Tasks 03-03 - Quadratic Functions & Basic Optimization

Section 03: Functions as Business Models

A: Skills

Problem 1: Vertex and Properties (x)

For each quadratic function, determine the vertex and the axis of symmetry. Decide whether the function has a maximum or a minimum value and give a reason.

  1. \(f(x) = x^2 + 6x - 7\)

  2. \(g(x) = -2x^2 + 8x - 3\)

  3. \(h(x) = 3x^2 + 12x + 15\)

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

  1. \(f(x) = x^2 + 6x - 7\):
    • Vertex x-coordinate: \(x_v = -\frac{6}{2 \cdot 1} = -3\)
    • Vertex y-coordinate: \(f(-3) = 9 - 18 - 7 = -16\)
    • Vertex: \((-3, -16)\)
    • Axis of symmetry: \(x = -3\)
    • \(a = 1 > 0\), the parabola opens upward, so the vertex is a minimum. Minimum value \(-16\)
  2. \(g(x) = -2x^2 + 8x - 3\):
    • Vertex x-coordinate: \(x_v = -\frac{8}{2 \cdot (-2)} = 2\)
    • Vertex y-coordinate: \(g(2) = -8 + 16 - 3 = 5\)
    • Vertex: \((2, 5)\)
    • Axis of symmetry: \(x = 2\)
    • \(a = -2 < 0\), the parabola opens downward, so the vertex is a maximum. Maximum value 5
  3. \(h(x) = 3x^2 + 12x + 15\):
    • Vertex x-coordinate: \(x_v = -\frac{12}{2 \cdot 3} = -2\)
    • Vertex y-coordinate: \(h(-2) = 12 - 24 + 15 = 3\)
    • Vertex: \((-2, 3)\)
    • Axis of symmetry: \(x = -2\)
    • \(a = 3 > 0\), the parabola opens upward, so the vertex is a minimum. Minimum value 3
  4. \(p(x) = -\frac{1}{2}x^2 + 4x - 6\):
    • Vertex x-coordinate: \(x_v = -\frac{4}{2 \cdot (-\frac{1}{2})} = 4\)
    • Vertex y-coordinate: \(p(4) = -8 + 16 - 6 = 2\)
    • Vertex: \((4, 2)\)
    • Axis of symmetry: \(x = 4\)
    • \(a = -\frac{1}{2} < 0\), the parabola opens downward, so the vertex is a maximum. Maximum value 2

Problem 2: Completing the Square (x)

In a) to d), convert the quadratic function from standard form to vertex form by completing the square. Give the vertex and decide whether it is a maximum or a minimum.

  1. \(f(x) = x^2 + 10x + 21\)

  2. \(g(x) = 2x^2 - 8x + 5\)

  3. \(h(x) = -x^2 + 6x - 7\)

  4. \(k(x) = 3x^2 - 18x + 24\)

  5. Check your vertex form from b) by expanding it with the perfect square formula. Explain with your vertex form from c) why \(h(x) \leq 2\) holds for every \(x\).

  1. \(f(x) = x^2 + 10x + 21\):

    • Complete the square: \((x^2 + 10x + 25) - 25 + 21\)
    • \(f(x) = (x + 5)^2 - 4\)
    • Vertex \((-5, -4)\). \(a = 1 > 0\), so it is a minimum

  2. \(g(x) = 2x^2 - 8x + 5\):

    • Factor out 2: \(2(x^2 - 4x) + 5\)
    • Complete the square: \(2(x^2 - 4x + 4 - 4) + 5\)
    • \(g(x) = 2((x - 2)^2 - 4) + 5\)
    • \(g(x) = 2(x - 2)^2 - 8 + 5\)
    • \(g(x) = 2(x - 2)^2 - 3\)
    • Vertex \((2, -3)\). \(a = 2 > 0\), so it is a minimum

  3. \(h(x) = -x^2 + 6x - 7\):

    • Factor out \(-1\): \(-(x^2 - 6x) - 7\)
    • Complete the square: \(-(x^2 - 6x + 9 - 9) - 7\)
    • \(h(x) = -((x - 3)^2 - 9) - 7\)
    • \(h(x) = -(x - 3)^2 + 9 - 7\)
    • \(h(x) = -(x - 3)^2 + 2\)
    • Vertex \((3, 2)\). \(a = -1 < 0\), so it is a maximum

  4. \(k(x) = 3x^2 - 18x + 24\):

    • Factor out 3: \(3(x^2 - 6x) + 24\)
    • Complete the square: \(3(x^2 - 6x + 9 - 9) + 24\)
    • \(k(x) = 3((x - 3)^2 - 9) + 24\)
    • \(k(x) = 3(x - 3)^2 - 27 + 24\)
    • \(k(x) = 3(x - 3)^2 - 3\)
    • Vertex \((3, -3)\). \(a = 3 > 0\), so it is a minimum

  5. Expanding the vertex form from b): \(2(x - 2)^2 - 3 = 2(x^2 - 4x + 4) - 3 = 2x^2 - 8x + 8 - 3 = 2x^2 - 8x + 5\) ✓

    For c): a square is never negative, so \((x - 3)^2 \geq 0\) and \(-(x - 3)^2 \leq 0\). Hence \(h(x) = -(x - 3)^2 + 2 \leq 2\) for every \(x\), with equality only for \(x = 3\). The maximum value 2 is reached at the vertex.

Problem 3: Three Forms of a Parabola (x)

A quadratic function can be written in standard form \(ax^2 + bx + c\), in vertex form \(a(x - h)^2 + k\) and, if it has zeros, in factored form \(a(x - r_1)(x - r_2)\).

