Tasks 06-04 - Area Between Curves

Section 06: Integral Calculus

A: Skills

Problem 1: Intersection Points and the Upper Graph (x)

Find the \(x\)-coordinates of all intersection points of the graphs of \(f\) and \(g\). Determine which graph lies above the other between neighbouring intersection points. In d), compare the graphs to the left and to the right of the intersection point.

  1. \(f(x) = 3x^2 + x - 6\) and \(g(x) = 2x + 4\)

  2. \(f(x) = 3x^2 - 13x\) and \(g(x) = 10\)

  3. \(f(x) = x^3 + 2x^2 - 5x\) and \(g(x) = 3x\)

  4. \(f(x) = e^{2x}\) and \(g(x) = 3e^x\)

  1. \(3x^2 + x - 6 = 2x + 4 \iff 3x^2 - x - 10 = 0\). Quadratic formula:

    \[x = \frac{1 \pm \sqrt{1 + 120}}{6} = \frac{1 \pm 11}{6}, \qquad x = 2 \text{ or } x = -\frac{5}{3}\]

    Test \(x = 0\) between the intersection points: \(f(0) = -6\) and \(g(0) = 4\). So \(g\) lies above \(f\) on \(\left(-\frac{5}{3}; 2\right)\).

  2. \(3x^2 - 13x = 10 \iff 3x^2 - 13x - 10 = 0\). Quadratic formula:

    \[x = \frac{13 \pm \sqrt{169 + 120}}{6} = \frac{13 \pm 17}{6}, \qquad x = 5 \text{ or } x = -\frac{2}{3}\]

    Test \(x = 0\): \(f(0) = 0 < 10 = g(0)\). So \(g\) lies above \(f\) on \(\left(-\frac{2}{3}; 5\right)\).

  3. \(x^3 + 2x^2 - 5x = 3x \iff x^3 + 2x^2 - 8x = 0 \iff x\left(x^2 + 2x - 8\right) = 0 \iff x(x + 4)(x - 2) = 0\)

    Intersection points at \(x = -4\), \(x = 0\) and \(x = 2\).

    • On \((-4; 0)\), test \(x = -1\): \(f(-1) = -1 + 2 + 5 = 6\) and \(g(-1) = -3\). \(f\) lies above \(g\).
    • On \((0; 2)\), test \(x = 1\): \(f(1) = 1 + 2 - 5 = -2\) and \(g(1) = 3\). \(g\) lies above \(f\).

  4. \(e^{2x} = 3e^x\). Since \(e^x > 0\), divide by \(e^x\): \(e^x = 3 \iff x = \ln 3 \approx 1.10\).

    • Left of \(\ln 3\), test \(x = 0\): \(f(0) = 1 < 3 = g(0)\). \(g\) lies above \(f\).
    • Right of \(\ln 3\), test \(x = 2\): \(f(2) = e^4 \approx 54.60 > 3e^2 \approx 22.17 = g(2)\). \(f\) lies above \(g\).

Typical mistake: in c), dividing \(x^3 + 2x^2 = 8x\) by \(x\). Then \(x^2 + 2x - 8 = 0\) gives only \(x = -4\) and \(x = 2\), and the intersection point at \(x = 0\) is lost. Bring every term to one side and factor out \(x\) instead.

Problem 2: Area Between Two Curves (xx)

Compute the area of the region enclosed by the graphs of \(f\) and \(g\). Use the three steps from the lecture: intersection points as limits, the upper graph, the integral of top minus bottom.

  1. \(f(x) = 2x^2 + 9x + 5\) and \(g(x) = -2x\)

  2. \(f(x) = 3x^2 - 10x - 4\) and \(g(x) = -3x^2 + 5x + 5\)

  1. Intersection points: \(2x^2 + 9x + 5 = -2x \iff 2x^2 + 11x + 5 = 0\)

    \[x = \frac{-11 \pm \sqrt{121 - 40}}{4} = \frac{-11 \pm 9}{4}, \qquad x = -\frac{1}{2} \text{ or } x = -5\]

    Upper graph: test \(x = -2\): \(f(-2) = 8 - 18 + 5 = -5\) and \(g(-2) = 4\). So \(g\) lies above \(f\) on \(\left[-5; -\frac{1}{2}\right]\).

    Area:

    \[\int_{-5}^{-1/2} \left[-2x - \left(2x^2 + 9x + 5\right)\right] dx = \int_{-5}^{-1/2} \left(-2x^2 - 11x - 5\right) dx = \left[-\frac{2}{3}x^3 - 5.5x^2 - 5x\right]_{-5}^{-1/2}\]

    \[= \left(\frac{1}{12} - 1.375 + 2.5\right) - \left(\frac{250}{3} - 137.5 + 25\right) = \frac{29}{24} - \left(-\frac{175}{6}\right) = \frac{29}{24} + \frac{700}{24} = \frac{729}{24} = \frac{243}{8} = 30.375\]

  2. Intersection points: \(3x^2 - 10x - 4 = -3x^2 + 5x + 5 \iff 6x^2 - 15x - 9 = 0 \iff 2x^2 - 5x - 3 = 0\)

    \[x = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}, \qquad x = 3 \text{ or } x = -\frac{1}{2}\]

    Upper graph: test \(x = 0\): \(f(0) = -4\) and \(g(0) = 5\). So \(g\) lies above \(f\) on \(\left[-\frac{1}{2}; 3\right]\).

    Area:

    \[\int_{-1/2}^{3} \left[\left(-3x^2 + 5x + 5\right) - \left(3x^2 - 10x - 4\right)\right] dx = \int_{-1/2}^{3} \left(-6x^2 + 15x + 9\right) dx = \left[-2x^3 + 7.5x^2 + 9x\right]_{-1/2}^{3}\]

    \[= (-54 + 67.5 + 27) - (0.25 + 1.875 - 4.5) = 40.5 - (-2.375) = 42.875\]

Typical mistake: integrating \(f - g\) although \(g\) is the upper graph. In a), this gives \(-30.375\). An area is never negative: find the upper graph with a test value first, then integrate top minus bottom.

