Tasks 05-02 - The Derivative as Rate of Change

Section 05: Differential Calculus

A: Skills

Problem 1: Derivatives from the Definition at a Point (x)

Use the limit definition \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\) to find the derivative at the given point.

  1. \(f(x) = 7 - 2x\) at \(a = 4\)

  2. \(f(x) = x^2 + 5x\) at \(a = -1\)

  3. \(f(x) = 10 - 0.5x^2\) at \(a = 3\)

  4. \(f(x) = \frac{6}{x}\) at \(a = 3\)

    Hint: Write \(\frac{6}{3 + h} - 2\) as one fraction first.

  5. Decide for each function whether it increases or decreases at the given point. State which of the four graphs is steepest there.

Every time the same four steps: set up the difference quotient, expand, cancel \(h\), let \(h \to 0\).

  1. \(f(4) = 7 - 8 = -1\) and \(f(4 + h) = 7 - 8 - 2h = -1 - 2h\):

\[f'(4) = \lim_{h \to 0} \frac{(-1 - 2h) - (-1)}{h} = \lim_{h \to 0} \frac{-2h}{h} = \lim_{h \to 0} (-2) = -2\]

  1. \(f(-1) = 1 - 5 = -4\) and \(f(-1 + h) = (1 - 2h + h^2) + (-5 + 5h) = -4 + 3h + h^2\):

\[f'(-1) = \lim_{h \to 0} \frac{(-4 + 3h + h^2) - (-4)}{h} = \lim_{h \to 0} \frac{h(3 + h)}{h} = \lim_{h \to 0} (3 + h) = 3\]

  1. \(f(3) = 10 - 4.5 = 5.5\) and \(f(3 + h) = 10 - 0.5(9 + 6h + h^2) = 5.5 - 3h - 0.5h^2\):

\[f'(3) = \lim_{h \to 0} \frac{(5.5 - 3h - 0.5h^2) - 5.5}{h} = \lim_{h \to 0} \frac{h(-3 - 0.5h)}{h} = \lim_{h \to 0} (-3 - 0.5h) = -3\]

  1. \(f(3) = 2\). One fraction in the numerator:

\[\frac{6}{3 + h} - 2 = \frac{6 - 2(3 + h)}{3 + h} = \frac{-2h}{3 + h}\]

\[f'(3) = \lim_{h \to 0} \frac{1}{h} \cdot \frac{-2h}{3 + h} = \lim_{h \to 0} \frac{-2}{3 + h} = -\frac{2}{3}\]

  1. A positive derivative means the function increases, a negative one that it decreases:
Function Derivative Behaviour at the point
a) \(7 - 2x\) \(-2\) decreases by 2 per unit of \(x\) (everywhere, it is a line)
b) \(x^2 + 5x\) \(3\) increases, about 3 per unit of \(x\)
c) \(10 - 0.5x^2\) \(-3\) decreases, about 3 per unit of \(x\)
d) \(\frac{6}{x}\) \(-\frac{2}{3}\) decreases, about \(\frac{2}{3}\) per unit of \(x\)

The graphs of b) and c) are steepest at their points: \(|3| = |-3| = 3\) is the largest absolute value. The sign only tells the direction.

Problem 2: Average Rate of Change (x)

A drugstore chain counts its branches at the end of each year. \(t\) is the number of years after the end of 2019.

\(t\) (years) 0 1 3 4 6
\(B(t)\) (branches) 80 110 200 215 275

  1. Compute the average rate of change of \(B\) on \([0; 1]\), \([1; 3]\), \([3; 4]\) and \([4; 6]\) with units. Decide in which period the chain grew fastest on average.

  2. Compute the average rate of change on \([0; 6]\). A trainee takes the mean of the four rates from a) instead. Explain why this gives a different value.

  3. Now let \(g(x) = 3x^2 - 2x\). Compute the average rate of change of \(g\) on \([1; 2]\), \([1; 1.5]\), \([1; 1.1]\) and \([1; 1.01]\).

  4. State the value that the average rates in c) approach. Confirm it by computing \(g'(1)\) with the limit definition.

  1. Average rate of change \(= \frac{B(t_2) - B(t_1)}{t_2 - t_1}\), in branches per year:
Period Computation Average rate
\([0; 1]\) \(\frac{110 - 80}{1}\) 30 branches per year
\([1; 3]\) \(\frac{200 - 110}{2}\) 45 branches per year
\([3; 4]\) \(\frac{215 - 200}{1}\) 15 branches per year
\([4; 6]\) \(\frac{275 - 215}{2}\) 30 branches per year

The chain grew fastest on average from the end of 2020 to the end of 2022 (\([1; 3]\)).

Typical mistake: taking \(200 - 110 = 90\) branches per year on \([1; 3]\). The period is two years long, so the change must be divided by 2.

  1. \(\frac{275 - 80}{6 - 0} = \frac{195}{6} = 32.5\) branches per year.

The trainee gets \(\frac{30 + 45 + 15 + 30}{4} = 30\) branches per year. The four periods have different lengths (1, 2, 1 and 2 years). The two-year periods count twice in the whole change: \(\frac{30 \cdot 1 + 45 \cdot 2 + 15 \cdot 1 + 30 \cdot 2}{6} = \frac{195}{6} = 32.5\).