  1. \(f(x) = -2(x - 1)(x - 7)\). Read off the zeros. Determine the vertex and the \(y\)-intercept. Give \(f\) in standard form and in vertex form.

  2. \(g(x) = 3(x - 2)^2 - 27\). Give the vertex. Expand \(g\) to standard form and factor it to find the zeros.

  3. \(h(x) = x^2 - 2x - 24\). Factor \(h\) and give the zeros. Determine the vertex and write \(h\) in vertex form.

  4. A student says: “The \(y\)-intercept of \(P(x) = -(x - 4)(x - 9)\) is 36, because \(4 \cdot 9 = 36\).” Find the mistake and give the correct \(y\)-intercept.

  1. Zero-product rule: \(f(x) = 0\) for \(x = 1\) and \(x = 7\).

    • The vertex lies halfway between the zeros: \(x_v = \frac{1 + 7}{2} = 4\), and \(f(4) = -2 \cdot 3 \cdot (-3) = 18\). Vertex \((4, 18)\)
    • \(y\)-intercept: \(f(0) = -2 \cdot (-1) \cdot (-7) = -14\)
    • Standard form: \(f(x) = -2(x^2 - 8x + 7) = -2x^2 + 16x - 14\)
    • Vertex form: \(f(x) = -2(x - 4)^2 + 18\). Check: \(-2(x^2 - 8x + 16) + 18 = -2x^2 + 16x - 14\) ✓

  2. Vertex \((2, -27)\).

    • Standard form: \(g(x) = 3(x^2 - 4x + 4) - 27 = 3x^2 - 12x - 15\)
    • Factor: \(g(x) = 3(x^2 - 4x - 5) = 3(x - 5)(x + 1)\). Two numbers with product \(-5\) and sum \(-4\): \(-5\) and \(1\)
    • Zeros: \(x = 5\) and \(x = -1\). Check: halfway between them lies \(\frac{-1 + 5}{2} = 2\), the \(x\)-coordinate of the vertex ✓

  3. Two numbers with product \(-24\) and sum \(-2\): \(-6\) and \(4\). So \(h(x) = (x - 6)(x + 4)\) with the zeros \(x = 6\) and \(x = -4\).

    • Vertex: \(x_v = \frac{-4 + 6}{2} = 1\), \(h(1) = 1 - 2 - 24 = -25\). Vertex \((1, -25)\)
    • Vertex form: \(h(x) = (x - 1)^2 - 25\). Check: \(x^2 - 2x + 1 - 25 = x^2 - 2x - 24\) ✓

  4. The \(y\)-intercept is the value at \(x = 0\), not the product of the numbers in the brackets: the sign in front of the brackets and the minus signs inside them count too.

    \(P(0) = -(0 - 4)(0 - 9) = -(-4) \cdot (-9) = -36\). The \(y\)-intercept is \(-36\). As a profit function, \(P\) would mean a loss of 36 at an output of 0, i.e. the fixed costs.

Problem 4: Parabola from Its Properties (x)

  1. A parabola has its vertex at \((3, 20)\) and passes through the point \((0, 2)\). Determine the function in vertex form and in standard form. Determine the zeros exactly from the vertex form.

  2. A parabola has the zeros \(x = -1\) and \(x = 5\) and passes through the point \((1, 24)\). Determine the function in factored form and in standard form. Give the vertex and decide whether it is a maximum or a minimum.

  3. A parabola has its vertex at \((-2, -5)\) and the \(y\)-intercept 3. Determine the function. Decide with the discriminant how many zeros it has and explain the result with the position of the vertex and the sign of \(a\).

  4. A parabola passes through the points \((1, 7)\) and \((9, 7)\), and its vertex lies at the height \(-9\). Determine the axis of symmetry and the function in vertex form.

  1. Vertex form with unknown \(a\): \(f(x) = a(x - 3)^2 + 20\).
    • Insert \((0, 2)\): \(2 = a \cdot 9 + 20 \iff 9a = -18 \iff a = -2\)
    • \(f(x) = -2(x - 3)^2 + 20 = -2(x^2 - 6x + 9) + 20 = -2x^2 + 12x + 2\)
    • Zeros: \(-2(x - 3)^2 + 20 = 0 \iff (x - 3)^2 = 10 \iff x - 3 = \pm\sqrt{10} \iff x = 3 \pm \sqrt{10}\), i.e. \(x_1 \approx -0.16\) and \(x_2 \approx 6.16\)
  2. Factored form with unknown \(a\): \(f(x) = a(x + 1)(x - 5)\).
    • Insert \((1, 24)\): \(24 = a \cdot 2 \cdot (-4) = -8a \iff a = -3\)
    • \(f(x) = -3(x + 1)(x - 5) = -3(x^2 - 4x - 5) = -3x^2 + 12x + 15\)
    • Vertex halfway between the zeros: \(x_v = \frac{-1 + 5}{2} = 2\), \(f(2) = -3 \cdot 3 \cdot (-3) = 27\). Vertex \((2, 27)\)
    • \(a = -3 < 0\), the parabola opens downward, so the vertex is a maximum
  3. Vertex form: \(f(x) = a(x + 2)^2 - 5\). The \(y\)-intercept is \(f(0) = 4a - 5 = 3\), so \(a = 2\).
    • \(f(x) = 2(x + 2)^2 - 5 = 2x^2 + 8x + 8 - 5 = 2x^2 + 8x + 3\)
    • Discriminant: \(\Delta = 8^2 - 4 \cdot 2 \cdot 3 = 64 - 24 = 40 > 0\), so \(f\) has two zeros
    • Explanation: the vertex lies below the \(x\)-axis (\(-5 < 0\)) and the parabola opens upward (\(a = 2 > 0\)), so it crosses the \(x\)-axis twice
  4. Both points have the same height 7. The axis of symmetry lies halfway between them: \(x = \frac{1 + 9}{2} = 5\). So the vertex is \((5, -9)\).
    • \(f(x) = a(x - 5)^2 - 9\). Insert \((1, 7)\): \(7 = 16a - 9 \iff a = 1\)
    • \(f(x) = (x - 5)^2 - 9\). Check with \((9, 7)\): \(16 - 9 = 7\) ✓

B: Set Up & Check

Problem 5: From Text to Functions (xx)

Translate each statement into a function, a formula or a condition. Then answer the check question: it tells you whether your translation works.