Problem 3: Curves That Cross (xx)

Compute the total area between the graphs of \(f\) and \(g\). The graphs cross, so split the interval at every intersection point and add the areas of the regions.

  1. \(f(x) = x^3 - 2x^2 - x\) and \(g(x) = 2x\), the region enclosed by both graphs. Compare the total area with \(\int_{-1}^{3} [f(x) - g(x)] \, dx\).

  2. \(f(x) = e^x\) and \(g(x) = 2\) on the interval \([0; 2]\). Compare the total area with \(\int_0^2 [f(x) - g(x)] \, dx\). Give exact values and decimals rounded to two decimal places.

  1. Intersection points: \(x^3 - 2x^2 - x = 2x \iff x^3 - 2x^2 - 3x = 0 \iff x\left(x^2 - 2x - 3\right) = 0 \iff x(x + 1)(x - 3) = 0\)

    So \(x = -1\), \(x = 0\) and \(x = 3\). Let \(d(x) = f(x) - g(x) = x^3 - 2x^2 - 3x\) with the antiderivative \(D(x) = \frac{x^4}{4} - \frac{2x^3}{3} - 1.5x^2\).

    • On \([-1; 0]\), test \(x = -0.5\): \(d(-0.5) = -0.125 - 0.5 + 1.5 = 0.875 > 0\), so \(f\) lies above \(g\).
    • On \([0; 3]\), test \(x = 1\): \(d(1) = 1 - 2 - 3 = -4 < 0\), so \(g\) lies above \(f\).

    With \(D(-1) = \frac{1}{4} + \frac{2}{3} - \frac{3}{2} = -\frac{7}{12}\), \(D(0) = 0\) and \(D(3) = \frac{81}{4} - 18 - \frac{27}{2} = -\frac{45}{4}\):

    \[A_1 = \int_{-1}^{0} [f(x) - g(x)] \, dx = D(0) - D(-1) = \frac{7}{12}, \qquad A_2 = \int_{0}^{3} [g(x) - f(x)] \, dx = -(D(3) - D(0)) = \frac{45}{4}\]

    Total area: \(\frac{7}{12} + \frac{45}{4} = \frac{7}{12} + \frac{135}{12} = \frac{142}{12} = \frac{71}{6} \approx 11.83\)

    Comparison: \(\int_{-1}^{3} [f(x) - g(x)] \, dx = D(3) - D(-1) = -\frac{45}{4} + \frac{7}{12} = -\frac{32}{3} \approx -10.67\). In the integral, the second region counts negative and the first positive. So the integral is neither the area nor even positive.

  2. Intersection point: \(e^x = 2 \iff x = \ln 2 \approx 0.69\), which lies in \([0; 2]\).

    • On \([0; \ln 2]\), test \(x = 0\): \(e^0 = 1 < 2\), so \(g\) lies above \(f\).
    • On \([\ln 2; 2]\), test \(x = 2\): \(e^2 \approx 7.39 > 2\), so \(f\) lies above \(g\).

    \[A_1 = \int_0^{\ln 2} \left(2 - e^x\right) dx = \left[2x - e^x\right]_0^{\ln 2} = (2\ln 2 - 2) - (0 - 1) = 2\ln 2 - 1 \approx 0.39\]

    \[A_2 = \int_{\ln 2}^{2} \left(e^x - 2\right) dx = \left[e^x - 2x\right]_{\ln 2}^{2} = \left(e^2 - 4\right) - (2 - 2\ln 2) = e^2 - 6 + 2\ln 2 \approx 2.78\]

    Total area: \(A_1 + A_2 = e^2 + 4\ln 2 - 7 \approx 3.16\)

    Comparison: \(\int_0^2 \left(e^x - 2\right) dx = \left[e^x - 2x\right]_0^2 = e^2 - 4 - 1 = e^2 - 5 \approx 2.39\). This is \(A_2 - A_1\): the region where \(g\) lies above \(f\) is subtracted instead of added.

Typical mistake: computing one integral over the whole interval and taking its absolute value. In a), \(\left|-\frac{32}{3}\right| \approx 10.67\) is still not the area \(\approx 11.83\), because the two regions cancel partly. Split at every intersection point first.

Problem 4: Bounded Regions (xx)

Sketch the region and compute its area. Give exact values and decimals rounded to two decimal places.

  1. The region enclosed by the graphs of \(y = \frac{3}{x}\) and \(y = 7 - 2x\)

  2. The region bounded by the graph of \(y = e^{0.5x}\), the line \(y = 5\) and the \(y\)-axis

  3. The region bounded by the graph of \(y = \sqrt{x}\), the line \(y = 20 - x\) and the \(x\)-axis

Hint: In c), the upper boundary of the region changes at the intersection point of the two graphs.

  1. Intersection points: \(\frac{3}{x} = 7 - 2x\). Multiply by \(x \neq 0\): \(3 = 7x - 2x^2 \iff 2x^2 - 7x + 3 = 0\)

    \[x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}, \qquad x = 3 \text{ or } x = \frac{1}{2}\]

    Upper graph: test \(x = 1\): \(\frac{3}{1} = 3 < 5 = 7 - 2\). The line lies above the hyperbola on \(\left[\frac{1}{2}; 3\right]\).

    \[\int_{1/2}^{3} \left(7 - 2x - \frac{3}{x}\right) dx = \left[7x - x^2 - 3\ln|x|\right]_{1/2}^{3} = (21 - 9 - 3\ln 3) - \left(3.5 - 0.25 - 3\ln\frac{1}{2}\right)\]

    \[= 8.75 - 3\ln 3 + 3\ln\frac{1}{2} = 8.75 - 3\ln 6 \approx 3.37\]

    (With \(\ln 3 - \ln\frac{1}{2} = \ln 6\).)

  2. Intersection point: \(e^{0.5x} = 5 \iff 0.5x = \ln 5 \iff x = 2\ln 5 \approx 3.22\). The \(y\)-axis is the line \(x = 0\). On \([0; 2\ln 5]\), \(e^{0.5x} \leq 5\), so the line lies on top.

    \[\int_0^{2\ln 5} \left(5 - e^{0.5x}\right) dx = \left[5x - 2e^{0.5x}\right]_0^{2\ln 5} = \left(10\ln 5 - 2 \cdot 5\right) - (0 - 2) = 10\ln 5 - 8 \approx 8.09\]

    (With \(e^{0.5 \cdot 2\ln 5} = e^{\ln 5} = 5\).)