  1. \(g(1) = 3 - 2 = 1\):
Interval \(g\) at the right end Average rate
\([1; 2]\) \(g(2) = 12 - 4 = 8\) \(\frac{8 - 1}{1} = 7\)
\([1; 1.5]\) \(g(1.5) = 6.75 - 3 = 3.75\) \(\frac{2.75}{0.5} = 5.5\)
\([1; 1.1]\) \(g(1.1) = 3.63 - 2.2 = 1.43\) \(\frac{0.43}{0.1} = 4.3\)
\([1; 1.01]\) \(g(1.01) = 3.0603 - 2.02 = 1.0403\) \(\frac{0.0403}{0.01} = 4.03\)
  1. The average rates approach 4. With the limit definition:

\[g(1 + h) = 3(1 + 2h + h^2) - 2(1 + h) = 1 + 4h + 3h^2\]

\[g'(1) = \lim_{h \to 0} \frac{(1 + 4h + 3h^2) - 1}{h} = \lim_{h \to 0} (4 + 3h) = 4\]

The difference quotient \(4 + 3h\) explains the table: \(h = 1\) gives 7, \(h = 0.5\) gives 5.5, \(h = 0.1\) gives 4.3, \(h = 0.01\) gives 4.03.

Problem 3: The Derivative as a Function (xx)

Let \(f(x) = 2x^2 - 12x + 7\).

  1. Use the limit definition to find \(f'(x)\) for every \(x\). Write the result in two notations.

  2. Show with the limit definition that \(g(x) = 5 - 3x\) has the derivative \(g'(x) = -3\) for every \(x\). Explain this result with the graph of \(g\).

  3. Compute \(f'(0)\), \(f'(2)\), \(f'(4)\) and \(f'(5)\). Decide at which of these \(x\)-values \(f\) increases and at which it decreases.

  4. Determine the point of the graph of \(f\) at which the tangent line is horizontal. Check your result with the vertex formula.

  5. Give the equation of the tangent line of \(f\) at \(x = 4\).

    Hint: The tangent line at \(x = 4\) is the line through \((4, f(4))\) with the slope \(f'(4)\).

  1. Replace \(a\) by \(x\) in the definition:

\[f(x + h) = 2(x^2 + 2xh + h^2) - 12(x + h) + 7 = 2x^2 + 4xh + 2h^2 - 12x - 12h + 7\]

\[f(x + h) - f(x) = 4xh + 2h^2 - 12h = h(4x + 2h - 12)\]

\[f'(x) = \lim_{h \to 0} \frac{h(4x + 2h - 12)}{h} = \lim_{h \to 0} (4x + 2h - 12) = 4x - 12\]

In two notations: \(f'(x) = 4x - 12\) and \(\frac{df}{dx} = 4x - 12\).

  1. \(g(x + h) - g(x) = (5 - 3x - 3h) - (5 - 3x) = -3h\), so

\[g'(x) = \lim_{h \to 0} \frac{-3h}{h} = -3\]

The graph of \(g\) is a line with slope \(-3\). A line has the same steepness everywhere, and its tangent line at every point is the line itself.

  1. \(f'(0) = -12\), \(f'(2) = -4\), \(f'(4) = 4\), \(f'(5) = 8\).

\(f\) decreases at \(x = 0\) and \(x = 2\) (negative derivative) and increases at \(x = 4\) and \(x = 5\) (positive derivative).

  1. Horizontal tangent line means slope 0:

\[f'(x) = 4x - 12 = 0 \iff x = 3, \qquad f(3) = 18 - 36 + 7 = -11\]

The tangent line is horizontal at \((3, -11)\).

Check with the vertex formula: \(x_V = -\frac{b}{2a} = -\frac{-12}{2 \cdot 2} = 3\) ✓. Since \(a = 2 > 0\), the parabola opens upward, and \((3, -11)\) is its lowest point.

  1. \(f(4) = 32 - 48 + 7 = -9\) and \(f'(4) = 4\). Point-slope form:

\[y - (-9) = 4(x - 4) \iff y = 4x - 25\]

Problem 4: Corners and Differentiability (xx)

  1. Let

    \[f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 3x - 2 & \text{if } x > 1 \end{cases}\]

    (1) Show that \(f\) is continuous at \(x = 1\).

    (2) Determine the one-sided limits \(\lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h}\) and \(\lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}\).

    (3) Decide whether \(f\) is differentiable at \(x = 1\).

    Hint: For \(h < 0\), the point \(1 + h\) lies left of 1, so use the first line of \(f\). For \(h > 0\), use the second line.

  2. Let

    \[g(x) = \begin{cases} x^2 + 1 & \text{if } x \leq 2 \\ 4x - 3 & \text{if } x > 2 \end{cases}\]

    Investigate in the same way whether \(g\) is differentiable at \(x = 2\).

  3. Let

    \[k(x) = \begin{cases} 2x & \text{if } x < 3 \\ 2x + 1 & \text{if } x \geq 3 \end{cases}\]

    A student says: “Both lines have the slope 2, so \(k'(3) = 2\).” Decide whether the student is right and give a reason.