  1. “A bowling centre rents out its lanes. At a price of \(p\) EUR per hour, the customers book \(Q(p) = 360 - 15p\) lane hours per week. The manager is looking for the price that maximizes the revenue.” Give the revenue \(R(p)\) and the formula for this price.

    Check: At 24 EUR per hour, nobody books a lane. Does the price with the highest revenue lie halfway between 0 and 24 EUR?

  2. “A joinery sells \(x\) raised garden beds per month at a price of \(p(x) = 250 - 5x\) EUR per bed. Its monthly profit is \(P(x) = -5x^2 + 150x - 800\) EUR.” An FSP task asks: ‘Compute the maximum profit and prove that it really is a maximum. Calculate the price per bed.’ Write down which function and which input each of the three answers needs.

    Check: A student answers: “The maximum lies at 15 beds, the price is 325 EUR per bed.” Is the price correct?

  3. “A tailor’s workshop sews uniforms for hotels in batches. For a batch of \(x\) uniforms, the costs per uniform (unit costs) are \(k(x) = 0.05x^2 - 6x + 250\) EUR. The workshop wants the batch size with the lowest unit costs.” Give the formula for this batch size and explain why the vertex is a minimum here.

    Check: Is a batch of 80 uniforms cheaper per uniform than a batch of 60 uniforms?

  4. “At an output of \(x\) tonnes per week, a crisps manufacturer makes a profit of \(P(x) = -2x^2 + 64x - 312\) (in 100 EUR). Its capacity (the maximum possible output) is 12 tonnes per week. The manager wants the output with the highest possible profit.” Give the domain of \(x\) in this context and the output the manager should choose.

    Hint: A parabola that opens downward rises up to its vertex and falls after it. If the whole domain lies left of the vertex, the largest value lies at the right end of the domain.

    Check: Should the manufacturer produce 16 tonnes per week, the \(x\)-coordinate of the vertex?

  5. “An umbrella manufacturer sells \(x\) thousand umbrellas per month at a price of \(p(x) = 18 - 0.5x\) EUR per umbrella. The variable costs (they grow with the output) are 4 EUR per umbrella, and the fixed costs are 30,000 EUR per month. The sales team says that the output with the highest revenue is also the output with the highest profit.” Give \(R(x)\) and \(P(x)\) in 1,000 EUR and the formulas for both outputs.

    Check: Is the sales team’s statement true?

  1. Revenue = price × quantity: \(R(p) = p \cdot Q(p) = p(360 - 15p) = -15p^2 + 360p\). Since \(a = -15 < 0\), the parabola opens downward and the vertex is a maximum:

    \(p_v = -\frac{360}{2 \cdot (-15)} = 12\) EUR per hour.

    Check: yes. \(Q(24) = 360 - 360 = 0\), and \(12 = \frac{0 + 24}{2}\). \(R\) has the zeros \(p = 0\) and \(p = 24\), and the vertex lies halfway between them. The revenue is \(R(12) = 12 \cdot 180 = 2{,}160\) EUR per week.

    Typical mistake: \(p_v = \frac{360}{2 \cdot (-15)} = -12\), without the minus sign of the vertex formula. A negative price cannot be right. The formula is \(x_v = -\frac{b}{2a}\).

  2. The three answers need three different steps:

    • Output with the maximum profit: the vertex of \(P\), \(x_v = -\frac{150}{2 \cdot (-5)} = 15\) beds. Maximum profit \(= P(x_v) = P(15)\)
    • Proof: \(a = -5 < 0\), the parabola opens downward, so the vertex is a maximum
    • Price: insert \(x_v\) into the price-demand function, \(p(15)\)

    Check: no. \(P(15) = -1125 + 2250 - 800 = 325\) EUR is the maximum profit per month. The price is \(p(15) = 250 - 75 = 175\) EUR per bed.

    Typical mistake: inserting \(x_v\) into \(P\) and calling the result the price. The price always comes from the price-demand function \(p(x)\), the profit from \(P(x)\).

  3. Batch size with the lowest unit costs: \(x_v = -\frac{-6}{2 \cdot 0.05} = 60\) uniforms. Here \(a = 0.05 > 0\), the parabola opens upward, so the vertex is a minimum: \(k(60) = 180 - 360 + 250 = 70\) EUR per uniform.

    Check: no. \(k(80) = 320 - 480 + 250 = 90\) EUR per uniform, 20 EUR more than with 60 uniforms.

    Typical mistake: “The larger the batch, the cheaper each uniform.” This only holds up to the vertex. Beyond 60 uniforms, the unit costs rise again (e.g. because of overtime), as the upward parabola shows.

  4. Domain: \(0 \leq x \leq 12\). The vertex lies at \(x_v = -\frac{64}{2 \cdot (-2)} = 16\), outside the domain. Since \(a = -2 < 0\), the parabola opens downward and the vertex is its maximum. Left of the axis of symmetry \(x = 16\), the profit still rises towards the vertex, and the whole domain lies left of it. So the best allowed output is the capacity: 12 tonnes per week, with \(P(12) = -288 + 768 - 312 = 168\), i.e. 16,800 EUR per week. Test value: \(P(10) = -200 + 640 - 312 = 128 < 168\) ✓

    Check: no. 16 tonnes lie above the capacity and cannot be produced.

