Tasks 03-03 - Quadratic Functions & Basic Optimization

Section 03: Functions as Business Models

Problem 1: Vertex and Properties (x)

For each quadratic function, find the vertex, axis of symmetry, and determine whether it has a maximum or minimum value:

  1. \(f(x) = x^2 - 6x + 5\)

  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. \(f(x) = x^2 - 6x + 5\):
    • Vertex x-coordinate: \(x = -\frac{-6}{2(1)} = 3\)
    • Vertex y-coordinate: \(f(3) = 9 - 18 + 5 = -4\)
    • Vertex: \((3, -4)\)
    • Axis of symmetry: \(x = 3\)
    • Since \(a = 1 > 0\), has minimum value of -4
  2. \(g(x) = -2x^2 + 8x - 3\):
    • Vertex x-coordinate: \(x = -\frac{8}{2(-2)} = 2\)
    • Vertex y-coordinate: \(g(2) = -8 + 16 - 3 = 5\)
    • Vertex: \((2, 5)\)
    • Axis of symmetry: \(x = 2\)
    • Since \(a = -2 < 0\), has maximum value of 5
  3. \(h(x) = 3x^2 + 12x + 15\):
    • Vertex x-coordinate: \(x = -\frac{12}{2(3)} = -2\)
    • Vertex y-coordinate: \(h(-2) = 12 - 24 + 15 = 3\)
    • Vertex: \((-2, 3)\)
    • Axis of symmetry: \(x = -2\)
    • Since \(a = 3 > 0\), has minimum value of 3
  4. \(p(x) = -\frac{1}{2}x^2 + 4x - 6\):
    • Vertex x-coordinate: \(x = -\frac{4}{2(-\frac{1}{2})} = 4\)
    • Vertex y-coordinate: \(p(4) = -8 + 16 - 6 = 2\)
    • Vertex: \((4, 2)\)
    • Axis of symmetry: \(x = 4\)
    • Since \(a = -\frac{1}{2} < 0\), has maximum value of 2

Problem 2: Completing the Square (x)

Convert each quadratic function from standard form to vertex form by completing the square:

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

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

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

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

  1. \(f(x) = x^2 + 10x + 21\):
    • Complete the square: \((x^2 + 10x + 25) - 25 + 21\)
    • \(f(x) = (x + 5)^2 - 4\)
    • Vertex form with vertex at \((-5, -4)\)
  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 at \((2, -3)\)
  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 at \((3, 2)\)
  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 at \((3, -3)\)

Problem 3: Revenue Optimization (xx)

A theater company is planning ticket prices for a new show. Market research indicates:

  • At €20 per ticket, they expect 500 attendees
  • For each €2 increase in price, attendance drops by 20 people
  • The theater has a capacity of 600 seats

  1. Express the attendance \(A\) as a function of ticket price \(p\).

  2. Write the revenue function \(R(p)\).

  3. Find the ticket price that maximizes revenue.

  4. What is the maximum revenue and how many tickets will be sold?

  5. Verify that the optimal solution respects the theater capacity.

  1. Attendance function:
    • Base: 500 attendees at €20
    • Change: -20 attendees per €2 increase = -10 per €1 increase
    • \(A(p) = 500 - 10(p - 20) = 500 - 10p + 200 = 700 - 10p\)
  2. Revenue function:
    • \(R(p) = p \cdot A(p) = p(700 - 10p) = 700p - 10p^2\)
  3. Revenue-maximizing price:
    • \(p = -\frac{700}{2(-10)} = \frac{700}{20} = 35\) euros
  4. Maximum revenue and tickets:
    • Attendance: \(A(35) = 700 - 350 = 350\) people
    • Revenue: \(R(35) = 35 \times 350 = €12,250\)
  5. Capacity check:
    • 350 attendees < 600 capacity ✓
    • The optimal solution is feasible

Problem 4: Profit Maximization (xx)

A smartphone accessories company manufactures premium cases. Their market analysis shows:

  • Demand: \(Q = 2000 - 25p\) (units per month)
  • Fixed costs: €15,000 per month
  • Variable cost: €20 per case