  3. Intersection point: \(\sqrt{x} = 20 - x\). Square both sides (Session 02-03): \(x = 400 - 40x + x^2 \iff x^2 - 41x + 400 = 0\)

    \[x = \frac{41 \pm \sqrt{1681 - 1600}}{2} = \frac{41 \pm 9}{2}, \qquad x = 25 \text{ or } x = 16\]

    Check in the original equation: \(\sqrt{16} = 4 = 20 - 16\) ✓, but \(\sqrt{25} = 5 \neq -5 = 20 - 25\). So the graphs intersect only at \((16, 4)\).

    Region: the line meets the \(x\)-axis at \(x = 20\). From \(x = 0\) to \(x = 16\), the region lies under \(y = \sqrt{x}\), and from \(x = 16\) to \(x = 20\) under the line.

    \[\int_0^{16} x^{1/2} \, dx + \int_{16}^{20} (20 - x) \, dx = \left[\frac{2}{3}x^{3/2}\right]_0^{16} + \left[20x - \frac{x^2}{2}\right]_{16}^{20} = \frac{128}{3} + (200 - 192) = \frac{152}{3} \approx 50.67\]

Typical mistake: in c), keeping \(x = 25\). Squaring can add a solution that does not fit the original equation, so check both values. The region also ends at \(x = 20\), where the line meets the \(x\)-axis, not at \(x = 25\).

Problem 5: Consumer and Producer Surplus (xx)

Prices are in EUR per unit, and \(q\) is the quantity in units. The consumer surplus is \(CS = \int_0^{q^*} [D(q) - p^*] \, dq\) and the producer surplus is \(PS = \int_0^{q^*} [p^* - S(q)] \, dq\).

  1. Compute the consumer surplus for the demand function \(D(q) = 130 - 2.5q\) at the market price \(p^* = 80\).

  2. Compute the producer surplus for the supply function \(S(q) = 28 + 0.9q\) at the market price \(p^* = 55\).

  3. Determine the equilibrium \((q^*, p^*)\) for \(D(q) = 104 - 2.4q\) and \(S(q) = 11 + 0.7q\). Compute the consumer surplus, the producer surplus and the total surplus. Check that the total surplus equals \(\int_0^{q^*} [D(q) - S(q)] \, dq\), the area between the demand and supply curves.

  4. Determine the equilibrium for \(D(q) = 61 - 0.9q\) and \(S(q) = 7 + 0.45q\). Compute the consumer surplus, the producer surplus and the total surplus.

  1. Quantity at the market price: \(130 - 2.5q = 80 \iff q^* = 20\).

    \[CS = \int_0^{20} [(130 - 2.5q) - 80] \, dq = \int_0^{20} (50 - 2.5q) \, dq = \left[50q - 1.25q^2\right]_0^{20} = 1000 - 500 = 500 \text{ EUR}\]

  2. Quantity at the market price: \(28 + 0.9q = 55 \iff 0.9q = 27 \iff q^* = 30\).

    \[PS = \int_0^{30} [55 - (28 + 0.9q)] \, dq = \int_0^{30} (27 - 0.9q) \, dq = \left[27q - 0.45q^2\right]_0^{30} = 810 - 405 = 405 \text{ EUR}\]

  3. Equilibrium: \(104 - 2.4q = 11 + 0.7q \iff 93 = 3.1q \iff q^* = 30\), and \(p^* = 104 - 72 = 32\) EUR. Check: \(S(30) = 11 + 21 = 32\) ✓

    \[CS = \int_0^{30} (72 - 2.4q) \, dq = \left[72q - 1.2q^2\right]_0^{30} = 2160 - 1080 = 1080 \text{ EUR}\]

    \[PS = \int_0^{30} (21 - 0.7q) \, dq = \left[21q - 0.35q^2\right]_0^{30} = 630 - 315 = 315 \text{ EUR}\]

    Total surplus: \(1080 + 315 = 1395\) EUR.

    Check: \(\int_0^{30} [(104 - 2.4q) - (11 + 0.7q)] \, dq = \int_0^{30} (93 - 3.1q) \, dq = \left[93q - 1.55q^2\right]_0^{30} = 2790 - 1395 = 1395\) ✓

  4. Equilibrium: \(61 - 0.9q = 7 + 0.45q \iff 54 = 1.35q \iff q^* = 40\), and \(p^* = 61 - 36 = 25\) EUR. Check: \(S(40) = 7 + 18 = 25\) ✓

    \[CS = \int_0^{40} (36 - 0.9q) \, dq = \left[36q - 0.45q^2\right]_0^{40} = 1440 - 720 = 720 \text{ EUR}\]

    \[PS = \int_0^{40} (18 - 0.45q) \, dq = \left[18q - 0.225q^2\right]_0^{40} = 720 - 360 = 360 \text{ EUR}\]

    Total surplus: \(720 + 360 = 1080\) EUR.

Typical mistake: integrating up to the quantity where the demand reaches 0 (in a), \(q = 52\)) instead of up to \(q^*\). Only the \(q^*\) units that are actually sold create a surplus.

B: Set Up & Check

Problem 6: From Words to Integrals (xx)

Translate each statement into one or more integrals. Then answer the check question: it tells you whether your translation works.

  1. A company offers guided bike tours through a city: “The \(q\)-th tourist of a week would pay up to \(D(q) = 52 - 0.4q\) EUR for a tour. The market price is 24 EUR per tour.” Determine the number of tours sold per week, write the consumer surplus as an integral and compute it.

    Check: A classmate computes \(\int_0^{q^*} D(q) \, dq\) and calls this the consumer surplus. Compute the classmate’s number. Explain what it measures and how it is related to your result.

  2. The tomato growers of a region: “The growers would sell the \(q\)-th tonne of tomatoes per week for at least \(S(q) = 0.9 + 0.025q\) EUR per kg. The market price is 1.90 EUR per kg.” Determine the quantity sold per week, write the producer surplus as an integral and compute it in EUR.