    Typical mistake: answering with the vertex without checking the domain. If the vertex lies outside the domain, the best output lies at the edge of the domain that is closest to the vertex.

  5. \(R(x) = x \cdot p(x) = 18x - 0.5x^2\) and \(C(x) = 4x + 30\), so \(P(x) = 18x - 0.5x^2 - (4x + 30) = -0.5x^2 + 14x - 30\) (all in 1,000 EUR).

    • Highest revenue: \(x_R = -\frac{18}{2 \cdot (-0.5)} = 18\), i.e. 18,000 umbrellas
    • Highest profit: \(x_P = -\frac{14}{2 \cdot (-0.5)} = 14\), i.e. 14,000 umbrellas

    Check: no. \(P(14) = -98 + 196 - 30 = 68\), but \(P(18) = -162 + 252 - 30 = 60\). At the output with the highest revenue, the profit is 8,000 EUR lower.

    Typical mistake: maximizing \(R\) when the task asks for the profit. The variable costs change the coefficient of \(x\) (from 18 to 14) and move the vertex to a smaller output. The fixed costs do not move it.

Problem 6: One Paragraph, All Conditions (xx)

A mustard manufactory sells jars of mustard to delicatessen shops (shops for fine food). At a price of 12 EUR per jar or more, the shops buy no mustard (maximum price). At a price of 0 EUR, they would take 6,000 jars per month (market saturation). In between, the demand is linear. The fixed costs (rent, machines, staff) are 3,000 EUR per month, and each jar costs 2 EUR for mustard seeds, vinegar, glass and label. The manufactory can fill at most 2,800 jars per month (capacity). The owner wants to know the number of jars with the highest profit and the price for it, from which number of jars on the manufactory makes a profit, and whether it can fill the number of jars with the highest revenue.

List the price-demand function \(p(x)\) (price in EUR per jar at \(x\) jars per month), the revenue \(R(x)\), the total costs \(C(x)\), the profit \(P(x)\), the domain of \(x\) in this context and a condition or formula for each of the owner’s questions. Do not solve anything.

Check: Can the manufactory fill the number of jars with the highest revenue? And the number of jars with the highest profit? Which of your conditions show it?

  1. Price-demand function: \(p(0) = 12\), and \(p(6000) = 0\) gives the slope \(\frac{12}{6000} = 0.002\) EUR per jar: \(p(x) = 12 - 0.002x\)
  2. Revenue: \(R(x) = x \cdot p(x) = 12x - 0.002x^2\)
  3. Total costs: \(C(x) = 3000 + 2x\)
  4. Profit: \(P(x) = R(x) - C(x) = -0.002x^2 + 10x - 3000\)
  5. Domain: \(0 \leq x \leq 2800\) (the capacity lies below the market saturation)
  6. Highest profit: \(x_P = -\frac{10}{2 \cdot (-0.002)}\), a maximum since \(a = -0.002 < 0\). Price: \(p(x_P)\)
  7. Profit from … jars on: the break-even point, the lower zero of \(P(x) = 0\)
  8. Highest revenue: \(x_R = -\frac{12}{2 \cdot (-0.002)}\). It can be filled if \(x_R \leq 2800\)

Check: \(x_R = \frac{12}{0.004} = 3000 > 2800\), so the manufactory cannot fill the number of jars with the highest revenue (conditions 5 and 8). \(x_P = \frac{10}{0.004} = 2500 \leq 2800\), so it can fill the number of jars with the highest profit (conditions 5 and 6).

Typical mistake: \(p(x) = 12 - 500x\), with the slope \(\frac{6000}{12}\). Then \(p(1) = -488\) EUR. The slope of \(p(x)\) is the maximum price divided by the market saturation: the price falls by 0.002 EUR for each additional jar.

C: Applied

Problem 7: Ticket Prices of a Theatre (xx)

A theatre company plans the ticket prices for a new show. Its hall has 600 seats (the capacity: at most 600 visitors per show). Market research shows: at a price of 20 EUR per ticket, 500 people come to a show. For each 2 EUR increase in price, 20 fewer people come. Let \(p\) be the ticket price in EUR.

  1. Explain why the number of visitors falls by 10 per EUR. Set up the number of visitors \(A(p)\) with the point-slope form. Determine the maximum price. Determine the prices at which more people want a ticket than there are seats, and give the prices for which \(A(p)\) describes the number of visitors.

(for verification: \(A(p) = 700 - 10p\))

  1. Give the revenue \(R(p)\) per show. Compute the price with the highest revenue and prove that it is a maximum. Compute the number of visitors and the maximum revenue, and check that the hall is large enough.

(for verification: 35 EUR)

  1. The director would like to keep the old price of 20 EUR. Determine with the axis of symmetry the other price that brings the same revenue as 20 EUR, and check your answer by computation.

  2. The theatre needs a revenue of at least 12,000 EUR per show to cover its costs. Determine the ticket prices for which this holds.

  3. For the Saturday shows, the theatre expects a larger audience: at 20 EUR per ticket, 1,100 people would like to come, and for each EUR more, 20 fewer. Determine the price with the highest revenue for a hall without a seat limit, and decide whether the 600 seats are enough at this price. Determine the best Saturday price for the real hall with 600 seats and compare its revenue with the revenue that the price without a seat limit brings in the real hall.

