{,}000\) EUR per month.

Profit: \(270 - 512 = -242\) MU, i.e. a loss of 24,200 EUR per month. Check: \(R(10) - C(10) = 520 - (250 - 600 + 600 + 512) = 520 - 762 = -242\) \(\checkmark\)

The price of 52 EUR lies between 24 EUR and 60 EUR. Without production, the company would lose its fixed costs of 51,200 EUR per month. With production it loses 27,000 EUR less: it should keep producing in the short term, but it cannot stay in the market at this price in the long run.

c) \(24 \le 30 < 60\): the offer covers the lowest variable cost per unit, but not all costs. In the short term, the order can be worth it.

\[C(13) = 0.25 \cdot 2197 - 6 \cdot 169 + 60 \cdot 13 + 512 = 549.25 - 1014 + 780 + 512 = 827.25\]

\[C(13) - C(10) = 827.25 - 762 = 65.25 \text{ MU}\]

The order brings \(3 \cdot 30 = 90\) MU and costs 65.25 MU more, i.e. 21.75 EUR per pair. The profit rises by \(90 - 65.25 = 24.75\) MU \(= 2{,}475\) EUR. Accept the order.

d) \(C(19) = 0.25 \cdot 6859 - 6 \cdot 361 + 60 \cdot 19 + 512 = 1714.75 - 2166 + 1140 + 512 = 1200.75\)

\[C(19) - C(13) = 1200.75 - 827.25 = 373.5 \text{ MU}\]

The second order brings \(6 \cdot 30 = 180\) MU but costs 373.5 MU more, i.e. 62.25 EUR per pair. The profit falls by 193.5 MU \(= 19{,}350\) EUR: reject the second order.

The short-term lower limit price, 24 EUR, is the lowest variable cost per unit, reached at 12 QU. Above 12 QU, the costs per unit rise again, and the extra units cost much more than the average: the second order costs \(\frac{373.5}{6} = 62.25\) MU per QU, far above the offer of 30 MU per QU. A special order must cover the additional costs it causes. The first order passed because 13 QU is still close to the cheapest output (\(\overline{C}_v(13) = 24.25\) MU per QU), so its additional costs were only 21.75 MU per QU.

Typical mistake: in c), rejecting the order because 30 EUR is below the long-term lower limit price of 60 EUR. The fixed costs are paid with or without the order, so it only has to cover its additional costs.

Problem 10: Fresh Pesto in Jars (xx)

A small company makes fresh pesto (a sauce of basil, nuts, cheese and olive oil) and sells it in jars to delicatessen shops. Its monthly total costs are

\[C(x) = 1{,}600 + 4.15x + 0.0004x^2\]

in EUR for \(x\) jars per month. The shops pay 5.60 EUR per jar.

  1. Determine the contribution margin per jar as a function of \(x\). Compute it at 1,000 and at 2,000 jars per month, and determine the output at which it becomes zero.

  2. Show that the company has no break-even point at this price.

(for verification: the discriminant is \(-0.4575\))

  1. Compute the long-term lower limit price and explain the result of b) with it.

  2. The owner wants to raise the price to the lowest price at which the company can at least break even. Determine this price with the discriminant of the break-even equation and compare it with c).

Hint: Write the break-even equation with the unknown price \(p\) and set its discriminant equal to 0.

a) \(\overline{C}_v(x) = 4.15 + 0.0004x\), so the contribution margin per jar is

\[5.60 - \overline{C}_v(x) = 1.45 - 0.0004x\]

At 1,000 jars: \(1.45 - 0.4 = 1.05\) EUR. At 2,000 jars: \(1.45 - 0.8 = 0.65\) EUR.