    Check: A classmate writes \(\int_0^{q^*} [S(q) - 1.9] \, dq = -20\) and answers: “The producer surplus is 20 EUR.” Correct both mistakes.

  3. The market for sugar beet in a small country: “Demand \(D(q) = 66 - 1.5q\) and supply \(S(q) = 13.5 + 0.25q\) in EUR per tonne, \(q\) in thousand tonnes per year. The government sets a minimum price of 30 EUR per tonne: the sugar factories may not pay less.” Determine the quantity traded at the minimum price, write the new producer surplus as an integral and compute it.

    Check: A classmate uses the quantity that the farms would like to sell at 30 EUR per tonne as the upper limit. Compute the classmate’s value. Why can this quantity not be sold?

  4. “Determine the area of the region enclosed by the graphs of \(f(x) = 1.5x^2 - 8x + 9\) and \(g(x) = 5 - x\).” Write the area as an integral with its limits and compute it.

    Check: A classmate computes \(\int_{2/3}^{4} [f(x) - g(x)] \, dx\) and gets a negative number. Explain what the sign tells you and give the area.

  5. “Determine the total area between the graphs of \(f(x) = x^3 + x^2 - 2x\) and \(g(x) = 4x\) on the interval \([-1; 3]\).” Write the area as a sum of integrals and compute it.

    Check: A classmate computes the single integral \(\int_{-1}^{3} [f(x) - g(x)] \, dx\). Compute it and explain why it is not the total area.

  1. Quantity: \(52 - 0.4q = 24 \iff 0.4q = 28 \iff q^* = 70\) tours per week.

    \[CS = \int_0^{70} [(52 - 0.4q) - 24] \, dq = \int_0^{70} (28 - 0.4q) \, dq = \left[28q - 0.2q^2\right]_0^{70} = 1960 - 980 = 980 \text{ EUR per week}\]

    Check: the classmate gets \(\int_0^{70} (52 - 0.4q) \, dq = \left[52q - 0.2q^2\right]_0^{70} = 3640 - 980 = 2660\). This is the most that all 70 tourists together would pay (the area under the demand curve), in EUR. They actually pay \(24 \cdot 70 = 1680\) EUR. The difference is the consumer surplus: \(2660 - 1680 = 980\) ✓.

    Typical mistake: forgetting to subtract the market price. The consumer surplus is only the part of the willingness to pay that lies above the price.

  2. Quantity: \(0.9 + 0.025q = 1.9 \iff 0.025q = 1 \iff q^* = 40\) tonnes per week.

    \[PS = \int_0^{40} [1.9 - (0.9 + 0.025q)] \, dq = \int_0^{40} (1 - 0.025q) \, dq = \left[q - 0.0125q^2\right]_0^{40} = 40 - 20 = 20\]

    Unit: EUR per kg times tonnes. One tonne is 1,000 kg, so the result counts thousand EUR: \(PS = 20{,}000\) EUR per week.

    Check: first mistake, the order. The producer surplus integrates price minus supply, \(p^* - S(q) \geq 0\), so it can never be negative. Second mistake, the unit: \(20\) means 20 thousand EUR, not 20 EUR.

    Typical mistake: reading the unit off the price alone. Check the units of both axes: the integral multiplies the price unit by the quantity unit.

  3. Equilibrium first: \(66 - 1.5q = 13.5 + 0.25q \iff 52.5 = 1.75q \iff q^* = 30\), \(p^* = 21\) EUR per tonne. The minimum price of 30 EUR lies above \(p^*\), so it changes the market.

    Quantity traded: the factories buy only what they demand at 30 EUR, \(66 - 1.5q = 30 \iff q_D = 24\) thousand tonnes.

    \[PS_{\text{new}} = \int_0^{24} [30 - (13.5 + 0.25q)] \, dq = \int_0^{24} (16.5 - 0.25q) \, dq = \left[16.5q - 0.125q^2\right]_0^{24} = 396 - 72 = 324\]

    The new producer surplus is 324 thousand EUR per year.

    Check: the farms would like to sell \(13.5 + 0.25q = 30 \iff q_S = 66\) thousand tonnes. The classmate’s value is \(\int_0^{66} (16.5 - 0.25q) \, dq = 1089 - 544.5 = 544.5\). But at 30 EUR per tonne, the factories buy only 24 thousand tonnes. The other 42 thousand tonnes find no buyer, so they bring the farms no surplus.

    Typical mistake: taking the quantity supplied at the minimum price as the quantity traded. The quantity traded is the quantity demanded at the minimum price.

  4. Intersection points: \(1.5x^2 - 8x + 9 = 5 - x \iff 1.5x^2 - 7x + 4 = 0 \iff 3x^2 - 14x + 8 = 0\), so \(x = \frac{14 \pm \sqrt{196 - 96}}{6} = \frac{14 \pm 10}{6}\), i.e. \(x = \frac{2}{3}\) or \(x = 4\). Test \(x = 2\): \(f(2) = -1 < 3 = g(2)\), so \(g\) is the upper graph.

    \[A = \int_{2/3}^{4} [g(x) - f(x)] \, dx = \int_{2/3}^{4} \left(-1.5x^2 + 7x - 4\right) dx = \left[-0.5x^3 + 3.5x^2 - 4x\right]_{2/3}^{4} = 8 - \left(-\frac{34}{27}\right) = \frac{250}{27} \approx 9.26\]

    Check: the classmate gets \(-\frac{250}{27} \approx -9.26\). The minus sign shows that \(f\) is the lower graph: the classmate has integrated bottom minus top. The area is \(\frac{250}{27} \approx 9.26\).

    Typical mistake: reporting the negative number as the area. An area is never negative. Decide the upper graph with a test value before integrating.