Hint: As long as the hall is sold out, the revenue is 600 times the price. Right of its vertex, a parabola that opens downward falls.

a) 20 fewer visitors per 2 EUR means \(\frac{20}{2} = 10\) fewer visitors per EUR. The slope is \(-10\) visitors per EUR, and the line passes through \((20, 500)\):

\[A(p) - 500 = -10(p - 20) \iff A(p) = -10p + 200 + 500 = 700 - 10p \quad \checkmark\]

Maximum price: \(700 - 10p = 0 \iff p = 70\). At 70 EUR or more, nobody comes.

More people than seats: \(700 - 10p > 600 \iff -10p > -100 \iff p < 10\) (the sign flips when dividing by \(-10\)). Below 10 EUR the show is sold out, and only 600 people get a seat. So \(A(p)\) describes the number of visitors for \(10 \leq p \leq 70\).

b) Revenue = price × visitors:

\[R(p) = p \cdot A(p) = p(700 - 10p) = -10p^2 + 700p\]

Vertex formula: \(p_v = -\frac{700}{2 \cdot (-10)} = 35\). Since \(a = -10 < 0\), the parabola opens downward, so the vertex is a maximum.

  • Visitors: \(A(35) = 700 - 350 = 350\)
  • Maximum revenue: \(R(35) = 35 \cdot 350 = 12{,}250\) EUR per show
  • Capacity: \(350 \leq 600\), and \(35\) lies in \([10; 70]\) ✓. The hall is large enough

c) The axis of symmetry is \(p = 35\). The price 20 EUR lies \(35 - 20 = 15\) EUR below it, so \(35 + 15 = 50\) EUR brings the same revenue.

Check: \(R(20) = 20 \cdot 500 = 10{,}000\) EUR and \(R(50) = 50 \cdot (700 - 500) = 50 \cdot 200 = 10{,}000\) EUR ✓

d) Condition: \(R(p) \geq 12000\).

\[-10p^2 + 700p \geq 12000 \iff -10p^2 + 700p - 12000 \geq 0 \;\big|: (-10) \iff p^2 - 70p + 1200 \leq 0\]

The sign flips when dividing by \(-10\). Factor: two numbers with product 1200 and sum \(-70\) are \(-30\) and \(-40\):

\[(p - 30)(p - 40) \leq 0\]

The parabola \(p^2 - 70p + 1200\) opens upward, so it is negative between its zeros 30 and 40. Test value: \(p = 35\) gives \(1225 - 2450 + 1200 = -25 \leq 0\) ✓

The revenue is at least 12,000 EUR for ticket prices from 30 EUR to 40 EUR, i.e. \(p \in [30; 40]\). Check: \(R(30) = 30 \cdot 400 = 12{,}000\) EUR ✓

e) Saturday demand through \((20, 1100)\) with slope \(-20\):

\[A_S(p) = 1100 - 20(p - 20) = 1500 - 20p, \qquad R_S(p) = p(1500 - 20p) = -20p^2 + 1500p\]

Without a seat limit: \(p_v = -\frac{1500}{2 \cdot (-20)} = 37.5\) EUR, a maximum since \(a = -20 < 0\). At this price, \(A_S(37.5) = 1500 - 750 = 750\) people would like to come. The 600 seats are not enough: only 600 tickets can be sold, and the revenue is \(37.5 \cdot 600 = 22{,}500\) EUR.

Best price with 600 seats: raise the price until the hall is exactly full.

\[1500 - 20p = 600 \iff 20p = 900 \iff p = 45\]

  • Below 45 EUR the hall is sold out. The revenue is \(600p\) and grows with every EUR more
  • Above 45 EUR the revenue is \(R_S(p) = -20(p - 37.5)^2 + 28125\). It falls, because \(p\) moves further away from the vertex at 37.5

So the best Saturday price is 45 EUR, with \(45 \cdot 600 = 27{,}000\) EUR per show. That is 4,500 EUR more than at 37.5 EUR. The seat limit overrides the vertex of the revenue parabola.

Problem 8: Handmade Leather Wallets (xx)

A small company makes handmade leather wallets and sells them online. At a price of \(p\) EUR per wallet, it sells \(Q(p) = 2000 - 25p\) wallets per month. The fixed costs (rent, staff, online shop) are 15,000 EUR per month, and each wallet costs 20 EUR in leather and material (variable costs).

  1. Give the revenue \(R(p)\). Explain why the total costs also depend on the price and show that \(C(p) = 55000 - 500p\). Show that the profit function (written \(\Pi\), since \(p\) is the price) is \(\Pi(p) = -25p^2 + 2500p - 55000\).

  2. Compute the price with the highest profit and prove that it is a maximum. Compute the number of wallets sold and the maximum profit per month.

(for verification: 50 EUR)

  1. Compute the price with the highest revenue and the profit at this price. Explain with the changes in revenue and costs why the profit-maximizing price is higher.

  2. Determine the break-even prices exactly and in EUR, rounded to cents (two decimal places). Give the prices at which the company makes a profit.

  3. The landlord raises the rent, so the fixed costs rise by 3,000 EUR per month. Explain without a new computation why the profit-maximizing price stays the same, and give the new maximum profit.

a) Revenue = price × quantity:

\[R(p) = p \cdot Q(p) = p(2000 - 25p) = 2000p - 25p^2\]

The total costs depend on the number of wallets. This number depends on the price: a higher price means fewer wallets and lower variable costs.

\[C(p) = 15000 + 20 \cdot Q(p) = 15000 + 20(2000 - 25p) = 15000 + 40000 - 500p = 55000 - 500p \quad \checkmark\]

\[\Pi(p) = R(p) - C(p) = 2000p - 25p^2 - (55000 - 500p) = -25p^2 + 2500p - 55000 \quad \checkmark\]

b) Vertex formula: \(p_v = -\frac{2500}{2 \cdot (-25)} = 50\) EUR. Since \(a = -25 < 0\), the parabola opens downward, so the vertex is a maximum.