  4. A metal workshop buys aluminium from a wholesaler. It pays 8 EUR per kg for the first 500 kg in a month and 6 EUR per kg for every kg above 500 kg. \(K(x)\) is the monthly cost in EUR for \(x\) kg.

    (1) Give \(K(x)\) as a piecewise function and show that \(K\) is continuous at \(x = 500\).

    (2) Determine the one-sided limits of the difference quotient of \(K\) at \(x = 500\) and decide whether \(K'(500)\) exists.

    (3) Explain what the two one-sided values mean for the workshop.

    1. \(f(1) = 1^2 = 1\). \(\lim_{x \to 1^-} f(x) = 1\) and \(\lim_{x \to 1^+} f(x) = 3 - 2 = 1\). The limit exists and equals \(f(1)\): \(f\) is continuous at \(x = 1\).

    2. From the left (\(h < 0\), first line):

    \[\lim_{h \to 0^-} \frac{(1 + h)^2 - 1}{h} = \lim_{h \to 0^-} \frac{2h + h^2}{h} = \lim_{h \to 0^-} (2 + h) = 2\]

    From the right (\(h > 0\), second line):

    \[\lim_{h \to 0^+} \frac{3(1 + h) - 2 - 1}{h} = \lim_{h \to 0^+} \frac{3h}{h} = 3\]

    1. The one-sided limits differ (\(2 \neq 3\)), so the limit of the difference quotient does not exist. \(f\) is not differentiable at \(x = 1\). The graph has a corner there: continuous, but not differentiable (like \(|x|\) at \(x = 0\)).

  2. \(g(2) = 5\), \(\lim_{x \to 2^-} g(x) = 5\) and \(\lim_{x \to 2^+} g(x) = 8 - 3 = 5\): \(g\) is continuous at \(x = 2\).

\[\lim_{h \to 0^-} \frac{(2 + h)^2 + 1 - 5}{h} = \lim_{h \to 0^-} \frac{4h + h^2}{h} = 4, \qquad \lim_{h \to 0^+} \frac{4(2 + h) - 3 - 5}{h} = \lim_{h \to 0^+} \frac{4h}{h} = 4\]

Both one-sided limits are 4. \(g\) is differentiable at \(x = 2\) with \(g'(2) = 4\). The line \(y = 4x - 3\) continues the parabola without a corner.

  1. The student is wrong. \(k(3) = 7\), but \(\lim_{x \to 3^-} k(x) = 6\). \(k\) has a jump at \(x = 3\) and is not continuous there. If \(k\) were differentiable at \(x = 3\), it would be continuous there (theorem). So \(k\) is not differentiable at \(x = 3\).

The difference quotient shows it, too: from the left, \(\frac{k(3 + h) - k(3)}{h} = \frac{2(3 + h) - 7}{h} = \frac{2h - 1}{h}\). The numerator approaches \(-1\) and \(h\) is a small negative number, so the quotient grows without bound. Equal slopes are not enough: the pieces must also meet.

    1. Up to 500 kg: \(K(x) = 8x\). Above: \(K(x) = 8 \cdot 500 + 6(x - 500) = 1000 + 6x\).

    \[K(x) = \begin{cases} 8x & \text{if } 0 \leq x \leq 500 \\ 1000 + 6x & \text{if } x > 500 \end{cases}\]

    \(K(500) = 4{,}000\) EUR, \(\lim_{x \to 500^-} K(x) = 4{,}000\) and \(\lim_{x \to 500^+} K(x) = 1000 + 3000 = 4{,}000\) EUR: continuous.

    1. From the left: \(\frac{8(500 + h) - 4000}{h} = \frac{8h}{h} = 8\). From the right: \(\frac{1000 + 6(500 + h) - 4000}{h} = \frac{6h}{h} = 6\). The one-sided limits are 8 and 6. They differ, so \(K'(500)\) does not exist (a corner).

    2. Just below 500 kg, one more kg costs 8 EUR. Just above 500 kg, one more kg costs only 6 EUR. The marginal cost jumps from 8 EUR per kg to 6 EUR per kg, and at exactly 500 kg there is no single marginal cost.

B: Set Up & Check

Problem 5: From Text to Derivatives (xx)

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

  1. “A furniture workshop builds bookshelves. Its monthly costs for \(x\) bookshelves are \(C(x) = 1800 + 45x + 0.25x^2\) EUR, with the derivative \(C'(x) = 45 + 0.5x\).” The owner says: “Our marginal cost at 100 bookshelves is 95 EUR.” Write this statement with the derivative.

    Check: Compute \(C'(100)\) and the exact cost of the 101st bookshelf, \(C(101) - C(100)\). Are both close to 95 EUR?

  2. “A podcast has paying supporters. \(S(t) = 1200 + 90t + 30t^2\) is their number \(t\) years after the start of 2022.” The host says: “From the start of 2022 to the start of 2026, we gained 210 supporters per year on average.” Write this statement as an average rate of change with units.

    Check: Compute the gain in each single year. Did the podcast gain 210 supporters in every year?

  3. “A dairy makes organic yoghurt. \(C(x)\) is its daily cost in EUR for \(x\) litres of yoghurt, and \(C'(x) = 0.9 + 0.004x\).” The production manager says: “At 250 litres per day, our costs rise by 1.90 EUR per litre.” Write this statement with the derivative and give its unit.