Zero: \(1.45 - 0.0004x = 0 \iff x = 3{,}625\) jars. From this output on, the price no longer covers the variable cost per jar.

b) Break-even: \(R(x) = C(x)\) with \(R(x) = 5.6x\):

\[5.6x = 1600 + 4.15x + 0.0004x^2 \iff 0.0004x^2 - 1.45x + 1600 = 0\]

Discriminant of the quadratic formula:

\[b^2 - 4ac = (-1.45)^2 - 4 \cdot 0.0004 \cdot 1600 = 2.1025 - 2.56 = -0.4575 < 0\]

The equation has no real solution: the revenue never reaches the costs, so there is no break-even point.

c) \(\overline{C}(x) = \frac{1600}{x} + 4.15 + 0.0004x\):

\[\overline{C}'(x) = -\frac{1600}{x^2} + 0.0004 = 0 \iff x^2 = 4{,}000{,}000 \iff x = 2{,}000\]

\(\overline{C}''(x) = \frac{3200}{x^3} > 0\), so this is a minimum: \(\overline{C}(2000) = 0.80 + 4.15 + 0.80 = 5.75\) EUR.

The long-term lower limit price is 5.75 EUR per jar. The price of 5.60 EUR is below it, so at every output the price is below the average total cost: each jar brings less than it costs on average. A break-even point is impossible.

d) With the price \(p\), the break-even equation is \(0.0004x^2 - (p - 4.15)x + 1600 = 0\). It has at least one solution if the discriminant is not negative. The lowest such price makes it zero:

\[(p - 4.15)^2 - 4 \cdot 0.0004 \cdot 1600 = 0 \iff (p - 4.15)^2 = 2.56 \iff p - 4.15 = \pm 1.6\]

The root \(p - 4.15 = -1.6\) would give the negative break-even output \(x = \frac{-1.6}{2 \cdot 0.0004} = -2{,}000\), so only \(p - 4.15 = 1.6\) is possible:

\[p = 5.75 \text{ EUR}\]

Then the only break-even point is \(x = \frac{1.6}{2 \cdot 0.0004} = 2{,}000\) jars. This is exactly the long-term lower limit price from c) and its output: at 5.75 EUR, the company covers all its costs at 2,000 jars and nowhere else.

Typical mistake: in b), stopping at the negative discriminant. Say what it means for the company: revenue and costs never meet, so it makes a loss at every output.

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: Dehumidifiers for DIY Stores [22 pts. total]

A company builds dehumidifiers (devices that remove moisture from the air in cellars and flats) and sells them to DIY stores. Its monthly total costs are

\[C(x) = x^3 - 29x^2 + 300x + 768\]

MU for an output of \(x\) QU. Many companies sell dehumidifiers, so the company cannot set its own price: the DIY stores pay the market price of 228 EUR per dehumidifier, however many the company sells. (1 QU = 100 dehumidifiers, 1 MU = 100 EUR) Round to two decimal places.

Part a)

The unit costs are the total costs per QU, \(\overline{C}(x) = \frac{C(x)}{x}\). Compute the output volume with the lowest unit costs. Also, determine the long term price floor (the long-term lower limit price). Give it in EUR per dehumidifier.

Hint: The cubic equation has a whole-number solution between 10 and 20.

(for verification: the long term price floor is 140 MU per QU)

[5 pts.]

Unit costs:

\[\overline{C}(x) = \frac{C(x)}{x} = x^2 - 29x + 300 + \frac{768}{x}\]

Necessary condition:

\[\overline{C}'(x) = 2x - 29 - \frac{768}{x^2} = 0 \quad \Big| \cdot x^2 \quad \Rightarrow \quad 2x^3 - 29x^2 - 768 = 0\]

Testing whole numbers (or the polynomial solver MENU → A → 2, degree 3, with \(a = 2\), \(b = -29\), \(c = 0\), \(d = -768\)): \(x = 16\) gives \(8192 - 7424 - 768 = 0\) \(\checkmark\).

Sufficient condition: \(\overline{C}''(x) = 2 + \frac{1536}{x^3} > 0\) for every \(x > 0\), so \(x = 16\) is a minimum.