  • Wallets sold: \(Q(50) = 2000 - 1250 = 750\) per month
  • Maximum profit: \(\Pi(50) = -62500 + 125000 - 55000 = 7{,}500\) EUR per month

c) Revenue maximum: \(p_R = -\frac{2000}{2 \cdot (-25)} = 40\) EUR, a maximum since \(a = -25 < 0\). \(Q(40) = 1000\) wallets and \(R(40) = 40{,}000\) EUR.

Profit at 40 EUR: \(\Pi(40) = -40000 + 100000 - 55000 = 5{,}000\) EUR, i.e. 2,500 EUR less than at 50 EUR.

From 40 EUR to 50 EUR, the company sells \(1000 - 750 = 250\) wallets fewer. The revenue falls from 40,000 EUR to \(50 \cdot 750 = 37{,}500\) EUR, i.e. by 2,500 EUR. The variable costs fall by \(250 \cdot 20 = 5{,}000\) EUR. The profit rises by \(5000 - 2500 = 2{,}500\) EUR. Every wallet costs 20 EUR to make, so selling fewer wallets at a higher price pays off, and the profit maximum lies at a higher price than the revenue maximum.

d) \(\Pi(p) = 0\). Divide by \(-25\):

\[-25p^2 + 2500p - 55000 = 0 \iff p^2 - 100p + 2200 = 0\]

Discriminant: \(\Delta = 100^2 - 4 \cdot 2200 = 10000 - 8800 = 1200\). With \(\sqrt{1200} = \sqrt{400 \cdot 3} = 20\sqrt{3}\):

\[p_{1,2} = \frac{100 \pm 20\sqrt{3}}{2} = 50 \pm 10\sqrt{3}\]

\[p_1 = 50 - 10\sqrt{3} \approx 32.68 \text{ EUR}, \qquad p_2 = 50 + 10\sqrt{3} \approx 67.32 \text{ EUR}\]

The parabola opens downward, so the profit is positive between the zeros. The company makes a profit for prices between 32.68 EUR and 67.32 EUR (\(50 - 10\sqrt{3} \approx 32.6795\) rounds up to 32.68, and \(50 + 10\sqrt{3} \approx 67.3205\) rounds down to 67.32, so both rounded prices still lie inside the profit zone).

e) The new profit function is \(\Pi_{\text{new}}(p) = \Pi(p) - 3000 = -25p^2 + 2500p - 58000\). Only the constant term changes. The vertex formula \(p_v = -\frac{b}{2a}\) uses only \(a = -25\) and \(b = 2500\), so the best price stays 50 EUR. The whole parabola moves down by 3,000.

New maximum profit: \(7500 - 3000 = 4{,}500\) EUR per month.

Problem 9: Fireworks for a Summer Festival (xx)

A fireworks company prepares the show for a town’s summer festival and tests a new rocket. The rocket is launched from a platform 5 m above the ground. Without a burst, its height after \(t\) seconds would be \(h(t) = -5t^2 + 40t + 5\) metres. The effect charge (the charge that makes the coloured stars) ignites at a time that is set on the rocket’s fuse.

  1. Compute when the rocket reaches its highest point and how high this point is. Prove that it is the highest point.

  2. The town wants the burst at a height of at least 65 m, above the church towers of the old town. Determine the times at which the rocket is at least 65 m high.

(for verification: from 2 s to 6 s)

  1. The fuse is not exact: the burst happens up to 0.5 s before or after the set time. Determine the set times for which the burst is at least 65 m high in any case. The company sets the fuse to 4 s. Explain why this is the best choice and give the lowest possible burst height with this setting.

Hint: A parabola that opens downward rises up to its vertex and falls after it, so on an interval its lowest value lies at one of the two ends.

  1. For the finale, the company wants a larger rocket, launched from the same platform, that reaches its highest point of 130 m after 5 seconds. Determine its height function \(h_2(t)\) in vertex form and in standard form. Compute for how many seconds this rocket is at least 65 m high, rounded to one decimal place.

a) Vertex formula: \(t_v = -\frac{40}{2 \cdot (-5)} = 4\) s and \(h(4) = -80 + 160 + 5 = 85\) m.

Since \(a = -5 < 0\), the parabola opens downward, so the vertex is a maximum. The rocket reaches its highest point of 85 m after 4 seconds. In vertex form, \(h(t) = -5(t - 4)^2 + 85\).

b) Condition: \(h(t) \geq 65\).

\[-5t^2 + 40t + 5 \geq 65 \iff -5t^2 + 40t - 60 \geq 0 \;\big|: (-5) \iff t^2 - 8t + 12 \leq 0\]

The sign flips when dividing by \(-5\). Factor: \((t - 2)(t - 6) \leq 0\). The parabola \(t^2 - 8t + 12\) opens upward and is negative between its zeros. Test value \(t = 4\): \(16 - 32 + 12 = -4 \leq 0\) ✓

The rocket is at least 65 m high from 2 s to 6 s after the launch, i.e. \(t \in [2; 6]\). Check: \(h(2) = -20 + 80 + 5 = 65\) ✓

c) With the set time \(s\), the burst happens between \(s - 0.5\) and \(s + 0.5\). This whole interval must lie in \([2; 6]\):

\[s - 0.5 \geq 2 \iff s \geq 2.5 \qquad \text{and} \qquad s + 0.5 \leq 6 \iff s \leq 5.5\]

Every set time from 2.5 s to 5.5 s is safe.