    Check: Compute \(C'(250)\). What would \(C(250) = 1.9\) mean instead?

  4. “A garden centre sells gas grills. Its monthly profit is \(P(x) = -2x^2 + 140x - 2600\) EUR for \(x\) grills, and \(P'(x) = 140 - 4x\).” The manager says: “At 30 grills per month, our profit still increases.” Write this statement with the derivative.

    Check: Compute \(P'(30)\) and \(P(30)\). Does the garden centre make a profit with 30 grills?

  5. “A company produces electric toothbrushes. At an output of \(x\) toothbrushes per day, its marginal cost is \(C'(x) = 14 + 0.03x\) EUR per toothbrush and its marginal revenue is \(R'(x) = 40 - 0.08x\) EUR per toothbrush.” The sales manager says: “At 200 toothbrushes per day, the next toothbrush costs more than it earns.” Write this statement with the derivatives.

    Check: Compute \(C'(200)\) and \(R'(200)\). Is the sales manager right? What should the company do?

  1. The marginal cost is the derivative of the cost function:

    \[C'(100) = 95 \text{ (EUR per bookshelf)}\]

    Check: yes. \(C'(100) = 45 + 0.5 \cdot 100 = 95\) EUR ✓. The 101st bookshelf costs exactly \(C(101) - C(100) = 45 + 0.25 \cdot (101^2 - 100^2) = 45 + 0.25 \cdot 201 = 95.25\) EUR. The marginal cost is approximately the cost of the next bookshelf.

    Typical mistake: computing the average cost \(\frac{C(100)}{100} = \frac{8800}{100} = 88\) EUR. The average cost spreads the fixed costs over all bookshelves. The marginal cost only looks at the next bookshelf.

  2. With \(t = 0\) at the start of 2022 and \(t = 4\) at the start of 2026:

    \[\frac{S(4) - S(0)}{4 - 0} = 210 \text{ supporters per year}\]

    Check: \(S(0) = 1{,}200\) and \(S(4) = 1200 + 360 + 480 = 2{,}040\), so \(\frac{840}{4} = 210\) ✓. The single years: \(S(1) = 1{,}320\), \(S(2) = 1{,}500\), \(S(3) = 1{,}740\), so the gains are 120, 180, 240 and 300 supporters. The podcast did not gain 210 in every year. The average rate hides that the growth speeds up.

    Typical mistake: dividing by 5 because the statement names five years (2022 to 2026). From the start of 2022 to the start of 2026 are only four years, \(t\) runs from 0 to 4.

  3. The rate at which the costs rise at 250 litres is the derivative at 250:

    \[C'(250) = 1.9, \quad \text{unit: EUR per litre}\]

    Check: \(C'(250) = 0.9 + 0.004 \cdot 250 = 0.9 + 1 = 1.9\) ✓. \(C(250) = 1.9\) would mean that 250 litres of yoghurt cost only 1.90 EUR in total.

    Typical mistake: writing \(C'(1.9) = 250\). The output (250 litres) goes into \(C'\), and the rate (1.90 EUR per litre) comes out.

  4. “The profit still increases” means a positive slope of \(P\):

    \[P'(30) > 0\]

    Check: \(P'(30) = 140 - 120 = 20 > 0\) ✓. But \(P(30) = -1800 + 4200 - 2600 = -200\) EUR: the garden centre makes a loss of 200 EUR with 30 grills. The loss gets smaller by about 20 EUR with every further grill, so the profit increases, but it is still negative.

    Typical mistake: writing \(P(30) > 0\). “The profit increases” is a statement about the change of \(P\), not about its sign.

  5. “The next toothbrush costs more than it earns” means the marginal cost is larger than the marginal revenue:

    \[C'(200) > R'(200) \quad \text{(equivalently } P'(200) = R'(200) - C'(200) < 0\text{)}\]

    Check: \(C'(200) = 14 + 6 = 20\) EUR and \(R'(200) = 40 - 16 = 24\) EUR per toothbrush. The sales manager is wrong: the next toothbrush costs about 20 EUR and brings about 24 EUR more revenue. Since \(MR > MC\), the company should produce more.

    Typical mistake: comparing the total costs \(C(200)\) with the total revenue \(R(200)\). \(C(200) > R(200)\) would mean a loss on the whole output. The statement is about the next toothbrush only.

Problem 6: One Paragraph, All Conditions (xx)

A tool manufacturer builds cordless drills. Its monthly total costs are a quadratic function \(C\) of the output \(x\) in QU, with \(C\) in MU, and its fixed costs are 30 MU. At an output of 4 QU, the marginal cost is 5 MU per QU. When the output grows from 4 QU to 10 QU, the costs rise on average by 8 MU per QU. The price per QU falls linearly with the output, from a maximum price of 26 MU per QU to 0 at the market saturation of 20 QU. The management looks for the output at which the marginal revenue equals the marginal cost, and it wants to know whether the profit still increases at 6 QU.

(1 QU = 100 drills, 1 MU = 1,000 EUR)

List every condition and every function in this paragraph as a mathematical statement. Do not solve anything.

Check: An assistant writes the condition for the marginal cost as \(C(4) = 5\). Name the mistake.

