Tasks 05-07 - Function Determination & Function Families

Section 05: Differential Calculus

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

A: Skills

Problem 1: Quadratic Functions from Conditions (x)

Each part asks for a quadratic function \(f(x) = ax^2 + bx + c\).

  1. Determine the quadratic function whose graph passes through the points \((-1, 1)\), \((1, -3)\) and \((2, 4)\).

  2. Determine the quadratic function whose graph passes through \((1, 6)\) and \((3, 4)\) and whose tangent line at \(x = 1\) has the slope 3.

  3. A parabola has its maximum at \((4, 3)\) and passes through \((1, -6)\). Write down the three conditions for \(f\). Then determine \(f\) with the vertex form and check all three conditions.

  1. Three unknowns, three point conditions:

\[\begin{cases} f(-1) = 1: & a - b + c = 1 \quad \text{(I)} \\ f(1) = -3: & a + b + c = -3 \quad \text{(II)} \\ f(2) = 4: & 4a + 2b + c = 4 \quad \text{(III)} \end{cases}\]

  1. − (I): \(2b = -4\), so \(b = -2\). Then (II) gives \(a + c = -1\) and (III) gives \(4a + c = 8\). Subtract: \(3a = 9\), so \(a = 3\) and \(c = -4\).

\[f(x) = 3x^2 - 2x - 4\]

Check: \(f(-1) = 3 + 2 - 4 = 1\) ✓, \(f(1) = 3 - 2 - 4 = -3\) ✓, \(f(2) = 12 - 4 - 4 = 4\) ✓

  1. Derivative first: \(f'(x) = 2ax + b\). The slope condition is a condition on \(f'\):

\[\begin{cases} f(1) = 6: & a + b + c = 6 \quad \text{(I)} \\ f'(1) = 3: & 2a + b = 3 \quad \text{(II)} \\ f(3) = 4: & 9a + 3b + c = 4 \quad \text{(III)} \end{cases}\]

  1. − (I): \(8a + 2b = -2\), so \(4a + b = -1\). Subtract (II): \(2a = -4\), so \(a = -2\). Then \(b = 3 - 2a = 7\) and \(c = 6 - a - b = 1\).

\[f(x) = -2x^2 + 7x + 1\]

Check: \(f(1) = -2 + 7 + 1 = 6\) ✓, \(f'(1) = -4 + 7 = 3\) ✓, \(f(3) = -18 + 21 + 1 = 4\) ✓

  1. A maximum at \((4, 3)\) gives two conditions, and the second point gives the third:

\[f(4) = 3, \qquad f'(4) = 0, \qquad f(1) = -6\]

Vertex form with the vertex \((h, k) = (4, 3)\): \(f(x) = a(x - 4)^2 + 3\). It fulfils the first two conditions already, so only \(a\) is unknown:

\[f(1) = a(1 - 4)^2 + 3 = 9a + 3 = -6 \iff a = -1\]

\[f(x) = -(x - 4)^2 + 3 = -x^2 + 8x - 13\]

Check: \(f(4) = -16 + 32 - 13 = 3\) ✓. \(f'(x) = -2x + 8\), so \(f'(4) = 0\) ✓. \(f(1) = -1 + 8 - 13 = -6\) ✓. \(a = -1 < 0\): the parabola opens downward, so the vertex is a maximum ✓.

Typical mistake: in c), writing only \(f(4) = 3\) and \(f(1) = -6\). Two equations cannot fix three unknowns. The word “maximum” hides the second condition \(f'(4) = 0\).

Problem 2: A Cubic from Two Extrema (xx)

The graph of a polynomial \(f\) of third degree has a local maximum at \((-2, 21)\) and a local minimum at \((1, -6)\).

  1. Write down the four conditions for \(f(x) = ax^3 + bx^2 + cx + d\) and determine \(f(x)\).

  2. Check with the second derivative test that the two points are really a maximum and a minimum.

Hint: Solve the system by hand with the elimination method. You can check your solution with the calculator (MENU → A → 1).

  1. Derivative first: \(f'(x) = 3ax^2 + 2bx + c\). Each extremum gives a point condition and a horizontal tangent:

\[\begin{cases} f(-2) = 21: & -8a + 4b - 2c + d = 21 \quad \text{(I)} \\ f'(-2) = 0: & 12a - 4b + c = 0 \quad \text{(II)} \\ f(1) = -6: & a + b + c + d = -6 \quad \text{(III)} \\ f'(1) = 0: & 3a + 2b + c = 0 \quad \text{(IV)} \end{cases}\]

    1. − (II): \(-9a + 6b = 0\), so \(b = 1.5a\).
  • From (IV): \(c = -3a - 2b = -3a - 3a = -6a\).
    1. − (I) removes \(d\): \(9a - 3b + 3c = -27\), so \(3a - b + c = -9\).
  • Substitute \(b\) and \(c\): \(3a - 1.5a - 6a = -4.5a = -9\), so \(a = 2\).
  • Back-substitute: \(b = 3\), \(c = -12\), and from (III) \(d = -6 - 2 - 3 + 12 = 1\).

\[f(x) = 2x^3 + 3x^2 - 12x + 1\]

Check: \(f(-2) = -16 + 12 + 24 + 1 = 21\) ✓, \(f(1) = 2 + 3 - 12 + 1 = -6\) ✓, and \(f'(x) = 6x^2 + 6x - 12 = 6(x + 2)(x - 1)\) is zero at \(x = -2\) and \(x = 1\) ✓. The calculator (4 unknowns) gives \(a = 2\), \(b = 3\), \(c = -12\), \(d = 1\).

