Tasks 03-01 - Function Concepts & Business Modeling

Section 03: Functions as Business Models

Problem 1: Creating Tables and Graphs (xx)

A streaming service charges according to the function \(S(h) = 10 + 2h\) where \(h\) is the number of hours watched per month and \(S(h)\) is the total cost in euros.

  1. Create a table showing the cost for \(h = 0, 5, 10, 15, 20, 25\) hours
  2. Sketch a graph of this function on graph paper or using graphing software
  3. What does 10 represent?
  4. If a customer has a budget of €30 per month, how many hours can they watch?
  1. Table of values:
Hours (h) Cost S(h)
0 €10
5 €20
10 €30
15 €40
20 €50
25 €60

  1. Graph: Linear function starting at (0, 10) with slope 2

    • Points: (0, 10), (5, 20), (10, 30), (15, 40), (20, 50), (25, 60)
    • Straight line increasing from left to right

  2. Y-intercept of 10: Represents the base monthly subscription fee (€10 fixed cost)

  3. Budget constraint:

    • Set \(S(h) = 30\)
    • \(10 + 2h = 30\)
    • \(2h = 20\)
    • \(h = 10\) hours

Problem 2: Domain and Range (xx)

Find the domain and range of each function:

  1. \(f(x) = \frac{2x + 3}{x - 4}\)
  2. \(g(x) = \sqrt{x + 5}\)
  3. \(h(x) = 3x - 8\)
  4. \(p(x) = \frac{1}{x^2 + 1}\)
  1. \(f(x) = \frac{2x + 3}{x - 4}\)
    • Domain: \(x - 4 \neq 0\), so \(x \neq 4\) Domain: \(\mathbb{R} \setminus \{4\}\) or \((-\infty, 4) \cup (4, \infty)\)
    • Range: All real numbers except the horizontal asymptote value As \(x \to \infty\), \(f(x) \to 2\) Range: \(\mathbb{R} \setminus \{2\}\)
  2. \(g(x) = \sqrt{x + 5}\)
    • Domain: \(x + 5 \geq 0\), so \(x \geq -5\) Domain: \([-5, \infty)\)
    • Range: Square root outputs are non-negative Minimum: \(g(-5) = 0\) Range: \([0, \infty)\)
  3. \(h(x) = 3x - 8\)
    • Domain: No restrictions for linear function Domain: \(\mathbb{R}\) or \((-\infty, \infty)\)
    • Range: Linear function covers all real values Range: \(\mathbb{R}\) or \((-\infty, \infty)\)
  4. \(p(x) = \frac{1}{x^2 + 1}\)
    • Domain: \(x^2 + 1\) is never zero (always \(\geq 1\)) Domain: \(\mathbb{R}\) or \((-\infty, \infty)\)
    • Range: Since \(x^2 \geq 0\), we have \(x^2 + 1 \geq 1\) So \(0 < \frac{1}{x^2 + 1} \leq 1\) Maximum when \(x = 0\): \(p(0) = 1\) Range: \((0, 1]\)

Problem 3: Bakery Business Functions (xx)

A local bakery has the following cost structure: - Monthly rent and utilities: €2,500 - Ingredients and materials per cake: €12 - Labor cost per cake: €8 - Each cake sells for €45

  1. Define the cost function \(C(x)\) where \(x\) is the number of cakes produced per month
  2. Define the revenue function \(R(x)\)
  3. Define the profit function \(P(x)\)
  4. Create a table showing \(C(x)\), \(R(x)\), and \(P(x)\) for \(x = 0, 50, 100, 150, 200\) cakes
  5. Sketch graphs of all three functions on the same axes
  6. How many cakes must be sold to break even? (Show this on your graph)
  7. If the bakery can produce a maximum of 200 cakes per month, what is the maximum possible profit?
  1. Cost function:
    • Fixed costs: €2,500
    • Variable costs per cake: €12 + €8 = €20
    • \(C(x) = 2500 + 20x\)
  2. Revenue function:
    • Price per cake: €45
    • \(R(x) = 45x\)
  3. Profit function:
    • \(P(x) = R(x) - C(x)\)
    • \(P(x) = 45x - (2500 + 20x)\)
    • \(P(x) = 45x - 2500 - 20x\)
    • \(P(x) = 25x - 2500\)
  4. Table of values:
Cakes (x) Cost C(x) Revenue R(x) Profit P(x)
0 €2,500 €0 -€2,500
50 €3,500 €2,250 -€1,250
100 €4,500 €4,500 €0
150 €5,500 €6,750 €1,250
200 €6,500 €9,000 €2,500
  1. Graph description:
    • Cost function: Starts at €2,500 (y-intercept), increases with slope 20
    • Revenue function: Starts at origin (0,0), increases with slope 45
    • Profit function: Starts at -€2,500, increases with slope 25
    • All three are straight lines
    • Cost and Revenue intersect at break-even point
  2. Break-even point:
    • Set \(P(x) = 0\) or \(C(x) = R(x)\)
    • \(25x - 2500 = 0\)
    • \(25x = 2500\)
    • \(x = 100\) cakes
    • On graph: where Revenue and Cost lines intersect, or where Profit crosses x-axis
  3. Maximum profit:
    • Since \(P(x) = 25x - 2500\) is linear with positive slope
    • Maximum occurs at maximum production: \(x = 200\)
    • \(P(200) = 25(200) - 2500 = 5000 - 2500 = €2,500\)