  1. Cost function: \(C(x) = ax^2 + bx + c\) (\(x\) in QU, \(C\) in MU)
  2. Fixed costs: \(C(0) = 30\), i.e. \(c = 30\)
  3. Marginal cost at 4 QU: \(C'(4) = 5\)
  4. Average rate of change from 4 QU to 10 QU: \(\frac{C(10) - C(4)}{10 - 4} = 8\)
  5. Linear demand: \(p(x) = mx + n\) with \(p(0) = 26\) (maximum price) and \(p(20) = 0\) (market saturation)
  6. Revenue: \(R(x) = x \cdot p(x)\)
  7. Profit: \(P(x) = R(x) - C(x)\)
  8. Output wanted: \(R'(x) = C'(x)\) (marginal revenue = marginal cost)
  9. Question: is \(P'(6) > 0\)?

Check: \(C(4) = 5\) says that the total costs of 4 QU are 5 MU, even less than the fixed costs of 30 MU. The marginal cost is the derivative, \(C'(4) = 5\).

Typical mistake: writing condition 4 as \(C(10) - C(4) = 8\). This is the whole increase of the costs. An average rate of change divides it by the length of the interval, here 6 QU.

C: Applied

Problem 7: Pre-Orders for a New E-Book Reader (xx)

A small electronics company takes pre-orders (orders placed before the product is on sale) for its new e-book reader in the 40 days before the launch. The total number of pre-orders \(t\) days after the start of the pre-order phase is

\[N(t) = -2t^2 + 160t, \qquad 0 \leq t \leq 40\]

  1. Compute \(N(10)\), \(N(20)\) and \(N(30)\). Determine the average rate of change of \(N\) on \([0; 10]\), \([10; 20]\) and \([20; 30]\) with units and describe how the pre-orders develop.

  2. Estimate the rate of change of \(N\) on day 15 with the difference quotient \(\frac{N(15 + h) - N(15)}{h}\) for \(h = 1\), \(h = 0.1\) and \(h = 0.01\).

  3. Use the limit definition to find \(N'(t)\). Compute \(N'(15)\) and compare it with b).

(for verification: \(N'(t) = 160 - 4t\))

  1. Compute \(N'(10)\) and interpret it with units. Explain why the average rate of change on \([10; 20]\) is smaller than \(N'(10)\).

  2. The marketing team starts an advertising campaign as soon as fewer than 40 new pre-orders per day come in. Determine when the campaign starts and the number of pre-orders up to then.

  3. Compute \(N'(40)\) and check with the vertex formula that \(N\) reaches its largest value on the launch day. Explain why the company uses the model only up to \(t = 40\).

a) Values:

  • \(N(10) = -200 + 1600 = 1{,}400\) pre-orders
  • \(N(20) = -800 + 3200 = 2{,}400\) pre-orders
  • \(N(30) = -1800 + 4800 = 3{,}000\) pre-orders

Average rates of change:

Period Computation Average rate
\([0; 10]\) \(\frac{1400 - 0}{10}\) 140 pre-orders per day
\([10; 20]\) \(\frac{2400 - 1400}{10}\) 100 pre-orders per day
\([20; 30]\) \(\frac{3000 - 2400}{10}\) 60 pre-orders per day

The number of pre-orders keeps growing, but more and more slowly: from one period to the next, the average rate falls by 40 pre-orders per day.

b) \(N(15) = -450 + 2400 = 1{,}950\).

\(h\) \(N(15 + h)\) Difference quotient
1 \(N(16) = -512 + 2560 = 2{,}048\) \(\frac{98}{1} = 98\)
0.1 \(N(15.1) = -456.02 + 2416 = 1{,}959.98\) \(\frac{9.98}{0.1} = 99.8\)
0.01 \(N(15.01) = -450.6002 + 2401.6 = 1{,}950.9998\) \(\frac{0.9998}{0.01} = 99.98\)

Estimate: on day 15, about 100 new pre-orders per day come in.

c) Difference quotient at any \(t\):

\[N(t + h) = -2(t^2 + 2th + h^2) + 160t + 160h = -2t^2 - 4th - 2h^2 + 160t + 160h\]

\[N(t + h) - N(t) = -4th - 2h^2 + 160h = h(160 - 4t - 2h)\]

\[N'(t) = \lim_{h \to 0} (160 - 4t - 2h) = 160 - 4t\]

\(N'(15) = 160 - 60 = 100\) pre-orders per day. This is the value the estimates in b) approach: the difference quotient at \(t = 15\) is \(100 - 2h\), which gives 98, 99.8 and 99.98.

d) \(N'(10) = 160 - 40 = 120\) pre-orders per day: on day 10, new pre-orders come in at a rate of about 120 per day.

The rate falls during \([10; 20]\), from \(N'(10) = 120\) to \(N'(20) = 80\) pre-orders per day. The average rate on \([10; 20]\) also contains the slower later days, so it is smaller than the rate at the start of the period: \(100 < 120\).

e) Fewer than 40 new pre-orders per day:

\[160 - 4t < 40 \iff 120 < 4t \iff t > 30\]

On day 30 the rate is exactly \(N'(30) = 40\), after that it is lower. The campaign starts after 30 days. Up to then, \(N(30) = 3{,}000\) pre-orders have come in.

f) \(N'(40) = 160 - 160 = 0\).