  1. \(f''(x) = 12x + 6\).
  • \(f''(-2) = -24 + 6 = -18 < 0\): concave down, local maximum ✓
  • \(f''(1) = 12 + 6 = 18 > 0\): concave up, local minimum ✓

Typical mistake: setting up only the two point conditions \(f(-2) = 21\) and \(f(1) = -6\). A cubic has four unknowns, and each extremum brings a second condition, \(f'(x_0) = 0\).

Problem 3: A Cubic with an Inflection Point and a Tangent (xx)

The graph of a polynomial \(f\) of third degree passes through the origin. It has an inflection point at \(x = 2\), and the inflectional tangent (the tangent at the inflection point) is the line \(y = 5x - 8\).

  1. Explain why the tangent line gives two conditions. Write down all four conditions for \(f(x) = ax^3 + bx^2 + cx + d\).

  2. Determine \(f(x)\).

  3. Check with the third derivative that \(x = 2\) really is an inflection point.

  1. The tangent touches the graph at \(x = 2\). So the graph and the tangent have the same point there, \(f(2) = 5 \cdot 2 - 8 = 2\), and the same slope, \(f'(2) = 5\). Together with the origin and the inflection point:

\[f(0) = 0, \qquad f''(2) = 0, \qquad f'(2) = 5, \qquad f(2) = 2\]

  1. Derivatives: \(f'(x) = 3ax^2 + 2bx + c\) and \(f''(x) = 6ax + 2b\).
  • \(f(0) = 0\): \(d = 0\)
  • \(f''(2) = 0\): \(12a + 2b = 0\), so \(b = -6a\)
  • \(f'(2) = 5\): \(12a + 4b + c = 5\), so \(12a - 24a + c = 5\) and \(c = 5 + 12a\)
  • \(f(2) = 2\): \(8a + 4b + 2c = 2\), so \(8a - 24a + 10 + 24a = 2\), i.e. \(8a = -8\) and \(a = -1\)

Then \(b = 6\) and \(c = -7\):

\[f(x) = -x^3 + 6x^2 - 7x\]

Check: \(f(2) = -8 + 24 - 14 = 2\) ✓, \(f'(x) = -3x^2 + 12x - 7\) with \(f'(2) = -12 + 24 - 7 = 5\) ✓, \(f''(x) = -6x + 12\) with \(f''(2) = 0\) ✓

  1. \(f'''(x) = -6 \neq 0\), so \(f''(2) = 0\) and \(f'''(2) \neq 0\): \(x = 2\) is an inflection point ✓. The graph changes from concave up to concave down there.

Typical mistake: taking only the slope from the tangent, \(f'(2) = 5\), and forgetting the point. Then only three conditions are left for four unknowns.

Problem 4: Function Families (xx)

A function family (German: Funktionsschar) is a set of functions that depend on a parameter. Each value of the parameter gives one function of the family.

  1. For which \(a\) does the graph of \(f_a(x) = ax^2 + 4x - 2a\) pass through the point \((3, -2)\)? Determine \(a\) and give the function.

  2. For which \(t\) does \(g_t(x) = x^3 + tx^2 - 16x + 1\) have a critical point at \(x = 4\)? Determine \(t\). Classify the critical point with the second derivative test and give its coordinates.

  3. Let \(h_k(x) = (x^2 - k) \cdot e^{x}\). Determine \(h_k'(x)\). For which \(k\) does \(h_k\) have a critical point at \(x = 1\)? For this \(k\), give all critical points of \(h_k\) and classify them.

  1. Function value condition \(f_a(3) = -2\):

\[9a + 12 - 2a = -2 \iff 7a = -14 \iff a = -2\]

\(f_{-2}(x) = -2x^2 + 4x + 4\). Check: \(f_{-2}(3) = -18 + 12 + 4 = -2\) ✓

  1. Critical point condition \(g_t'(4) = 0\) with \(g_t'(x) = 3x^2 + 2tx - 16\):

\[48 + 8t - 16 = 0 \iff 8t = -32 \iff t = -4\]

For \(t = -4\): \(g_{-4}(x) = x^3 - 4x^2 - 16x + 1\), \(g_{-4}'(x) = 3x^2 - 8x - 16 = (3x + 4)(x - 4)\) and \(g_{-4}''(x) = 6x - 8\).

\(g_{-4}''(4) = 16 > 0\): local minimum, \(g_{-4}(4) = 64 - 64 - 64 + 1 = -63\), point \((4, -63)\).

  1. Product rule with \(u = x^2 - k\), \(u' = 2x\) and \(v = e^{x}\), \(v' = e^{x}\):

\[h_k'(x) = 2x \cdot e^{x} + (x^2 - k) \cdot e^{x} = (x^2 + 2x - k) \cdot e^{x}\]

\(e^{x} > 0\), so \(h_k'(1) = 0 \iff 1 + 2 - k = 0 \iff k = 3\).

For \(k = 3\): \(h_3'(x) = (x^2 + 2x - 3) \cdot e^{x} = (x + 3)(x - 1) \cdot e^{x}\). Critical points at \(x = -3\) and \(x = 1\). The sign of \(h_3'\) is the sign of \((x + 3)(x - 1)\):

\(x\) \(x < -3\) \(-3 < x < 1\) \(x > 1\)
\(h_3'(x)\) \(+\) \(-\) \(+\)
  • \(x = -3\): \(h_3'\) changes from \(+\) to \(-\), local maximum, \(h_3(-3) = 6e^{-3} \approx 0.30\)
  • \(x = 1\): \(h_3'\) changes from \(-\) to \(+\), local minimum, \(h_3(1) = -2e \approx -5.44\)

Typical mistake: in b), solving \(g_t(4) = 0\) instead of \(g_t'(4) = 0\). A critical point is a condition on the derivative, not on the function value.