  5. Intersection points: \(x^3 + x^2 - 2x = 4x \iff x^3 + x^2 - 6x = 0 \iff x(x + 3)(x - 2) = 0\). In \([-1; 3]\) lie \(x = 0\) and \(x = 2\), while \(x = -3\) lies outside. With \(d(x) = f(x) - g(x)\) and \(D(x) = \frac{x^4}{4} + \frac{x^3}{3} - 3x^2\): \(D(-1) = -\frac{37}{12}\), \(D(0) = 0\), \(D(2) = -\frac{16}{3}\) and \(D(3) = \frac{9}{4}\). Signs: \(d(-0.5) > 0\), \(d(1) = -4 < 0\), \(d(2.5) > 0\).

    \[A = \int_{-1}^{0} d(x) \, dx + \int_{0}^{2} [-d(x)] \, dx + \int_{2}^{3} d(x) \, dx = \frac{37}{12} + \frac{16}{3} + \frac{91}{12} = \frac{192}{12} = 16\]

    Check: \(\int_{-1}^{3} d(x) \, dx = D(3) - D(-1) = \frac{9}{4} + \frac{37}{12} = \frac{16}{3} \approx 5.33\). The middle region, where \(g\) lies above \(f\), counts negative and cancels part of the other two. The total area is 16.

    Typical mistake: splitting at \(x = -3\) as well. That intersection point lies outside the interval, so it does not split the region.

Problem 7: One Paragraph, All Conditions (xx)

In a region, the market for raw milk (fresh milk that farms deliver to dairies) is modelled with linear functions, with \(q\) in million litres per month and prices in EUR per 100 litres. The dairies pay at most 57 EUR per 100 litres (maximum price), and at a price of 0 they would take 19 million litres per month (market saturation). The farms deliver nothing below 21 EUR per 100 litres, and for every additional million litres per month, the lowest price at which they deliver rises by 2 EUR per 100 litres. The market is in equilibrium. The farmers’ association wants to know the consumer surplus and the producer surplus, and whether a minimum price of 39 EUR per 100 litres would raise the producer surplus.

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

Check: An assistant writes the market saturation as \(D(0) = 19\). Explain what is wrong and give the correct condition.

  1. Demand: \(D(q) = m q + b\), linear, with the maximum price \(D(0) = 57\)
  2. Market saturation: \(D(19) = 0\), so \(D(q) = 57 - 3q\)
  3. Supply: \(S(0) = 21\) and slope 2, so \(S(q) = 21 + 2q\)
  4. Equilibrium: \(D(q^*) = S(q^*)\) and \(p^* = D(q^*)\)
  5. Consumer surplus: \(CS = \int_0^{q^*} [D(q) - p^*] \, dq\)
  6. Producer surplus: \(PS = \int_0^{q^*} [p^* - S(q)] \, dq\)
  7. Minimum price: quantity traded \(q_m\) with \(D(q_m) = 39\)
  8. New producer surplus: \(PS_{\text{new}} = \int_0^{q_m} [39 - S(q)] \, dq\)
  9. The question of the association: is \(PS_{\text{new}} > PS\)?

Check: the saturation is a quantity, reached when the price is 0. On the graph of \(D\) it is the point \((19, 0)\), so the condition is \(D(19) = 0\). \(D(0)\) is the price at quantity 0, the maximum price: \(D(0) = 57\).

Typical mistake: swapping price and quantity. \(D\) takes a quantity and gives a price. Write each point as (quantity, price) before setting up the condition.

C: Applied

Problem 8: Used Textbooks on a University Marketplace (xx)

At the start of each semester, students buy and sell used textbooks on their university’s online marketplace. For \(q\) thousand books per semester, the demand function

\[D(q) = 64 - 2.7q\]

gives the highest price in EUR per book that the buyers of the \(q\)-th thousand books would pay, and the supply function

\[S(q) = 0.05q^2 + 0.9q + 23\]

gives the lowest price in EUR per book at which the sellers of the \(q\)-th thousand books would sell. Round to two decimal places.

  1. Determine the market equilibrium.

(for verification: \(q^* = 10\), \(p^* = 37\) EUR per book)

  1. Compute the consumer surplus and the producer surplus in EUR. Explain what each surplus means for the buyers and for the sellers. Decide which group gains more from the marketplace.

  2. Show that the total surplus equals the area of the region between the graphs of \(D\) and \(S\) for \(0 \leq q \leq q^*\).

  3. A horizontal line \(p = k\) divides the region from c) into two parts. Explain why \(k\) must lie above \(p^*\) if both parts are to have equal area. Set up the equation that \(k\) must satisfy so that the line \(p = k\) divides the region into two parts of equal area, and solve it.

Hint: For \(k > p^*\), the part above the line lies between the line \(p = k\) and the graph of \(D\). It starts at \(q = 0\) and ends where \(D(q) = k\).

a) Equilibrium: \(S(q) = D(q)\).

\[0.05q^2 + 0.9q + 23 = 64 - 2.7q \iff 0.05q^2 + 3.6q - 41 = 0 \iff q^2 + 72q - 820 = 0\]

\[q = \frac{-72 \pm \sqrt{5184 + 3280}}{2} = \frac{-72 \pm 92}{2}, \qquad q = 10 \text{ or } q = -82\]

Only \(q^* = 10\) is a possible quantity. Price: \(p^* = D(10) = 64 - 27 = 37\) EUR per book. Check: \(S(10) = 5 + 9 + 23 = 37\) ✓

At the equilibrium, 10,000 books change hands at 37 EUR each.

b) Unit: EUR per book times thousand books gives thousand EUR.

\[CS = \int_0^{10} [(64 - 2.7q) - 37] \, dq = \int_0^{10} (27 - 2.7q) \, dq = \left[27q - 1.35q^2\right]_0^{10} = 270 - 135 = 135\]

\[PS = \int_0^{10} \left[37 - \left(0.05q^2 + 0.9q + 23\right)\right] dq = \int_0^{10} \left(14 - 0.9q - 0.05q^2\right) dq = \left[14q - 0.45q^2 - \frac{0.05}{3}q^3\right]_0^{10} = 95 - \frac{50}{3} = \frac{235}{3} \approx 78.33\]

  • Consumer surplus 135,000 EUR: the buyers would have paid 135,000 EUR more in total than they actually pay at 37 EUR per book.
  • Producer surplus about 78,333.33 EUR: the sellers receive about 78,333.33 EUR more in total than the lowest prices at which they would have sold.