Vertex of \(N(t) = -2t^2 + 160t\): \(a = -2 < 0\), so the parabola opens downward and its vertex is a maximum.

\[t_V = -\frac{b}{2a} = -\frac{160}{2 \cdot (-2)} = 40, \qquad N(40) = -3200 + 6400 = 3{,}200 \text{ pre-orders}\]

The largest value lies on the launch day, and the tangent line is horizontal there (\(N'(40) = 0\)) ✓.

For \(t > 40\), the model would give \(N'(t) < 0\): the total number of pre-orders would fall, as if customers cancelled their orders. The model only describes the pre-order phase. After the launch, the reader is sold normally and the pre-orders end.

Problem 8: Marginal Cost in a Guitar Workshop (xx)

A workshop builds electric guitars and sells them to music stores. Its monthly total costs for \(x\) guitars are

\[C(x) = 4000 + 60x + 0.5x^2 \text{ EUR}, \qquad 0 \leq x \leq 80\]

where 80 guitars per month is its capacity (the maximum possible output).

  1. Compute the total costs and the average cost per guitar (total costs divided by the number of guitars) for 40 guitars per month.

  2. Show with the limit definition that \(C'(x) = 60 + x\).

  3. Compute the marginal cost at 40 guitars and the exact cost of the 41st guitar, \(C(41) - C(40)\). Explain what the marginal cost means for the workshop and where the small difference comes from.

  4. Compute \(C'(20)\) and \(C'(60)\) and describe how the marginal cost changes with the output. Explain why the marginal cost at 40 guitars is much lower than the average cost from a).

  5. The workshop builds 40 guitars per month for its regular customers. A music store chain offers to buy additional guitars for 130 EUR each, less than the average cost from a). Determine the output up to which the next guitar costs less than 130 EUR. Compute the extra profit if the workshop builds 30 more guitars for the chain and decide whether it should accept the offer.

(for verification: 70 guitars)

a) \(C(40) = 4000 + 2400 + 800 = 7{,}200\) EUR. Average cost: \(\frac{7200}{40} = 180\) EUR per guitar.

b) Difference quotient:

\[C(x + h) - C(x) = 60h + 0.5\left((x + h)^2 - x^2\right) = 60h + 0.5(2xh + h^2) = h(60 + x + 0.5h)\]

\[C'(x) = \lim_{h \to 0} (60 + x + 0.5h) = 60 + x \quad \checkmark\]

c) \(C'(40) = 60 + 40 = 100\) EUR per guitar. At an output of 40 guitars, the costs rise by about 100 EUR per further guitar: the 41st guitar costs approximately 100 EUR.

Exact cost of the 41st guitar:

\[C(41) - C(40) = 60 + 0.5 \cdot (41^2 - 40^2) = 60 + 0.5 \cdot 81 = 100.50 \text{ EUR}\]

The difference of 0.50 EUR comes from the term \(0.5h\) in the difference quotient: one more guitar means \(h = 1\), which gives \(100 + 0.5 = 100.5\). The derivative lets \(h \to 0\) and drops this term.

d) \(C'(20) = 80\) EUR and \(C'(60) = 120\) EUR per guitar. The marginal cost rises by 1 EUR with every further guitar: the more guitars the workshop builds, the more each further guitar costs.

The average cost \(\frac{C(40)}{40} = \frac{4000}{40} + 60 + 0.5 \cdot 40 = 100 + 60 + 20 = 180\) EUR contains 100 EUR of fixed costs per guitar. The fixed costs arise anyway, so the next guitar does not add to them. It only adds about 100 EUR.

e) The next guitar costs less than 130 EUR as long as

\[C'(x) = 60 + x < 130 \iff x < 70\]

Up to 70 guitars, a further guitar costs less than 130 EUR. Check with exact values: the 70th guitar costs \(C(70) - C(69) = 60 + 0.5 \cdot 139 = 129.50\) EUR, the 71st guitar \(C(71) - C(70) = 60 + 0.5 \cdot 141 = 130.50\) EUR.

Extra profit for 30 more guitars (from 40 to 70, below the capacity of 80):

  • Extra revenue: \(30 \cdot 130 = 3{,}900\) EUR
  • Extra costs: \(C(70) - C(40) = (4000 + 4200 + 2450) - 7200 = 10{,}650 - 7{,}200 = 3{,}450\) EUR
  • Extra profit: \(3{,}900 - 3{,}450 = 450\) EUR

The workshop should accept the offer. The price of 130 EUR lies below the average cost of 180 EUR, but the fixed costs are already covered by the regular business. For the decision, only the extra costs count, and they are lower than the extra revenue.

Problem 9: Revenue and Profit of a Laptop Refurbisher (xx)

A company buys used laptops from businesses, refurbishes them (repairs, cleans and tests them) and sells them online. If it wants to sell \(x\) laptops per week, it can ask the price

\[p(x) = 480 - 2x \text{ EUR per laptop}, \qquad 0 \leq x \leq 240\]

Its weekly costs are \(C(x) = 6000 + 120x + x^2\) EUR. You may use the derivatives \(R'(x) = 480 - 4x\) and \(C'(x) = 120 + 2x\).