  1. Write the revenue function \(R(p)\) in terms of price.

  2. Write the cost function \(C(p)\) in terms of price.

  3. Write the profit function \(\Pi(p)\) and identify its form.

  4. Find the profit-maximizing price and quantity.

  5. Calculate the maximum monthly profit.

  6. Determine the break-even prices.

  1. Revenue function:
    • \(R(p) = p \cdot Q = p(2000 - 25p) = 2000p - 25p^2\)
  2. Cost function:
    • Variable cost: €20 per unit × quantity
    • \(C(p) = 15000 + 20(2000 - 25p) = 15000 + 40000 - 500p = 55000 - 500p\)
  3. Profit function:
    • \(\Pi(p) = R(p) - C(p) = 2000p - 25p^2 - (55000 - 500p)\)
    • \(\Pi(p) = -25p^2 + 2500p - 55000\)
    • This is a quadratic function opening downward
  4. Profit-maximizing price:
    • \(p = -\frac{2500}{2(-25)} = \frac{2500}{50} = 50\) euros
    • Quantity: \(Q = 2000 - 25(50) = 750\) units
  5. Maximum profit:
    • \(\Pi(50) = -25(2500) + 2500(50) - 55000\)
    • \(= -62500 + 125000 - 55000 = €7,500\)
  6. Break-even prices (\(\Pi(p) = 0\)):
    • \(-25p^2 + 2500p - 55000 = 0\)
    • Divide by -25: \(p^2 - 100p + 2200 = 0\)
    • Using quadratic formula: \(p = \frac{100 \pm \sqrt{10000 - 8800}}{2} = \frac{100 \pm \sqrt{1200}}{2}\)
    • \(p = \frac{100 \pm 34.64}{2}\)
    • \(p_1 \approx 32.68\) euros, \(p_2 \approx 67.32\) euros

Problem 5: Area Optimization (xxx)

A farmer has 400 meters of fencing to create a rectangular grazing area along a straight river. The side along the river needs no fence.

  1. Let \(x\) be the length perpendicular to the river. Express the area \(A\) as a function of \(x\).

  2. Find the dimensions that maximize the grazing area.

  3. What is the maximum area?

  4. If the farmer needs at least 8,000 m² for the animals, what range of widths \(x\) will work?

  5. Due to terrain, the length parallel to the river cannot exceed 150 meters. How does this affect the optimal dimensions?

  1. Area function:
    • Let \(x\) = width (perpendicular to river), \(y\) = length (parallel to river)
    • Fencing constraint: \(2x + y = 400\) (no fence on river side)
    • So \(y = 400 - 2x\)
    • Area: \(A(x) = x \cdot y = x(400 - 2x) = 400x - 2x^2\)
  2. Optimal dimensions:
    • \(x = -\frac{400}{2(-2)} = \frac{400}{4} = 100\) meters
    • \(y = 400 - 2(100) = 200\) meters
    • Dimensions: 100m × 200m
  3. Maximum area:
    • \(A(100) = 400(100) - 2(10000) = 40000 - 20000 = 20,000\) m²
  4. For at least 8,000 m²:
    • Solve: \(400x - 2x^2 \geq 8000\)
    • \(-2x^2 + 400x - 8000 \geq 0\)
    • \(x^2 - 200x + 4000 \leq 0\)
    • Using quadratic formula: \(x = \frac{200 \pm \sqrt{40000 - 16000}}{2} = \frac{200 \pm \sqrt{24000}}{2}\)
    • \(x = \frac{200 \pm 155}{2}\)
    • \(x_1 = 22.5\), \(x_2 = 177.5\)
    • Width must be between 22.5m and 177.5m
  5. With length constraint (\(y \leq 150\)):
    • Need: \(400 - 2x \leq 150\)
    • \(250 \leq 2x\)
    • \(x \geq 125\)
    • Since optimal was \(x = 100\), must use \(x = 125\)
    • New dimensions: 125m × 150m
    • New area: \(125 \times 150 = 18,750\) m²

Problem 6: Projectile Motion Application (xxx)

A company is testing delivery drones that follow parabolic flight paths. One drone’s height \(h\) (in meters) after \(t\) seconds is given by: \[h(t) = -2t^2 + 16t + 10\]