The buyers gain more (\(135 > 78.33\)).

c) Total surplus: \(135 + \frac{235}{3} = \frac{640}{3} \approx 213.33\) thousand EUR. The area between the graphs (\(D\) lies above \(S\) for \(q < 10\), e.g. \(D(0) = 64 > 23 = S(0)\)):

\[\int_0^{10} [D(q) - S(q)] \, dq = \int_0^{10} \left(41 - 3.6q - 0.05q^2\right) dq = \left[41q - 1.8q^2 - \frac{0.05}{3}q^3\right]_0^{10} = 410 - 180 - \frac{50}{3} = \frac{640}{3} \approx 213.33 \; \checkmark\]

The line \(p = p^*\) splits this region into the consumer surplus above it and the producer surplus below it.

d) Why \(k > p^*\): for equal areas, each part must have half the size, \(\frac{1}{2} \cdot \frac{640}{3} = \frac{320}{3} \approx 106.67\). At \(k = p^* = 37\), the part above the line is the consumer surplus, \(135 > 106.67\). It is too large, so the line has to move up: \(k > 37\).

Set-up: the line meets the graph of \(D\) where \(64 - 2.7q = k\), i.e. at \(q_k = \frac{64 - k}{2.7}\). The part above the line lies between \(p = k\) and \(D\) for \(0 \leq q \leq q_k\):

\[\int_0^{q_k} [D(q) - k] \, dq = \frac{320}{3}, \qquad q_k = \frac{64 - k}{2.7}\]

Solve:

\[\int_0^{q_k} (64 - k - 2.7q) \, dq = (64 - k)q_k - 1.35q_k^2 = \frac{(64 - k)^2}{2.7} - \frac{(64 - k)^2}{5.4} = \frac{(64 - k)^2}{5.4}\]

\[\frac{(64 - k)^2}{5.4} = \frac{320}{3} \iff (64 - k)^2 = 576 \iff 64 - k = 24 \iff k = 40\]

(\(64 - k = -24\) would give \(k = 88\), above the maximum price 64.) The line \(p = 40\) divides the region into two parts of equal area.

Typical mistake: in d), integrating up to \(q^* = 10\) for the part above the line. The part above \(p = k\) ends where the line meets the demand curve, at \(q_k = \frac{24}{2.7} \approx 8.89\).

Problem 9: Wooden Sledges (xx)

A small workshop builds wooden sledges for snowy winter days. At an output of \(x\) sledges per week, the marginal revenue and the marginal costs are

\[MR(x) = 110 - 1.5x \qquad \text{and} \qquad MC(x) = 0.25x^2 - 12x + 164\]

EUR per sledge. The fixed costs are 650 EUR per week, and the workshop can build at most 40 sledges per week (capacity). Round to two decimal places.

  1. Determine the outputs at which the marginal revenue equals the marginal costs.

  2. Determine on which parts of \([0; 40]\) the marginal revenue is larger than the marginal costs. Explain what this means for the profit when the workshop builds one more sledge.

  3. Compute the area between the graphs of \(MR\) and \(MC\) on \([0; 6]\) and on \([6; 36]\). Explain the business meaning of both areas.

(for verification: 153 and 1,125)

  1. Determine the maximum profit per week from the areas in c) and the fixed costs, without setting up the profit function. Prove that no output in \([0; 40]\) brings a higher profit. Determine the price per sledge at this output.

  2. The owner says: “At 6 sledges per week, the marginal revenue equals the marginal costs. So 6 sledges is the best output.” Explain with the areas what really happens at 6 sledges.

a) \(MR(x) = MC(x)\):

\[110 - 1.5x = 0.25x^2 - 12x + 164 \iff 0.25x^2 - 10.5x + 54 = 0 \iff x^2 - 42x + 216 = 0\]

\[x = \frac{42 \pm \sqrt{1764 - 864}}{2} = \frac{42 \pm 30}{2}, \qquad x = 6 \text{ or } x = 36\]

b) \(MR(x) - MC(x) = -0.25x^2 + 10.5x - 54 = -0.25(x - 6)(x - 36)\). Test values: \(MR(0) - MC(0) = 110 - 164 = -54 < 0\), \(MR(10) - MC(10) = 95 - 69 = 26 > 0\), \(MR(40) - MC(40) = 50 - 84 = -34 < 0\).

  • On \([0; 6)\): \(MR < MC\). One more sledge adds more to the costs than to the revenue, so the profit falls.
  • On \((6; 36)\): \(MR > MC\). One more sledge adds more to the revenue, so the profit rises.
  • On \((36; 40]\): \(MR < MC\) again, and the profit falls.

c) On \([0; 6]\), \(MC\) is the upper graph:

\[\int_0^6 [MC(x) - MR(x)] \, dx = \int_0^6 \left(0.25x^2 - 10.5x + 54\right) dx = \left[\frac{x^3}{12} - 5.25x^2 + 54x\right]_0^6 = 18 - 189 + 324 = 153\]

On \([6; 36]\), \(MR\) is the upper graph:

\[\int_6^{36} [MR(x) - MC(x)] \, dx = \left[-\frac{x^3}{12} + 5.25x^2 - 54x\right]_6^{36} = (-3888 + 6804 - 1944) - (-18 + 189 - 324) = 972 + 153 = 1125\]

Meaning: \(MR - MC\) is the marginal profit \(P'\), and its integral is the change in profit (net change theorem). From 0 to 6 sledges, the profit falls by 153 EUR per week. From 6 to 36 sledges, it rises by 1,125 EUR per week.

d) Without sledges, the workshop only pays the fixed costs: \(P(0) = -650\) EUR. With c):

\[P(6) = -650 - 153 = -803, \qquad P(36) = -803 + 1125 = 322 \text{ EUR per week}\]

Proof: by b), \(P' = MR - MC\) changes its sign from \(+\) to \(-\) at \(x = 36\), so \(P\) has a local maximum there, and at \(x = 6\) a local minimum. Compare with the ends of \([0; 40]\): \(P(0) = -650\) and

\[P(40) = P(36) - \int_{36}^{40} [MC(x) - MR(x)] \, dx = 322 - \frac{196}{3} \approx 256.67\]

The maximum profit is 322 EUR per week at 36 sledges. No output in \([0; 40]\) brings more.