  1. Show that the weekly revenue is \(R(x) = 480x - 2x^2\). Compute the marginal revenue at 100 and at 150 laptops and interpret both values.

  2. Determine the number of laptops with the highest revenue with the vertex of \(R\) and give the price. Check that the marginal revenue is 0 there.

  3. Give the profit function \(P(x)\). Determine the number of laptops with the highest profit with the vertex of \(P\). Check that \(MR = MC\) there and give the price and the maximum profit.

(for verification: 60 laptops)

  1. Compute the marginal profit \(P'(x) = R'(x) - C'(x)\) at 50 and at 70 laptops. Explain what the company should do at each of these outputs.

  2. Determine the break-even point and the profit limit (upper break-even point). Compute the profit at the output with the highest revenue from b) and explain why the company should not aim for the highest revenue.

a) Revenue = number of laptops · price:

\[R(x) = x \cdot p(x) = x(480 - 2x) = 480x - 2x^2 \quad \checkmark\]

  • \(R'(100) = 480 - 400 = 80\) EUR per laptop: at 100 laptops, one more sold laptop brings about 80 EUR more revenue.
  • \(R'(150) = 480 - 600 = -120\) EUR per laptop: at 150 laptops, one more laptop lowers the revenue by about 120 EUR. To sell it, the company must lower the price for all laptops, and this loss is larger than the price of the extra laptop.

b) \(R(x) = -2x^2 + 480x\) with \(a = -2 < 0\): the parabola opens downward, so its vertex is a maximum.

\[x_V = -\frac{480}{2 \cdot (-2)} = 120, \qquad R(120) = 57{,}600 - 28{,}800 = 28{,}800 \text{ EUR}\]

The revenue is highest at 120 laptops per week, with the price \(p(120) = 480 - 240 = 240\) EUR per laptop.

Check: \(R'(120) = 480 - 480 = 0\) ✓. At the highest point of \(R\), the tangent line is horizontal.

c) Profit function:

\[P(x) = R(x) - C(x) = 480x - 2x^2 - 6000 - 120x - x^2 = -3x^2 + 360x - 6000\]

\(a = -3 < 0\): the parabola opens downward, the vertex is a maximum.

\[x_V = -\frac{360}{2 \cdot (-3)} = 60\]

Check: \(R'(60) = 480 - 240 = 240\) and \(C'(60) = 120 + 120 = 240\), so \(MR = MC\) ✓.

Price: \(p(60) = 480 - 120 = 360\) EUR per laptop. Maximum profit: \(P(60) = -10{,}800 + 21{,}600 - 6{,}000 = 4{,}800\) EUR per week.

d) Marginal profit:

  • \(P'(50) = R'(50) - C'(50) = 280 - 220 = 60\) EUR per laptop. \(MR > MC\): the next laptop adds about 60 EUR of profit, so the company should sell more.
  • \(P'(70) = R'(70) - C'(70) = 200 - 260 = -60\) EUR per laptop. \(MR < MC\): the next laptop reduces the profit by about 60 EUR, so the company should sell fewer.

Below 60 laptops the profit still increases, above 60 it decreases. This fits the maximum at 60 laptops from c).

e) Zeros of the profit function:

\[-3x^2 + 360x - 6000 = 0 \;\big|\; : (-3) \iff x^2 - 120x + 2000 = 0\]

Quadratic formula with \(\Delta = 120^2 - 4 \cdot 2000 = 6400\) and \(\sqrt{6400} = 80\):

\[x_{1,2} = \frac{120 \pm 80}{2}, \qquad x_1 = 20, \quad x_2 = 100\]

\(P(0) = -6{,}000 < 0\), so the loss turns into a profit at 20 laptops. Break-even point: 20 laptops per week. Profit limit: 100 laptops per week.

Profit at the highest revenue: \(P(120) = -43{,}200 + 43{,}200 - 6{,}000 = -6{,}000\) EUR. With 120 laptops, the company makes a loss of 6,000 EUR per week, because 120 lies above the profit limit.

At 120 laptops, the marginal revenue is 0, but the marginal cost is \(C'(120) = 360\) EUR. The last laptops cost about 360 EUR each and bring no extra revenue. The highest revenue ignores the costs. The company should aim for \(MR = MC\), not for \(MR = 0\).

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: Electric Outboard Motors [20 pts. total]

A young company builds electric outboard motors (motors that are attached to the back of a small boat) and sells them to boat dealers. Its market research shows that the price-demand relation is linear. The monthly total costs for \(x\) motors are

\[C(x) = 40000 + 600x + 3x^2 \text{ EUR}, \qquad 0 \leq x \leq 400\]

and the marginal cost is \(C'(x) = 600 + 6x\) EUR per motor.

Part a)

Determine the linear demand function \(p(x)\) assuming a maximum price of 2,400 EUR per motor and a market saturation at 400 motors per month. Show that the revenue function is given by \(R(x) = 2400x - 6x^2\). [4 pts.]

Two points of the demand line:

  • Maximum price: at 2,400 EUR nobody buys, so \(p(0) = 2400\)
  • Market saturation: even at a price of 0 EUR, the dealers take only 400 motors, so \(p(400) = 0\)

Slope:

\[m = \frac{0 - 2400}{400 - 0} = -6\]

Each further motor per month lowers the price by 6 EUR.