Problem 5: Cost Function from Marginal Cost (x)

A company knows its marginal costs \(C'(x) = 6x^2 - 30x + 50\) MU per QU for an output of \(x\) QU. Its fixed costs are 90 MU.

  1. Determine the cost function \(C(x)\) and check it by differentiating.

  2. Explain why you need the fixed costs to find \(C(x)\).

  3. The marginal revenue is \(R'(x) = 120 - 8x\) MU per QU. Without sales, there is no revenue. Determine the revenue function \(R(x)\) and the profit function \(P(x)\).

Hint: Undo the power rule term by term. For example, \(x^4\) has the derivative \(4x^3\), so \(4x^3\) comes from \(x^4\).

  1. Find a function whose derivative is \(C'(x)\) (integration, the reverse of differentiation). Term by term:
  • \(6x^2\) is the derivative of \(2x^3\)
  • \(-30x\) is the derivative of \(-15x^2\)
  • \(50\) is the derivative of \(50x\)

\[C(x) = 2x^3 - 15x^2 + 50x + k\]

with a constant \(k\). The fixed costs are the costs without production: \(C(0) = k = 90\).

\[C(x) = 2x^3 - 15x^2 + 50x + 90\]

Check: \(C'(x) = 6x^2 - 30x + 50\) ✓ and \(C(0) = 90\) ✓

  1. A constant disappears when you differentiate. So every function \(2x^3 - 15x^2 + 50x + k\) has the same marginal costs, whatever \(k\) is. The marginal costs only say how fast the costs grow, not where they start. The fixed costs give the starting value \(C(0) = 90\) and fix \(k\).

  2. Reverse the power rule again: \(R(x) = 120x - 4x^2 + k\). Without sales, there is no revenue: \(R(0) = k = 0\).

\[R(x) = 120x - 4x^2\]

\[P(x) = R(x) - C(x) = 120x - 4x^2 - (2x^3 - 15x^2 + 50x + 90) = -2x^3 + 11x^2 + 70x - 90\]

Typical mistake: forgetting the constant and writing \(C(x) = 2x^3 - 15x^2 + 50x\). This cost function has no fixed costs. The constant is where the fixed costs go.

B: Set Up & Check

Problem 6: From FSP Sentences to Equations (xx)

In the FSP, the conditions for a function are written as sentences. Translate each sentence into mathematics. Then answer the check question: it tells you whether your translation works. The cost functions are polynomials of third degree, \(C(x) = ax^3 + bx^2 + cx + d\) in MU for \(x\) QU, and the revenue functions are \(R(x) = ax^3 + bx^2 + cx\).

  1. A company makes doormats: “The fixed costs are 240 MU.” Write down the condition and name the coefficient it fixes.

    Check: The same text says: “The variable costs (the part of the costs that grows with the output) of 5 QU are 110 MU.” Write down this condition as well. How high are the total costs of 5 QU?

  2. A company makes coffee grinders: “At 3 QU, the marginal costs are 9 MU per QU.” Write down the condition as an equation in \(a\), \(b\) and \(c\).

    Check: Does \(C(x) = x^3 - 8x^2 + 30x + 140\) fulfil your condition? Compute \(C(3)\) as well. What would the condition \(C(3) = 9\) mean?

  3. A company makes bicycle locks: “At 5 QU, the curvature of the cost function changes sign.” Write down the condition as an equation in \(a\) and \(b\).

    Check: Which of \(C_1(x) = 0.2x^3 - 3x^2 + 20x + 50\) and \(C_2(x) = 0.2x^3 - 2x^2 + 5x + 50\) fulfils your condition? Why is the other one not a sensible cost function?

  4. A company sells picnic baskets: “The revenue has a local maximum at 7 QU.” Write down the condition. How many equations does this sentence give?

    Check: \(R(x) = -x^3 + 9x^2 + 21x\) fulfils your condition. Compute \(R(7)\). Why may you not add the equation \(R(7) = 0\)?

  5. A company sells garden lights: “The revenue function has an inflection point at \((3, 144)\).” Write down the conditions and explain why this sentence gives two equations.

    Check: Which of \(R_1(x) = -2x^3 + 18x^2 + 12x\) and \(R_2(x) = -x^3 + 6x^2 + 39x\) has an inflection point at \((3, 144)\)? Which condition does the other one miss?

  1. The fixed costs are the costs without production:

    \[C(0) = 240 \iff d = 240\]

    Check: the variable costs are the total costs minus the fixed costs, so \(C(5) - C(0) = 110\), i.e. \(C(5) = 350\) or \(125a + 25b + 5c + 240 = 350\). The total costs of 5 QU are 350 MU.

    Typical mistake: \(C(1) = 240\), as if the fixed costs were the costs of the first QU. They arise at \(x = 0\). In the check, \(C(5) = 110\) forgets the fixed costs.

  2. The marginal costs are the values of \(C'(x) = 3ax^2 + 2bx + c\):

    \[C'(3) = 9 \iff 27a + 6b + c = 9\]

    Check: \(C'(x) = 3x^2 - 16x + 30\), so \(C'(3) = 27 - 48 + 30 = 9\) ✓. \(C(3) = 27 - 72 + 90 + 140 = 185\): the total costs of 3 QU are 185 MU. \(C(3) = 9\) would say that 3 QU cost only 9 MU in total, less than the fixed costs.