Long term price floor:

\[\overline{C}(16) = 256 - 464 + 300 + 48 = 140 \text{ MU per QU}\]

The unit costs are lowest at 16 QU, i.e. 1,600 dehumidifiers per month. 1 MU per QU is 100 EUR per 100 dehumidifiers, i.e. 1 EUR per dehumidifier: the long term price floor is 140 EUR per dehumidifier.

Grading: 1 Pt. \(\overline{C}(x)\) and \(\overline{C}'(x)\), 1 Pt. \(2x^3 - 29x^2 - 768 = 0\), 1 Pt. \(x = 16\) with \(\overline{C}''(16) > 0\), 1 Pt. \(\overline{C}(16) = 140\) MU per QU, 1 Pt. 140 EUR per dehumidifier.

Part b)

Determine the lowest variable costs per unit and the short-term lower limit price. Explain what the company should do if the market price ever falls below this price. [3 pts.]

Variable costs per unit: \(C_f = 768\), so

\[\overline{C}_v(x) = \frac{C(x) - 768}{x} = x^2 - 29x + 300\]

\[\overline{C}_v'(x) = 2x - 29 = 0 \iff x = 14.5, \qquad \overline{C}_v''(x) = 2 > 0 \;\Rightarrow\; \text{minimum}\]

\[\overline{C}_v(14.5) = 210.25 - 420.5 + 300 = 89.75 \text{ MU per QU}\]

The variable costs per unit are lowest at 14.5 QU (1,450 dehumidifiers). The short-term lower limit price is 89.75 EUR per dehumidifier.

Below this price, the company does not even cover its variable costs: every dehumidifier sold makes the loss larger. It should stop production.

Grading: 1 Pt. \(\overline{C}_v(x)\) and \(x = 14.5\), 1 Pt. 89.75 EUR per dehumidifier, 1 Pt. stop production with the reason.

Part c)

Give a term for the profit function \(P(x)\). At 8 QU, the unit costs are exactly as high as the market price. Show this and explain why 8 QU is therefore a break-even point. Then determine the profit limit (upper break-even point) and decide in which output range the company makes a profit.

Hint: Divide \(P(x)\) by \((x - 8)\).

[6 pts.]

Profit function: \(R(x) = 228x\), so

\[P(x) = 228x - (x^3 - 29x^2 + 300x + 768) = -x^3 + 29x^2 - 72x - 768\]

Break-even point:

\[\overline{C}(8) = 64 - 232 + 300 + \frac{768}{8} = 64 - 232 + 300 + 96 = 228 \text{ MU per QU} \quad \checkmark\]

At 8 QU, each QU costs on average exactly the price of 228 MU. So the revenue \(228 \cdot 8\) equals the total costs \(\overline{C}(8) \cdot 8 = C(8)\), and \(P(8) = 0\). Check: \(P(8) = -512 + 1856 - 576 - 768 = 0\) \(\checkmark\)

Polynomial long division by \((x - 8)\):

\[\begin{array}{l} (-x^3 + 29x^2 - 72x - 768) : (x - 8) = -x^2 + 21x + 96 \\ \underline{-(-x^3 + 8x^2)} \\ \phantom{(-x^3}\;\; 21x^2 - 72x \\ \phantom{(-x^3}\;\; \underline{-(21x^2 - 168x)} \\ \phantom{(-x^3 + 29x^2}\;\; 96x - 768 \\ \phantom{(-x^3 + 29x^2}\;\; \underline{-(96x - 768)} \\ \phantom{(-x^3 + 29x^2 - 72x}\;\; 0 \end{array}\]

Profit limit: \(-x^2 + 21x + 96 = 0 \iff x^2 - 21x - 96 = 0\). Quadratic formula:

\[x = \frac{21 \pm \sqrt{441 + 384}}{2} = \frac{21 \pm \sqrt{825}}{2}, \qquad x_1 \approx 24.86, \quad x_2 \approx -3.86\]

\(x_2\) is negative and has no meaning. The profit limit lies at about 24.86 QU, i.e. about 2,486 dehumidifiers per month. Since \(P(0) = -768 < 0\), the company makes a loss below 8 QU, so 8 QU is the (lower) break-even point.