Demand function: \(p(x) = 2400 - 6x\) EUR per motor, \(0 \leq x \leq 400\).

Revenue:

\[R(x) = x \cdot p(x) = x(2400 - 6x) = 2400x - 6x^2 \quad \checkmark\]

Grading: 1 Pt. the points \((0, 2400)\) and \((400, 0)\), 1 Pt. slope \(-6\), 1 Pt. \(p(x) = 2400 - 6x\), 1 Pt. \(R(x) = x \cdot p(x) = 2400x - 6x^2\).

Part b)

Compute the average rate of change of the revenue when the sales grow from 100 to 200 motors per month and when they grow from 200 to 300 motors per month. Give the units and interpret both values. [3 pts.]

Revenues:

  • \(R(100) = 240{,}000 - 60{,}000 = 180{,}000\) EUR
  • \(R(200) = 480{,}000 - 240{,}000 = 240{,}000\) EUR
  • \(R(300) = 720{,}000 - 540{,}000 = 180{,}000\) EUR

Average rates of change:

\[\frac{R(200) - R(100)}{200 - 100} = \frac{60{,}000}{100} = 600 \text{ EUR per motor}, \qquad \frac{R(300) - R(200)}{300 - 200} = \frac{-60{,}000}{100} = -600 \text{ EUR per motor}\]

Interpretation: from 100 to 200 motors, each further motor brings on average 600 EUR more revenue. From 200 to 300 motors, the revenue falls on average by 600 EUR per further motor. To sell more motors, the company must lower the price for all motors, and above 200 motors the lower price weighs more than the larger number.

Grading: 1 Pt. the three revenues, 1 Pt. both average rates with the unit EUR per motor, 1 Pt. interpretation including the negative value.

Part c)

Show with the limit definition that the marginal revenue is \(R'(x) = 2400 - 12x\). [4 pts.]

Difference quotient:

\[R'(x) = \lim_{h \to 0} \frac{R(x + h) - R(x)}{h}\]

Expand:

\[R(x + h) = 2400(x + h) - 6(x + h)^2 = 2400x + 2400h - 6x^2 - 12xh - 6h^2\]

\[R(x + h) - R(x) = 2400h - 12xh - 6h^2 = h(2400 - 12x - 6h)\]

Cancel \(h\) and take the limit:

\[R'(x) = \lim_{h \to 0} (2400 - 12x - 6h) = 2400 - 12x \quad \checkmark\]

Grading: 1 Pt. difference quotient set up, 1 Pt. expanding \((x + h)^2\), 1 Pt. factoring out and cancelling \(h\), 1 Pt. limit for \(h \to 0\).

Part d)

The company currently sells 80 motors per month. Compute the marginal cost and the marginal revenue at 80 motors and explain their meaning. Compare the marginal cost with the exact cost of the 81st motor and decide whether the company should sell more motors. [4 pts.]

Marginal cost: \(C'(80) = 600 + 480 = 1{,}080\) EUR per motor. The 81st motor costs approximately 1,080 EUR.

Marginal revenue: \(R'(80) = 2400 - 960 = 1{,}440\) EUR per motor. The 81st motor brings approximately 1,440 EUR more revenue.

Exact cost of the 81st motor:

\[C(81) - C(80) = 600 + 3 \cdot (81^2 - 80^2) = 600 + 3 \cdot 161 = 1{,}083 \text{ EUR}\]

This is very close to the marginal cost of 1,080 EUR.

Decision: \(MR = 1{,}440 > 1{,}080 = MC\). The next motor adds about \(1440 - 1080 = 360\) EUR to the profit, so the company should sell more motors.

Grading: 1 Pt. \(C'(80) = 1{,}080\) EUR with meaning, 1 Pt. \(R'(80) = 1{,}440\) EUR with meaning, 1 Pt. \(C(81) - C(80) = 1{,}083\) EUR and comparison, 1 Pt. decision with \(MR > MC\).

Part e)

Determine the number of motors at which the marginal revenue equals the marginal cost. Show with the vertex of the profit function that the profit is maximal there. Calculate the price and the maximum profit. [5 pts.]

\(MR = MC\):

\[2400 - 12x = 600 + 6x \iff 1800 = 18x \iff x = 100\]

Profit function:

\[P(x) = R(x) - C(x) = 2400x - 6x^2 - 40000 - 600x - 3x^2 = -9x^2 + 1800x - 40000\]

Vertex: \(a = -9 < 0\), so the parabola opens downward and its vertex is a maximum.

\[x_V = -\frac{1800}{2 \cdot (-9)} = 100 \quad \checkmark\]

The profit is maximal at 100 motors per month, where \(MR = MC\).

Price: \(p(100) = 2400 - 600 = 1{,}800\) EUR per motor.

Maximum profit: \(P(100) = -90{,}000 + 180{,}000 - 40{,}000 = 50{,}000\) EUR per month.

Grading: 1 Pt. equation \(MR = MC\), 1 Pt. \(x = 100\) motors, 1 Pt. profit function and vertex at 100 with the reason \(a < 0\), 1 Pt. price 1,800 EUR, 1 Pt. maximum profit 50,000 EUR.