    Typical mistake: \(C(3) = 9\). The marginal costs tell you how fast the costs grow, so the condition is on \(C'\), not on \(C\).

  3. The curvature changes sign at an inflection point, so the condition is on \(C''(x) = 6ax + 2b\):

    \[C''(5) = 0 \iff 30a + 2b = 0\]

    Check: \(C_1''(x) = 1.2x - 6\), so \(C_1''(5) = 0\) ✓: \(C_1\) fulfils the condition. \(C_2''(x) = 1.2x - 4\), so \(C_2''(5) = 2 \neq 0\). Instead, \(C_2'(5) = 15 - 20 + 5 = 0\): \(C_2\) fulfils the wrong condition \(C'(5) = 0\). It is not a sensible cost function, because \(C_2'(x) = 0.6x^2 - 4x + 5 = 0.2(3x - 5)(x - 5)\) is negative for \(\frac{5}{3} < x < 5\): there, the costs would fall when the output grows. (For \(C_1\), the marginal costs \(C_1'(x) = 0.6x^2 - 6x + 20\) have the discriminant \(36 - 48 < 0\), so they are always positive, and they are lowest at 5 QU.)

    Typical mistake: \(C'(5) = 0\). “Curvature” belongs to the second derivative. \(C'(5) = 0\) would mean that the costs do not grow at all at 5 QU.

  4. A local maximum needs a horizontal tangent:

    \[R'(7) = 0 \iff 147a + 14b + c = 0\]

    This is one equation. The sentence does not give the value of the maximum, so there is no second equation.

    Check: \(R'(x) = -3x^2 + 18x + 21 = -3(x - 7)(x + 1)\), so \(R'(7) = 0\) ✓. \(R(7) = -343 + 441 + 147 = 245\): the highest revenue is 245 MU, not 0. \(R(7) = 0\) would mean no revenue at all at 7 QU. \(R''(7) = -42 + 18 = -24 < 0\) confirms the maximum. This inequality is a check afterwards, not a condition for the system.

    Typical mistake: writing two equations, \(R'(7) = 0\) and \(R(7) = 0\). A maximum at \((x_0, y_0)\) gives two conditions only if the value \(y_0\) is known.

  5. The point lies on the graph, and the curvature changes there:

    \[R(3) = 144 \iff 27a + 9b + 3c = 144, \qquad R''(3) = 0 \iff 18a + 2b = 0\]

    Two equations, because “inflection point at \((3, 144)\)” contains two pieces of information: where the inflection point is (\(x = 3\)) and how high the graph is there (144).

    Check: \(R_1(3) = -54 + 162 + 36 = 144\) ✓ and \(R_1''(x) = -12x + 36\), so \(R_1''(3) = 0\) ✓ (\(R_1'''(x) = -12 \neq 0\)). \(R_1\) has the inflection point \((3, 144)\). \(R_2(3) = -27 + 54 + 117 = 144\) ✓, but \(R_2''(x) = -6x + 12\), so \(R_2''(3) = -6 \neq 0\): the point \((3, 144)\) lies on the graph of \(R_2\), but it is no inflection point. \(R_2\) misses \(R''(3) = 0\).

    Typical mistake: writing only \(R''(3) = 0\). This says where the inflection point is, but it loses the value 144.

Problem 7: One Paragraph, All Conditions (xx)

A company builds garden shredders (machines that chop branches into small pieces) and sells them to garden centres. Its monthly total costs in MU are a polynomial of third degree in the output \(x\) in QU. The fixed costs are 120 MU. At the start of the production, the marginal costs are 35 MU per QU. The marginal costs are the same at 1 QU and at 7 QU. At 6 QU, the average costs (the total costs per QU) are 37 MU per QU. The garden centres buy no shredder at a price of 115 MU per QU or more (maximum price), and even for free they would take at most 23 QU (market saturation). In between, the demand is linear. The owner wants to know the output with the highest profit and a proof that it really is a maximum.

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

Check: An assistant writes the sentence about 1 QU and 7 QU as \(C'(1) = 0\) and \(C'(7) = 0\). Explain what is wrong. Show that the correct condition means that the curvature of \(C\) changes sign at 4 QU.

  1. Cost function: \(C(x) = ax^3 + bx^2 + cx + d\), so \(C'(x) = 3ax^2 + 2bx + c\) and \(C''(x) = 6ax + 2b\)
  2. Fixed costs: \(C(0) = 120\), i.e. \(d = 120\)
  3. Marginal costs at the start: \(C'(0) = 35\), i.e. \(c = 35\)
  4. Same marginal costs at 1 QU and 7 QU: \(C'(1) = C'(7)\)
  5. Average costs at 6 QU: \(\frac{C(6)}{6} = 37\), i.e. \(C(6) = 222\)
  6. Linear demand through \((0, 115)\) and \((23, 0)\): \(p(x) = mx + 115\) with \(p(23) = 0\)
  7. Revenue and profit: \(R(x) = x \cdot p(x)\) and \(P(x) = R(x) - C(x)\)
  8. Highest profit: \(P'(x_0) = 0\) and \(P''(x_0) < 0\) (or \(P'\) changes its sign from \(+\) to \(-\) at \(x_0\))

Conditions 2 to 5 are the four conditions for the four unknowns \(a\), \(b\), \(c\), \(d\).

Check: the sentence compares two marginal costs. It does not say that they are zero. \(C'(1) = 0\) would mean that the costs do not grow at all at 1 QU. The correct condition is

\[C'(1) = C'(7) \iff 3a + 2b + c = 147a + 14b + c \iff 144a + 12b = 0 \iff b = -12a\]

Then \(C''(4) = 24a + 2b = 24a - 24a = 0\): the curvature of \(C\) changes sign at 4 QU. The reason: \(C'\) is a parabola, and it has the same value at 1 and at 7. So its axis of symmetry lies in the middle, at \(x = 4\), where \(C'\) has its vertex (the lowest marginal costs) and \(C''(4) = 0\).