With \(s = 4\), the burst happens between 3.5 s and 4.5 s. These times lie symmetric to the axis of symmetry \(t = 4\), so they give the same height:

\[h(3.5) = -5(3.5 - 4)^2 + 85 = -5 \cdot 0.25 + 85 = 83.75 \text{ m} = h(4.5)\]

Between these two times, the rocket rises to its highest point and falls again, so it is never lower than at the two ends: the burst is at least 83.75 m high. 4 s is the best setting: the burst happens as close as possible to the highest point, and the lowest possible burst height is as large as possible. Any other set time moves one end of the burst interval further away from \(t = 4\), where the rocket is lower.

d) Vertex form with the vertex \((5, 130)\): \(h_2(t) = a(t - 5)^2 + 130\). The launch point is \((0, 5)\):

\[5 = a \cdot 25 + 130 \iff 25a = -125 \iff a = -5\]

\[h_2(t) = -5(t - 5)^2 + 130 = -5(t^2 - 10t + 25) + 130 = -5t^2 + 50t + 5\]

At least 65 m:

\[-5(t - 5)^2 + 130 \geq 65 \iff -5(t - 5)^2 \geq -65 \iff (t - 5)^2 \leq 13 \iff |t - 5| \leq \sqrt{13}\]

So \(5 - \sqrt{13} \leq t \leq 5 + \sqrt{13}\), i.e. about 1.39 s to 8.61 s. (The same bounds come from the standard form as in b): \(-5t^2 + 50t + 5 \geq 65 \iff t^2 - 10t + 12 \leq 0\), and the quadratic formula gives \(t = 5 \pm \sqrt{13}\).) Duration: \(2\sqrt{13} \approx 7.2\) s. The finale rocket stays above 65 m almost twice as long as the test rocket (4 s).

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: Wallboxes for Electric Cars [23 pts. total]

A young company builds wallboxes (charging stations for electric cars that are mounted on a garage wall) and sells them directly to homeowners in its region. Its fixed costs (development, staff, rent) are 324,000 EUR per year. It models its total costs per year by

\[C(x) = x^2 + 8x + 324\]

with \(C\) in MU and \(x\) in QU per year. The term \(x^2\) describes costs per wallbox that rise with the output, for example because a second, more expensive supplier delivers the chips and the staff work overtime.

(QU = quantity units, MU = monetary units. 1 QU = 100 wallboxes, 1 MU = 1,000 EUR, so a price of \(p\) MU per QU is a price of \(10p\) EUR per wallbox. Revenue = quantity × price. Profit = revenue − total costs.)

Part a)

Determine the linear price-demand function \(p(x)\) (price in MU per QU at a demand of \(x\) QU per year) assuming a maximum price of 800 EUR per wallbox and a market saturation at 4,000 wallboxes per year. Give the revenue function \(R(x)\). Compute the output with the highest revenue and the corresponding price per wallbox in EUR. [5 pts.]

(for verification: \(p(x) = 80 - 2x\))

Units. A price of 800 EUR per wallbox means \(100 \cdot 800 = 80{,}000\) EUR \(= 80\) MU per QU. The market saturation is 4,000 wallboxes \(= 40\) QU.

Linear price-demand function \(p(x) = c - dx\). At the maximum price the demand is 0, so \(p(0) = 80\) and \(c = 80\). At the market saturation the price is 0:

\[p(40) = 80 - 40d = 0 \iff d = 2 \qquad \Rightarrow \qquad p(x) = 80 - 2x, \quad 0 \leq x \leq 40\]

Revenue = quantity × price:

\[R(x) = x \cdot p(x) = x(80 - 2x) = -2x^2 + 80x\]

Highest revenue. Vertex formula: \(x_R = -\frac{80}{2 \cdot (-2)} = 20\). Since \(a = -2 < 0\), the parabola opens downward, so the vertex is a maximum. The revenue is highest at 20 QU \(= 2{,}000\) wallboxes per year, with \(R(20) = -800 + 1600 = 800\) MU \(= 800{,}000\) EUR.

Price: \(p(20) = 80 - 40 = 40\) MU per QU \(= 400\) EUR per wallbox.

Grading: 1 Pt. conversion of maximum price and market saturation to MU per QU and QU, 1 Pt. \(p(x)\), 1 Pt. \(R(x) = x \cdot p(x)\), 1 Pt. vertex at 20 QU with the reason \(a < 0\), 1 Pt. price of 400 EUR per wallbox.

Part b)

Show that the profit function is given by \(P(x) = -3x^2 + 72x - 324\). [2 pts.]

\[P(x) = R(x) - C(x) = (-2x^2 + 80x) - (x^2 + 8x + 324)\]

The minus sign changes every sign in the second bracket:

\[P(x) = -2x^2 + 80x - x^2 - 8x - 324 = -3x^2 + 72x - 324 \quad \checkmark\]

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

Part c)

Determine the (lower) break-even point and the profit limit (upper break-even point). Decide in which output range the company makes a profit and state this profit zone as an interval. [4 pts.]

(for verification: break-even point at 6 QU)

Zero form. At a break-even point the profit is zero, \(P(x) = 0\). Divide by \(-3\):

\[-3x^2 + 72x - 324 = 0 \;\big|: (-3) \iff x^2 - 24x + 108 = 0\]

Factor: two numbers with product 108 and sum \(-24\) are \(-6\) and \(-18\):

\[(x - 6)(x - 18) = 0\]

Zero-product rule: \(x = 6\) or \(x = 18\). Check: \(P(6) = -108 + 432 - 324 = 0\) ✓

Both zeros lie in the domain \([0; 40]\). Break-even point: \(x = 6\) QU \(= 600\) wallboxes. Profit limit: \(x = 18\) QU \(= 1{,}800\) wallboxes.