Price: revenue from the marginal revenue with \(R(0) = 0\): \(R(x) = 110x - 0.75x^2\), so the price is \(p(x) = \frac{R(x)}{x} = 110 - 0.75x\) and \(p(36) = 110 - 27 = 83\) EUR per sledge.

e) At 6 sledges, the first region ends: the profit has fallen by 153 EUR since \(x = 0\) and starts to rise only now. So \(P(6) = -803\) EUR is the lowest profit, not the highest. \(MR = MC\) is only the necessary condition. At \(x = 6\), \(MR - MC\) changes from \(-\) to \(+\), which marks a minimum.

Typical mistake: taking the first solution of \(MR = MC\) as the best output. Check the sign change of \(MR - MC\), or look at the areas: the profit only reaches its maximum after the large region on \([6; 36]\).

Problem 10: Smart Speakers (xxx)

A company launches a smart speaker (a loudspeaker that you control with your voice) and plans to sell it for 24 months. \(t\) months after the launch, the revenue from the speaker comes in at a rate of

\[r(t) = 1 + 3.4t - 0.1t^2\]

thousand EUR per month, and the costs for production, support and advertising arise at a rate of \(c(t) = 13 + 0.8t\) thousand EUR per month, for \(t\) between 0 and 24. Round to two decimal places.

  1. Determine the times at which both rates are equal. Decide in which periods the company makes a loss and in which it makes a profit.

  2. Compute the loss of the first 6 months as the area between the graphs of \(r\) and \(c\).

  3. Compute the profit from month 6 to month 20 in the same way. Give the net result of the first 20 months.

(for verification: the profit from month 6 to month 20 is about 45.73 thousand EUR)

  1. The accumulated result after \(T\) months is \(G(T) = \int_0^T [r(t) - c(t)] \, dt\). Show that the company has made up for its early losses after exactly 15 months, i.e. \(G(15) = 0\). Determine the second time in \((0; 24]\) with \(G(T) = 0\) and explain its meaning.

  2. The management wants to take the speaker off the market at the time with the highest accumulated result. Determine this time and the result, and prove that no other time in \([0; 24]\) gives a higher result.

a) \(r(t) = c(t)\):

\[1 + 3.4t - 0.1t^2 = 13 + 0.8t \iff -0.1t^2 + 2.6t - 12 = 0 \iff t^2 - 26t + 120 = 0\]

\[t = \frac{26 \pm \sqrt{676 - 480}}{2} = \frac{26 \pm 14}{2}, \qquad t = 6 \text{ or } t = 20\]

Test values for \(r(t) - c(t)\): at \(t = 0\), \(1 - 13 = -12 < 0\), at \(t = 10\), \(25 - 21 = 4 > 0\), and at \(t = 24\), \(25 - 32.2 = -7.2 < 0\).

  • \(0 \leq t < 6\) and \(20 < t \leq 24\): \(c > r\), the company makes a loss.
  • \(6 < t < 20\): \(r > c\), the company makes a profit.

b) On \([0; 6]\), \(c\) is the upper graph:

\[\int_0^6 [c(t) - r(t)] \, dt = \int_0^6 \left(0.1t^2 - 2.6t + 12\right) dt = \left[\frac{t^3}{30} - 1.3t^2 + 12t\right]_0^6 = 7.2 - 46.8 + 72 = 32.4\]

The loss of the first 6 months is 32.4 thousand EUR, i.e. 32,400 EUR.

c) On \([6; 20]\), \(r\) is the upper graph. With \(G(t) = -\frac{t^3}{30} + 1.3t^2 - 12t\) as antiderivative of \(r - c\):

\[\int_6^{20} [r(t) - c(t)] \, dt = G(20) - G(6) = \left(-\frac{800}{3} + 520 - 240\right) - (-7.2 + 46.8 - 72) = \frac{40}{3} + 32.4 = \frac{686}{15} \approx 45.73\]

The profit from month 6 to month 20 is about 45.73 thousand EUR. Net result of the first 20 months: \(45.73 - 32.4 \approx 13.33\) thousand EUR, i.e. \(G(20) = \frac{40}{3}\).

d) From c), \(G(T) = -\frac{T^3}{30} + 1.3T^2 - 12T\) with \(G(0) = 0\).

\[G(15) = -\frac{3375}{30} + 1.3 \cdot 225 - 180 = -112.5 + 292.5 - 180 = 0 \; \checkmark\]

After 15 months, the profit since month 6 exactly makes up for the loss of the first 6 months.

Zeros of \(G\): factor out \(T\):

\[G(T) = -\frac{T}{30}\left(T^2 - 39T + 360\right) = 0, \qquad T = \frac{39 \pm \sqrt{1521 - 1440}}{2} = \frac{39 \pm 9}{2}\]

So \(T = 0\), \(T = 15\) or \(T = 24\). The second time is \(T = 24\): after month 20, the company makes a loss again, and by the end of month 24, these new losses have used up the whole accumulated profit. Over the planned 24 months, the speaker brings a net result of 0.

e) \(G'(T) = r(T) - c(T)\). By a), \(G'\) changes its sign from \(-\) to \(+\) at \(T = 6\) (minimum) and from \(+\) to \(-\) at \(T = 20\) (maximum). Compare the local maximum with the ends of \([0; 24]\):

\[G(0) = 0, \qquad G(20) = \frac{40}{3} \approx 13.33, \qquad G(24) = 0\]

The accumulated result is highest after 20 months, at about 13.33 thousand EUR. No other time in \([0; 24]\) gives more, so the management should take the speaker off the market after 20 months.

Typical mistake: in e), stopping at the time with the highest revenue rate (\(r\) has its vertex at \(t = 17\)) or at the payback time 15. The accumulated result keeps growing as long as \(r(t) > c(t)\), i.e. until \(t = 20\).

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: Ride-Hailing in a City [21 pts. total]

In a large city, people book rides with ride-hailing services: a driver with a private car picks them up and takes them to their destination, and they book and pay the ride with an app. The city models the market for these rides with \(q\) in thousand rides per week and prices in EUR per ride. Round to two decimal places.