Typical mistake: \(C(6) = 37\) for the average costs. 37 MU is the cost per QU, so the total costs of 6 QU are \(6 \cdot 37 = 222\) MU.

C: Applied

Problem 8: Revenue of a Telescope Maker (xx)

A company builds telescopes for hobby astronomers and sells them to specialist shops. Its monthly revenue \(R(x)\) in MU is a polynomial of third degree in the output \(x\) in QU (1 QU = 100 telescopes, 1 MU = 1,000 EUR). The sales department knows the following: 200 telescopes bring a revenue of 164,000 EUR. At this output, the marginal revenue is 72 MU per QU. 600 telescopes bring a revenue of 324,000 EUR. Round to two decimal places.

  1. Explain why the approach \(R(x) = ax^3 + bx^2 + cx\) has no constant term. Set up the system of equations and determine \(R(x)\).

(for verification: \(R(x) = -0.5x^3 - 3x^2 + 90x\))

  1. Determine the output with the highest revenue and prove that it is a maximum. Give the highest revenue in EUR.

  2. The revenue is \(R(x) = x \cdot p(x)\). Determine the price-demand function \(p(x)\). Give the maximum price in EUR per telescope and compute the market saturation.

  3. Calculate the price per telescope at which the revenue is highest. The sales manager wants to sell 800 telescopes per month “to make even more revenue”. Decide whether the sales manager is right and explain your decision.

  4. Show that the graph of \(R\) has no inflection point for \(x > 0\). Explain what this means for the marginal revenue.

a) Without sales, there is no revenue: \(R(0) = 0\), so the constant term \(d\) is 0. Derivative: \(R'(x) = 3ax^2 + 2bx + c\). The data in QU and MU: \(R(2) = 164\), \(R'(2) = 72\) and \(R(6) = 324\).

\[\begin{cases} R(2) = 164: & 8a + 4b + 2c = 164 \iff 4a + 2b + c = 82 \quad \text{(I)} \\ R'(2) = 72: & 12a + 4b + c = 72 \quad \text{(II)} \\ R(6) = 324: & 216a + 36b + 6c = 324 \iff 36a + 6b + c = 54 \quad \text{(III)} \end{cases}\]

    1. − (I): \(8a + 2b = -10\), so \(4a + b = -5\)
    1. − (II): \(24a + 2b = -18\), so \(12a + b = -9\)
  • Subtract: \(8a = -4\), so \(a = -0.5\), then \(b = -5 + 2 = -3\) and from (I) \(c = 82 + 2 + 6 = 90\)

\[R(x) = -0.5x^3 - 3x^2 + 90x\]

b) Power rule, then factor out \(-1.5\):

\[R'(x) = -1.5x^2 - 6x + 90 = -1.5(x^2 + 4x - 60) = -1.5(x - 6)(x + 10)\]

\(R'(x) = 0\) at \(x = 6\) (\(x = -10\) has no meaning). \(R''(x) = -3x - 6\), so \(R''(6) = -24 < 0\): maximum.

\(R(6) = -108 - 108 + 540 = 324\) MU. The highest revenue is 324,000 EUR per month at 600 telescopes.

c) For \(x > 0\):

\[p(x) = \frac{R(x)}{x} = -0.5x^2 - 3x + 90\]

Maximum price: \(p(0) = 90\) MU per QU, i.e. 90,000 EUR per 100 telescopes or 900 EUR per telescope.

Market saturation: \(p(x) = 0 \iff x^2 + 6x - 180 = 0\). Quadratic formula:

\[x = -3 \pm \sqrt{9 + 180} = -3 \pm \sqrt{189}, \qquad x_1 \approx 10.75, \quad x_2 \approx -16.75\]

Only \(x_1\) has a meaning: even for free, the shops would take at most about 10.75 QU, i.e. about 1,075 telescopes per month.

d) \(p(6) = -18 - 18 + 90 = 54\) MU per QU, i.e. 540 EUR per telescope.

The sales manager is wrong. \(R(8) = -256 - 192 + 720 = 272\) MU, less than 324 MU. To sell 800 telescopes, the company has to lower the price to \(p(8) = -32 - 24 + 90 = 34\) MU per QU, i.e. 340 EUR per telescope. For \(x > 6\), \(R'(x) < 0\): the lower price costs more revenue than the extra telescopes bring.

e) \(R''(x) = -3x - 6 = 0 \iff x = -2\), which is not an output. For \(x > 0\), \(R''(x) < 0\): the graph is concave down on the whole domain, and its curvature never changes. So there is no inflection point for \(x > 0\).

Meaning: \(R''\) is the rate of change of the marginal revenue \(R'\). \(R''(x) < 0\) means that the marginal revenue falls with every further QU. Every further 100 telescopes bring less extra revenue than the 100 before. At 6 QU, the extra revenue has fallen to 0, and after that it is negative.

Problem 9: Launch of a Sparkling Water Maker (xx)

A start-up plans to sell a sparkling water maker (a device that adds bubbles to tap water). Before the launch, an analyst models the monthly profit \(P(x)\) in MU as a polynomial of third degree in the output \(x\) in QU (1 QU = 1,000 water makers, 1 MU = 1,000 EUR). The requirements are:

  • The fixed costs are 12 MU. Without sales, the loss equals the fixed costs.
  • The start-up breaks even at 2 QU.
  • The profit is highest at 8 QU, and there it is 180 MU.