  1. Find the maximum height reached by the drone.

  2. At what time does it reach maximum height?

  3. When does the drone return to its starting height of 10 meters?

  4. If the drone must maintain a minimum height of 30 meters for safety, during what time interval is it safe?

  5. The drone’s horizontal speed is 5 m/s. How far does it travel horizontally while above 30 meters?

  1. Maximum height:
    • Time at maximum: \(t = -\frac{16}{2(-2)} = 4\) seconds
    • Maximum height: \(h(4) = -32 + 64 + 10 = 42\) meters
  2. Time at maximum height:
    • 4 seconds (calculated above)
  3. Return to starting height (h = 10):
    • \(-2t^2 + 16t + 10 = 10\)
    • \(-2t^2 + 16t = 0\)
    • \(t(-2t + 16) = 0\)
    • \(t = 0\) (start) or \(t = 8\) seconds (return)
  4. Time interval above 30 meters:
    • Solve: \(-2t^2 + 16t + 10 \geq 30\)
    • \(-2t^2 + 16t - 20 \geq 0\)
    • \(t^2 - 8t + 10 \leq 0\)
    • Using quadratic formula: \(t = \frac{8 \pm \sqrt{64 - 40}}{2} = \frac{8 \pm \sqrt{24}}{2} = \frac{8 \pm 4.90}{2}\)
    • \(t_1 \approx 1.55\) seconds, \(t_2 \approx 6.45\) seconds
    • Safe interval: approximately 1.55 to 6.45 seconds
  5. Horizontal distance while above 30m:
    • Duration above 30m: \(6.45 - 1.55 = 4.90\) seconds
    • Horizontal distance: \(5 \text{ m/s} \times 4.90 \text{ s} = 24.5\) meters

Problem 7: Multi-Product Optimization (xxxx)

TechGear sells two related products: wireless earbuds and charging cases. Market research reveals:

Earbuds:

  • Demand: \(Q_e = 1000 - 5p_e\) when sold alone
  • Production cost: €30 per unit
  • Can be sold with or without case

Charging Cases:

  • Only bought by earbud customers
  • 60% of earbud buyers will buy a case if priced at €20
  • For each €5 increase in case price, 10% fewer earbud buyers purchase a case
  • Production cost: €8 per case

Bundle Option:

  • The company is considering bundling both products
  • Bundle demand: \(Q_b = 800 - 4p_b\) where \(p_b\) is bundle price
  • Same production costs apply

  1. Write the function for case demand \(Q_c\) as a percentage of earbud sales, depending on case price \(p_c\).

  2. If earbuds are priced at €80 and cases at €25, calculate:

    • Number of earbuds sold
    • Number of cases sold
    • Total profit from this pricing strategy

  3. Find the optimal bundle price and calculate the profit from bundling.

  4. The company can only produce 500 units total (earbuds + cases) per month. Which strategy should they pursue:

    • Separate pricing at €80 for earbuds and €25 for cases
    • Bundling at the optimal bundle price
    • Another pricing strategy you determine
    • Justify your answer with calculations.
  1. Case demand function:
    • Base: 60% buy at €20
    • Change: -10% per €5 increase = -2% per €1 increase
    • Percentage buying case: \(P(p_c) = 60 - 2(p_c - 20) = 100 - 2p_c\) percent
    • Case quantity: \(Q_c = Q_e \times \frac{P(p_c)}{100} = Q_e \times \frac{100 - 2p_c}{100}\)
  2. At \(p_e = 80\), \(p_c = 25\):
    • Earbuds sold: \(Q_e = 1000 - 5(80) = 600\) units
    • Case purchase rate: \(P(25) = 100 - 50 = 50\%\)
    • Cases sold: \(Q_c = 600 \times 0.50 = 300\) units
    • Profit from earbuds: \(600 \times (80 - 30) = €30,000\)
    • Profit from cases: \(300 \times (25 - 8) = €5,100\)
    • Total profit: €35,100
  3. Bundle optimization:
    • Revenue: \(R(p_b) = p_b(800 - 4p_b) = 800p_b - 4p_b^2\)
    • Cost: \(C = (30 + 8)(800 - 4p_b) = 38(800 - 4p_b) = 30400 - 152p_b\)
    • Profit: \(\Pi(p_b) = 800p_b - 4p_b^2 - 30400 + 152p_b = -4p_b^2 + 952p_b - 30400\)
    • Optimal price: \(p_b = -\frac{952}{2(-4)} = 119\) euros
    • Quantity: \(Q_b = 800 - 4(119) = 324\) bundles
    • Profit: \(\Pi(119) = -4(14161) + 952(119) - 30400 = €26,092\)
  4. Production constraint analysis:
    • Separate pricing (€80/€25):
      • Total units: 600 earbuds + 300 cases = 900 units
      • Exceeds 500 unit capacity - not feasible
    • Bundling at €119:
      • Total units: 324 bundles = 648 units (324 earbuds + 324 cases)
      • Exceeds 500 unit capacity - not feasible
    • Optimal strategy with constraint:

      • For bundles: Maximum 250 bundles (500 units total)

      • Bundle profit at higher price to sell exactly 250:

      • \(250 = 800 - 4p_b\), so \(p_b = 137.50\)

      • Profit: \(250 \times (137.50 - 38) = €24,875\)

      • For separate: Need \(Q_e + Q_c \leq 500\)

      • If \(p_e = 100\): \(Q_e = 500\), no capacity for cases

      • Profit: \(500 \times (100 - 30) = €35,000\)

    • Best strategy: Sell 500 earbuds only at €100 each for €35,000 profit