Profit range: \(P(x) = -(x - 8)(x - 24.86)(x + 3.86)\) approximately. For \(8 < x < 24.86\), the factors have the signs \((+)\), \((-)\) and \((+)\), so \(P(x) > 0\). The company makes a profit between 8 QU and about 24.86 QU, i.e. between 800 and about 2,486 dehumidifiers per month.

Grading: 1 Pt. \(P(x)\), 1 Pt. \(\overline{C}(8) = 228\) with the reason that revenue equals costs, 2 Pts. long division, 1 Pt. profit limit \(\approx 24.86\) QU, 1 Pt. profit range \(8 < x < 24.86\).

Part d)

(for verification: \(P(x) = -x^3 + 29x^2 - 72x - 768\))

Compute the maximum profit and prove it really is a maximum. Give the output volume and the maximum profit in EUR. [4 pts.]

Necessary condition:

\[P'(x) = -3x^2 + 58x - 72 = 0 \iff 3x^2 - 58x + 72 = 0\]

\[x = \frac{58 \pm \sqrt{3364 - 864}}{6} = \frac{58 \pm 50}{6}, \qquad x_1 = 18, \quad x_2 = \frac{4}{3}\]

\(x_2 = \frac{4}{3}\) lies below the break-even point, in the loss range.

Sufficient condition: \(P''(x) = -6x + 58\), so \(P''(18) = -108 + 58 = -50 < 0\): \(P\) has a maximum at \(x = 18\) (\(P''(\frac{4}{3}) = 50 > 0\) is a minimum). The profit is 0 at both ends of the profit range from c), so no other output brings more.

Maximum profit:

\[P(18) = -5832 + 9396 - 1296 - 768 = 1500 \text{ MU} = 1500 \cdot 100 \text{ EUR} = 150{,}000 \text{ EUR}\]

The maximum profit is 150,000 EUR per month at 18 QU, i.e. 1,800 dehumidifiers.

Grading: 1 Pt. \(P'(x)\), 1 Pt. \(x = 18\) with the quadratic formula, 1 Pt. \(P''(18) < 0\) as proof, 1 Pt. \(P(18) = 1500\) MU \(= 150{,}000\) EUR.

Part e)

A large competitor enters the market, and the market price falls to 112 EUR per dehumidifier. Decide with your results from a) and b) whether the company should keep producing. At this price, the company plans to produce 15 QU per month. Compute the contribution margin per dehumidifier and in total, and the monthly profit or loss in EUR. Compare it with the loss if the company stops production. [4 pts.]

Decision: \(89.75 \le 112 < 140\) (EUR per dehumidifier). The price covers the variable costs, but not all costs. The company should keep producing in the short term, but it cannot stay in the market at this price in the long run.

Contribution margin:

\[\overline{C}_v(15) = 225 - 435 + 300 = 90 \text{ MU per QU}\]

Per dehumidifier: \(112 - 90 = 22\) EUR. In total: \(15 \cdot 22 = 330\) MU \(= 33{,}000\) EUR per month.

Loss: \(330 - 768 = -438\) MU, i.e. a loss of 43,800 EUR per month. Check: \(112 \cdot 15 - C(15) = 1680 - 2118 = -438\) MU \(\checkmark\)

If the company stops, it still has to pay the fixed costs of 768 MU and loses 76,800 EUR per month. Producing loses 33,000 EUR less.

Grading: 1 Pt. decision with both lower limit prices, 1 Pt. \(\overline{C}_v(15) = 90\) and 22 EUR per dehumidifier, 1 Pt. 33,000 EUR and the loss of 43,800 EUR, 1 Pt. comparison with 76,800 EUR.