Round to two decimal places.

  1. Translate the requirements into four conditions and determine \(P(x)\).

(for verification: \(P(x) = -x^3 + 13x^2 - 16x - 12\))

  1. Prove that the profit really is highest at 8 QU. Determine the other critical point and explain its meaning for the start-up.

  2. Determine the profit limit (upper break-even point) with polynomial long division. Decide in which output range the start-up makes a profit.

  3. Determine the inflection point of \(P\) and explain its economical significance.

a) Approach \(P(x) = ax^3 + bx^2 + cx + d\) with \(P'(x) = 3ax^2 + 2bx + c\). The highest profit gives two conditions:

\[\begin{cases} P(0) = -12: & d = -12 \\ P(2) = 0: & 8a + 4b + 2c - 12 = 0 \iff 4a + 2b + c = 6 \quad \text{(I)} \\ P'(8) = 0: & 192a + 16b + c = 0 \quad \text{(II)} \\ P(8) = 180: & 512a + 64b + 8c - 12 = 180 \iff 64a + 8b + c = 24 \quad \text{(III)} \end{cases}\]

    1. − (I): \(60a + 6b = 18\), so \(10a + b = 3\)
    1. − (III): \(128a + 8b = -24\), so \(16a + b = -3\)
  • Subtract: \(6a = -6\), so \(a = -1\), then \(b = 13\) and from (I) \(c = 6 + 4 - 26 = -16\)

\[P(x) = -x^3 + 13x^2 - 16x - 12\]

b) \(P'(x) = -3x^2 + 26x - 16 = -(3x - 2)(x - 8)\), so the critical points are \(x = 8\) and \(x = \frac{2}{3}\). Second derivative: \(P''(x) = -6x + 26\).

  • \(P''(8) = -22 < 0\): maximum at 8 QU with \(P(8) = -512 + 832 - 128 - 12 = 180\) MU ✓
  • \(P''\left(\frac{2}{3}\right) = -4 + 26 = 22 > 0\): local minimum with \(P\left(\frac{2}{3}\right) = -\frac{464}{27} \approx -17.19\) MU

Meaning: at about 667 water makers per month, the loss is largest, about 17,190 EUR per month. It is larger than the fixed costs, because \(P'(x) < 0\) for \(0 < x < \frac{2}{3}\): there, each further water maker makes the loss larger. From about 667 water makers on, every further water maker makes the loss smaller, and from 2,000 on, the start-up makes a profit.

c) \(P(2) = 0\), so \((x - 2)\) is a factor of \(P(x)\). Polynomial long division:

\[\begin{array}{l} (-x^3 + 13x^2 - 16x - 12) : (x - 2) = -x^2 + 11x + 6 \\ \underline{-(-x^3 + 2x^2)} \\ \phantom{(-x^3}\;\; 11x^2 - 16x \\ \phantom{(-x^3}\;\; \underline{-(11x^2 - 22x)} \\ \phantom{(-x^3 + 13x^2}\;\; 6x - 12 \\ \phantom{(-x^3 + 13x^2}\;\; \underline{-(6x - 12)} \\ \phantom{(-x^3 + 13x^2 - 16x}\;\; 0 \end{array}\]

Profit limit: \(-x^2 + 11x + 6 = 0 \iff x^2 - 11x - 6 = 0\). Quadratic formula:

\[x = \frac{11 \pm \sqrt{121 + 24}}{2} = \frac{11 \pm \sqrt{145}}{2}, \qquad x_1 \approx 11.52, \quad x_2 \approx -0.52\]

The profit limit lies at about 11.52 QU, i.e. about 11,520 water makers per month.

Profit range: \(P(x) \approx -(x - 2)(x - 11.52)(x + 0.52)\). For \(2 < x < 11.52\), the factors have the signs \((+)\), \((-)\) and \((+)\), so \(P(x) > 0\). The start-up makes a profit between 2,000 and about 11,520 water makers per month.

d) \(P''(x) = -6x + 26 = 0 \iff x = \frac{13}{3} \approx 4.33\), and \(P'''(x) = -6 \neq 0\): inflection point at \(x = \frac{13}{3}\) with \(P\left(\frac{13}{3}\right) = \frac{2198}{27} \approx 81.41\), point \((4.33, 81.41)\).

Economical significance: \(P'\left(\frac{13}{3}\right) = \frac{121}{3} \approx 40.33\) MU per QU is the highest marginal profit. At about 4,333 water makers per month, the profit grows fastest, by about 40.33 EUR per further water maker. Below this output, the profit grows faster and faster, above it more and more slowly, until it stops growing at 8 QU.

Problem 10: A Production Line for Bicycle Frames (xxx)

A company plans a new production line for aluminium bicycle frames. The engineers can choose how strongly the line is automated. This setting is described by a parameter \(t > 0\). With the setting \(t\), the monthly costs are

\[C_t(x) = x^3 - tx^2 + 48x + 100\]

MU for an output of \(x\) QU (1 QU = 100 frames, 1 MU = 1,000 EUR). The functions \(C_t\) form a function family with the parameter \(t\).

  1. Explain why all cost functions of the family have the same fixed costs. Determine \(C_t'(x)\) and \(C_t''(x)\). Show that the marginal costs are lowest at \(x = \frac{t}{3}\) and that the lowest marginal costs are \(48 - \frac{t^2}{3}\) MU per QU.

  2. A cost function must never fall. Decide with the discriminant for which settings \(t\) the equation \(C_t'(x) = 0\) has no solution, exactly one solution or two solutions. Decide which settings \(t\) are sensible.