Part a)

Determine the linear demand function assuming a maximum price of 44 EUR per ride and a market saturation at 55,000 rides per week. Give \(D(q)\) with \(q\) in thousand rides per week. [4 pts.]

(for verification: \(D(q) = 44 - 0.8q\))

Units: 55,000 rides per week are \(q = 55\) thousand rides per week.

  • Maximum price: at the quantity 0, the price is 44 EUR per ride: point \((0, 44)\)
  • Market saturation: at the price 0, the quantity is 55 thousand rides: point \((55, 0)\)

Slope:

\[m = \frac{0 - 44}{55 - 0} = -0.8\]

Demand function with the \(y\)-intercept 44:

\[D(q) = 44 - 0.8q\]

Check: \(D(0) = 44\) ✓, \(D(55) = 44 - 44 = 0\) ✓

Grading: 1 Pt. conversion into thousand rides, 1 Pt. both points, 1 Pt. slope, 1 Pt. \(D(q)\).

Part b)

The drivers would offer the \(q\)-th thousand rides per week for at least

\[S(q) = 14 + 0.4q\]

EUR per ride. Determine the market equilibrium and interpret it. [3 pts.]

Equilibrium: \(D(q) = S(q)\)

\[44 - 0.8q = 14 + 0.4q \iff 30 = 1.2q \iff q^* = 25\]

\[p^* = D(25) = 44 - 20 = 24, \qquad \text{check: } S(25) = 14 + 10 = 24 \; \checkmark\]

At the equilibrium, 25,000 rides per week are booked at a price of 24 EUR per ride. At this price, the riders book exactly as many rides as the drivers offer.

Grading: 1 Pt. \(D(q) = S(q)\), 1 Pt. \(q^* = 25\), 1 Pt. \(p^* = 24\) with the meaning.

Part c)

Compute the consumer surplus and the producer surplus at the equilibrium. State the businesslike meaning of both quantities. [5 pts.]

Unit: EUR per ride times thousand rides per week gives thousand EUR per week.

Consumer surplus:

\[CS = \int_0^{25} [D(q) - 24] \, dq = \int_0^{25} (20 - 0.8q) \, dq = \left[20q - 0.4q^2\right]_0^{25} = 500 - 250 = 250\]

Producer surplus:

\[PS = \int_0^{25} [24 - S(q)] \, dq = \int_0^{25} (10 - 0.4q) \, dq = \left[10q - 0.2q^2\right]_0^{25} = 250 - 125 = 125\]

Meaning:

  • \(CS = 250{,}000\) EUR per week: the riders would have paid 250,000 EUR more per week than they pay at 24 EUR per ride.
  • \(PS = 125{,}000\) EUR per week: the drivers receive 125,000 EUR more per week than the lowest prices at which they would have offered the rides.

Grading: 1 Pt. set-up of \(CS\) with the limits 0 and 25, 1 Pt. \(CS = 250\), 1 Pt. set-up of \(PS\), 1 Pt. \(PS = 125\), 1 Pt. both meanings with the unit.

Part d)

To protect the taxi companies, the city sets a minimum price of 28 EUR per ride: no ride may cost less. Determine the number of rides that are booked per week and the number of rides that the drivers would like to offer at this price. Compute the new consumer surplus and the new producer surplus. Decide whether the drivers as a group gain from the minimum price. [5 pts.]

The minimum price lies above \(p^* = 24\), so it changes the market.

Rides booked (quantity demanded, which is the quantity traded): \(44 - 0.8q = 28 \iff 0.8q = 16 \iff q_D = 20\), i.e. 20,000 rides per week.

Rides offered: \(14 + 0.4q = 28 \iff 0.4q = 14 \iff q_S = 35\), i.e. 35,000 rides per week. So 15,000 offered rides per week find no rider.

New surpluses (up to the quantity traded \(q_D = 20\)):

\[CS_{\text{new}} = \int_0^{20} (44 - 0.8q - 28) \, dq = \int_0^{20} (16 - 0.8q) \, dq = \left[16q - 0.4q^2\right]_0^{20} = 320 - 160 = 160\]

\[PS_{\text{new}} = \int_0^{20} [28 - (14 + 0.4q)] \, dq = \int_0^{20} (14 - 0.4q) \, dq = \left[14q - 0.2q^2\right]_0^{20} = 280 - 80 = 200\]

Decision: the producer surplus rises from 125,000 to 200,000 EUR per week. Although fewer rides are booked, the drivers as a group gain 75,000 EUR per week from the higher price. The riders lose \(250{,}000 - 160{,}000 = 90{,}000\) EUR per week.

Grading: 1 Pt. \(q_D = 20\) as the quantity traded, 1 Pt. \(q_S = 35\), 1 Pt. \(CS_{\text{new}} = 160\), 1 Pt. \(PS_{\text{new}} = 200\), 1 Pt. decision with the comparison \(200 > 125\).

Part e)

The minimum price lowers the total surplus of the market. This loss (the deadweight loss) equals the area of the region between the graphs of \(D\) and \(S\) from the quantity traded to the equilibrium quantity. Compute the deadweight loss with an integral and verify your result with the total surpluses from c) and d). [4 pts.]

Set-up: on \([20; 25]\), \(D\) lies above \(S\) (e.g. \(D(20) = 28 > 22 = S(20)\)), and the region runs from the quantity traded \(q_D = 20\) to \(q^* = 25\):

\[DWL = \int_{20}^{25} [D(q) - S(q)] \, dq = \int_{20}^{25} (30 - 1.2q) \, dq = \left[30q - 0.6q^2\right]_{20}^{25} = (750 - 375) - (600 - 240) = 375 - 360 = 15\]

The deadweight loss is 15,000 EUR per week.

Verification: total surplus without the minimum price \(250 + 125 = 375\), with the minimum price \(160 + 200 = 360\). The difference is \(375 - 360 = 15\) ✓. The rides between 20,000 and 25,000 per week no longer take place, and the benefit they brought to riders and drivers is lost.

Grading: 1 Pt. set-up with \(D - S\) and the limits 20 and 25, 1 Pt. antiderivative, 1 Pt. \(DWL = 15\) thousand EUR, 1 Pt. verification \(375 - 360 = 15\).