Profit zone. The parabola opens downward (\(a = -3 < 0\)), so the profit is positive between the zeros. Test value: \(P(10) = -300 + 720 - 324 = 96 > 0\) ✓. The company makes a profit for outputs in \((6; 18)\), i.e. between 600 and 1,800 wallboxes per year. The interval is open, since the profit is zero at both ends.

Grading: 1 Pt. zero form, 1 Pt. factoring and zero-product rule, 1 Pt. naming break-even point and profit limit, 1 Pt. open interval with reason.

Part d)

Compute the maximum profit in EUR and prove that it really is a maximum. Calculate the price per wallbox in EUR that the company must ask in order to sell exactly the profit-maximizing output. [5 pts.]

(for verification: 1,200 wallboxes per year)

Vertex formula:

\[x_v = -\frac{72}{2 \cdot (-3)} = 12 \text{ QU} = 1{,}200 \text{ wallboxes per year}\]

(The vertex lies halfway between the zeros 6 and 18 ✓.)

Maximum profit:

\[P(12) = -3 \cdot 144 + 72 \cdot 12 - 324 = -432 + 864 - 324 = 108 \text{ MU} = 108{,}000 \text{ EUR}\]

Proof. \(a = -3 < 0\), the parabola opens downward, so the vertex is a maximum. With the vertex form: \(P(x) = -3(x - 12)^2 + 108\). A square is never negative, so \(-3(x - 12)^2 \leq 0\) and \(P(x) \leq 108\) for every \(x\), with equality only for \(x = 12\).

Price: \(p(12) = 80 - 24 = 56\) MU per QU \(= 10 \cdot 56 = 560\) EUR per wallbox.

Grading: 1 Pt. \(x_v = 12\), 1 Pt. maximum profit of 108,000 EUR, 1 Pt. proof with \(a < 0\) (or the vertex form), 1 Pt. \(p(12) = 56\) MU per QU, 1 Pt. 560 EUR per wallbox.

Part e)

The sales manager wants to sell 2,000 wallboxes per year, since the revenue is highest there. Compute the profit at this output and explain with the changes in revenue and costs why the profit maximum lies at a smaller output. [3 pts.]

Profit at 20 QU:

\[P(20) = -1200 + 1440 - 324 = -84 \text{ MU}\]

The company would make a loss of 84,000 EUR per year. 20 QU lies above the profit limit of 18 QU, outside the profit zone from c).

Changes from 12 QU to 20 QU:

  • Revenue: \(R(12) = -288 + 960 = 672\) MU and \(R(20) = 800\) MU, i.e. \(+128\) MU
  • Costs: \(C(12) = 144 + 96 + 324 = 564\) MU and \(C(20) = 400 + 160 + 324 = 884\) MU, i.e. \(+320\) MU

The 800 additional wallboxes bring 128,000 EUR more revenue but cost 320,000 EUR more, so the profit falls by 192,000 EUR (from 108,000 EUR to a loss of 84,000 EUR). To sell more, the company has to lower the price for all wallboxes, while the costs per wallbox rise with the output. So the profit maximum lies at a smaller output than the revenue maximum.

Grading: 1 Pt. loss of 84,000 EUR, 1 Pt. changes of revenue and costs, 1 Pt. explanation.

Part f)

The company’s bank grants a loan for a new production hall only if the average cost per wallbox (total costs divided by the number of wallboxes) is at most 530 EUR. Give the average cost \(AC(x) = \frac{C(x)}{x}\) in MU per QU and its domain restriction. Determine the outputs at which the average cost is exactly 530 EUR per wallbox. Decide for which outputs the bank’s condition holds and whether it holds at the profit-maximizing output. [4 pts.]

Hint: The denominator is positive, so you can multiply both sides of the inequality by it without changing the sign.

Average cost:

\[AC(x) = \frac{x^2 + 8x + 324}{x} = x + 8 + \frac{324}{x} \text{ MU per QU}\]

Domain restriction: \(x \neq 0\) (in the context \(0 < x \leq 40\)).

Exactly 530 EUR per wallbox \(= 100 \cdot 530 = 53{,}000\) EUR per QU \(= 53\) MU per QU. Multiplying by \(x\) is allowed, since \(x \neq 0\):

\[\frac{x^2 + 8x + 324}{x} = 53 \iff x^2 + 8x + 324 = 53x \iff x^2 - 45x + 324 = 0\]

Two numbers with product 324 and sum \(-45\) are \(-9\) and \(-36\): \((x - 9)(x - 36) = 0\), so \(x = 9\) or \(x = 36\). Both lie in \((0; 40]\). The average cost is exactly 530 EUR at 900 and 3,600 wallboxes per year. Check: \(AC(9) = 9 + 8 + 36 = 53\) ✓

Bank’s condition \(AC(x) \leq 53\). Multiplying by \(x > 0\) keeps the sign: \(x^2 - 45x + 324 \leq 0\). This upward parabola is negative between its zeros 9 and 36. Test values: \(AC(4) = 4 + 8 + 81 = 93 > 53\) and \(AC(12) = 12 + 8 + 27 = 47 \leq 53\) ✓

The condition holds for 900 to 3,600 wallboxes per year, i.e. \(x \in [9; 36]\). At the profit-maximizing output of 1,200 wallboxes, the average cost is 47 MU per QU \(= 470\) EUR per wallbox \(\leq 530\) EUR. The bank’s condition holds.

Grading: 1 Pt. \(AC(x)\) with \(x \neq 0\), 1 Pt. conversion to 53 MU per QU and zero form, 1 Pt. outputs 900 and 3,600 wallboxes, 1 Pt. outputs \([9; 36]\) with test value and decision at 1,200 wallboxes.