  3. The material for one frame costs 210 EUR, so the marginal costs can never be lower than 21 MU per QU. Determine the setting \(t\) for which the lowest marginal costs are exactly 21 MU per QU, and the output at which they occur.

(for verification: \(t = 9\))

  1. The company plans to produce 500 frames per month. Show that \(C_t(5) = 465 - 25t\). Among all settings \(t\) with lowest marginal costs of at least 21 MU per QU, determine the setting with the lowest monthly costs and give these costs in EUR.

a) \(C_t(0) = 0 - 0 + 0 + 100 = 100\) for every \(t\): the parameter only appears in the term \(-tx^2\), which is 0 at \(x = 0\). All graphs of the family pass through \((0, 100)\), and the fixed costs are 100 MU = 100,000 EUR per month for every setting.

\[C_t'(x) = 3x^2 - 2tx + 48, \qquad C_t''(x) = 6x - 2t\]

The lowest marginal costs are a minimum of \(C_t'\), so \(C_t''(x) = 0 \iff 6x = 2t \iff x = \frac{t}{3}\). The graph of \(C_t'\) is a parabola that opens upward (\(3 > 0\)), so this is its vertex and a minimum. (Also: \(C_t'''(x) = 6 > 0\).) Value:

\[C_t'\left(\frac{t}{3}\right) = 3 \cdot \frac{t^2}{9} - 2t \cdot \frac{t}{3} + 48 = \frac{t^2}{3} - \frac{2t^2}{3} + 48 = 48 - \frac{t^2}{3}\]

b) \(C_t'(x) = 0 \iff 3x^2 - 2tx + 48 = 0\). Discriminant:

\[\Delta = (-2t)^2 - 4 \cdot 3 \cdot 48 = 4t^2 - 576 = 4(t^2 - 144)\]

  • \(\Delta < 0 \iff t^2 < 144 \iff 0 < t < 12\): no solution. The parabola \(C_t'\) opens upward and stays above the \(x\)-axis, so \(C_t'(x) > 0\): the costs always grow.
  • \(\Delta = 0 \iff t = 12\): exactly one solution, \(x = \frac{2t}{6} = 4\). \(C_{12}'(x) = 3x^2 - 24x + 48 = 3(x - 4)^2 \geq 0\): the costs never fall, but at 4 QU a further frame costs nothing.
  • \(\Delta > 0 \iff t > 12\): two solutions. Between them, \(C_t'(x) < 0\): the costs would fall when the output grows.

Sensible are the settings \(0 < t \leq 12\) (with positive marginal costs everywhere: \(0 < t < 12\)). This agrees with a): the lowest marginal costs \(48 - \frac{t^2}{3}\) are not negative exactly for \(t^2 \leq 144\).

c) 210 EUR per frame are 21,000 EUR per 100 frames, i.e. 21 MU per QU.

\[48 - \frac{t^2}{3} = 21 \iff \frac{t^2}{3} = 27 \iff t^2 = 81 \iff t = 9 \quad (t > 0)\]

The lowest marginal costs lie at \(x = \frac{9}{3} = 3\) QU, i.e. at 300 frames. Check: \(C_9'(3) = 27 - 54 + 48 = 21\) ✓

d) \(C_t(5) = 125 - 25t + 240 + 100 = 465 - 25t\) ✓

Lowest marginal costs of at least 21 MU per QU: \(48 - \frac{t^2}{3} \geq 21 \iff t^2 \leq 81 \iff 0 < t \leq 9\).

\(C_t(5) = 465 - 25t\) falls when \(t\) grows. So the largest allowed setting, \(t = 9\), gives the lowest costs:

\[C_9(5) = 465 - 225 = 240 \text{ MU} = 240{,}000 \text{ EUR per month}\]

(For comparison: with \(t = 6\), the costs are \(C_6(5) = 315\) MU.)

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: Patio Heaters for Garden Centres [23 pts. total]

A company produces patio heaters (gas heaters for terraces and beer gardens) and sells them to garden centres and DIY stores. An analysis of its monthly costs gives the following information:

The fixed costs are 160 MU. At 6 QU, the marginal costs reach their lowest value, 18 MU per QU. At 8 QU, the total costs amount to 416 MU.

At a price of 1,320 EUR per heater, the stores buy 400 heaters per month. At a price of 1,140 EUR per heater, they buy 1,000 heaters per month. The demand is linear.

(1 QU = 100 patio heaters, 1 MU = 1,000 EUR)

Part a)

Determine the cost function \(C(x)\), assuming it is a polynomial of third degree. [11 pts.]

(for verification: \(C(x) = 0.5x^3 - 9x^2 + 72x + 160\))

Approach: \(C(x) = ax^3 + bx^2 + cx + d\), so \(C'(x) = 3ax^2 + 2bx + c\) and \(C''(x) = 6ax + 2b\).

Conditions: the lowest marginal costs are a minimum of \(C'\). So this sentence gives two conditions: \(C''(6) = 0\) (where) and \(C'(6) = 18\) (how high).

  1. Fixed costs: \(C(0) = 160\), so \(d = 160\)
  2. Lowest marginal costs at 6 QU: \(C''(6) = 0\), so \(36a + 2b = 0\)
  3. Their value: \(C'(6) = 18\), so \(108a + 12b + c = 18\)
  4. Total costs at 8 QU: \(C(8) = 416\), so \(512a + 64b + 8c + 160 = 416\), i.e. \(512a + 64b + 8c = 256\)

Solving (with \(d = 160\) a 3×3 system):

  • From 2: \(b = -18a\)
  • Into 3: \(108a - 216a + c = 18\), so \(c = 18 + 108a\)
  • Into 4: \(512a - 1152a + 8(18 + 108a) = 256\), so \(512a - 1152a + 144 + 864a = 256\), i.e. \(224a = 112\) and \(a = 0.5\)
  • Back-substitute: \(b = -9\), \(c = 18 + 54 = 72\)

\[C(x) = 0.5x^3 - 9x^2 + 72x + 160\]

Check: \(C'(x) = 1.5x^2 - 18x + 72 = 1.5(x - 6)^2 + 18\). This parabola opens upward and has its vertex at \((6, 18)\): the marginal costs are lowest at 6 QU, with 18 MU per QU ✓. \(C(8) = 256 - 576 + 576 + 160 = 416\) ✓

Grading: 1 Pt. approach with \(C'\) and \(C''\), 1 Pt. \(d = 160\), 2 Pts. \(C''(6) = 0\) and \(C'(6) = 18\), 1 Pt. \(C(8) = 416\), 4 Pts. solving the system, 1 Pt. \(C(x)\), 1 Pt. check.

Part b)

Determine the linear price-demand function \(p(x)\), with \(p\) in MU per QU. Give the maximum price in EUR per heater and the market saturation in heaters per month. [4 pts.]

(for verification: \(p(x) = 144 - 3x\))

Units: 1 MU per QU is 1,000 EUR per 100 heaters, i.e. 10 EUR per heater.

  • 1,320 EUR per heater = 132 MU per QU, 400 heaters = 4 QU: point \((4, 132)\)
  • 1,140 EUR per heater = 114 MU per QU, 1,000 heaters = 10 QU: point \((10, 114)\)

Slope:

\[m = \frac{114 - 132}{10 - 4} = \frac{-18}{6} = -3\]

Point-slope form with \((4, 132)\):

\[p(x) = -3(x - 4) + 132 = 144 - 3x\]

Maximum price: \(p(0) = 144\) MU per QU, i.e. 1,440 EUR per heater. At this price or more, the stores buy no heater.

Market saturation: \(p(x) = 0 \iff x = 48\) QU, i.e. 4,800 heaters per month. Even for free, the stores would not take more.

Grading: 1 Pt. both points in QU and MU per QU, 1 Pt. slope, 1 Pt. \(p(x)\), 1 Pt. maximum price and saturation with units.

Part c)

Show that the monthly profit is given by \(P(x) = -0.5x^3 + 6x^2 + 72x - 160\). [2 pts.]

Revenue: \(R(x) = x \cdot p(x) = x(144 - 3x) = 144x - 3x^2\)

Profit:

\[P(x) = R(x) - C(x) = 144x - 3x^2 - (0.5x^3 - 9x^2 + 72x + 160) = -0.5x^3 + 6x^2 + 72x - 160 \quad \checkmark\]

Grading: 1 Pt. \(R(x) = 144x - 3x^2\), 1 Pt. \(P(x)\) with the bracket resolved correctly.

Part d)

To reduce the gas consumption on terraces, the government plans a levy of \(t\) MU per QU that the company has to pay for every QU it sells (\(t \geq 0\)). The monthly profit is then \(P_t(x) = P(x) - t \cdot x\), a function family with the parameter \(t\).

Determine the levy \(t\) for which the profit \(P_t\) is highest at 11 QU. Prove that it really is a maximum. [3 pts.]

Derivative: \(P_t(x) = -0.5x^3 + 6x^2 + 72x - 160 - tx\), so

\[P_t'(x) = -1.5x^2 + 12x + 72 - t\]

Necessary condition at 11 QU:

\[P_t'(11) = -181.5 + 132 + 72 - t = 22.5 - t = 0 \iff t = 22.5\]

The levy is 22.5 MU per QU, i.e. 225 EUR per heater.

Sufficient condition: \(P_t''(x) = -3x + 12\) does not depend on \(t\), and \(P_t''(11) = -33 + 12 = -21 < 0\): maximum. With \(t = 22.5\), \(P_t'(x) = -1.5x^2 + 12x + 49.5 = -1.5(x - 11)(x + 3)\), so \(x = 11\) is the only critical point for \(x \geq 0\).

Grading: 1 Pt. \(P_t'(x)\), 1 Pt. \(t = 22.5\), 1 Pt. \(P_t''(11) < 0\).

Part e)

Without the levy, the maximum profit is 704 MU at 12 QU. Compute the maximum profit with the levy from part d) and the government’s monthly income from the levy, both in EUR. Explain why the company’s maximum profit falls by more than the government’s income. [3 pts.]

Maximum profit with the levy: \(P(11) = -665.5 + 726 + 792 - 160 = 692.5\), so

\[P_{22.5}(11) = P(11) - 22.5 \cdot 11 = 692.5 - 247.5 = 445 \text{ MU} = 445{,}000 \text{ EUR per month}\]

Income from the levy: \(22.5 \cdot 11 = 247.5\) MU, i.e. 247,500 EUR per month.

Comparison: the maximum profit falls by \(704 - 445 = 259\) MU, i.e. 259,000 EUR, which is 11,500 EUR more than the levy income. The company pays the levy, and in addition it now sells 100 heaters fewer than at its old optimum. Even without the levy, 11 QU bring less profit than 12 QU: \(P(12) - P(11) = 704 - 692.5 = 11.5\) MU. This part of the loss goes to nobody.

Grading: 1 Pt. 445,000 EUR, 1 Pt. 247,500 EUR, 1 Pt. explanation with the smaller output (\(P(11) < P(12)\)).