Problem 4: Manufacturing Constraints (xxx)

A small electronics manufacturer produces two types of devices: tablets and smartphones. The production process has the following constraints:

  • Each tablet requires 3 hours of assembly time and 2 hours of testing
  • Each smartphone requires 2 hours of assembly time and 1 hour of testing
  • The factory has 120 hours of assembly time available per week
  • The factory has 60 hours of testing time available per week
  • Tablets sell for €300 with a production cost of €180
  • Smartphones sell for €200 with a production cost of €110
  1. Let \(t\) represent tablets produced and \(s\) represent smartphones produced. Write the constraint inequalities.
  2. Express the total revenue \(R\) as a function of \(t\) and \(s\)
  3. Express the total profit \(P\) as a function of \(t\) and \(s\)
  4. If the company decides to produce only tablets, what is the maximum number they can produce per week? What would be the profit?
  5. Can the company produce 30 tablets and 20 smartphones in one week? Justify your answer.
  1. Constraint inequalities:
    • Assembly time: \(3t + 2s \leq 120\)
    • Testing time: \(2t + s \leq 60\)
    • Non-negativity: \(t \geq 0, s \geq 0\)
  2. Revenue function:
    • \(R(t, s) = 300t + 200s\)
  3. Profit function:
    • Profit per tablet: €300 - €180 = €120
    • Profit per smartphone: €200 - €110 = €90
    • \(P(t, s) = 120t + 90s\)
  4. Maximum tablets only (\(s = 0\)):
    • Assembly constraint: \(3t \leq 120 \Rightarrow t \leq 40\)
    • Testing constraint: \(2t \leq 60 \Rightarrow t \leq 30\)
    • Limiting constraint is testing: maximum 30 tablets
    • Profit: \(P(30, 0) = 120(30) + 90(0) = €3,600\)
  5. Checking feasibility of \(t = 30, s = 20\):
    • Assembly time: \(3(30) + 2(20) = 90 + 40 = 130\) hours
    • This exceeds the available 120 hours
    • Testing time: \(2(30) + 1(20) = 60 + 20 = 80\) hours
    • This exceeds the available 60 hours
    • No, this production plan is not feasible as it violates both constraints

Problem 5: Investment Portfolio Analysis (xxxx)

An investment advisor is creating a model for a client’s portfolio. The client can invest in three options:

  • Bonds: Fixed return of 4% per year
  • Stocks: Variable return modeled by \(r_s(x) = 0.12 - 0.0001x\) where \(x\) is the amount invested in thousands of euros
  • Real Estate Fund: Return rate of \(r_e(x) = \frac{8}{100 + 0.01x}\) where \(x\) is the amount in thousands of euros

The client has €50,000 to invest total.

  1. If the client invests €20,000 in bonds, €20,000 in stocks, and €10,000 in real estate, calculate the expected annual return in euros.

  2. Find the domain for the stock return function \(r_s(x)\) that ensures a positive return rate.

  3. Explain why the real estate return function \(r_e(x)\) shows diminishing returns as investment increases.

  4. If the client wants to split the investment equally between bonds and stocks only (€25,000 each), compare this to investing all €50,000 in bonds. Which strategy yields higher returns?

  1. Expected annual returns:
    • Bonds: €20,000 × 0.04 = €800
    • Stocks: For \(x = 20\) (thousands): \(r_s(20) = 0.12 - 0.0001(20) = 0.12 - 0.002 = 0.118 = 11.8\%\) Return: €20,000 × 0.118 = €2,360
    • Real Estate: For \(x = 10\) (thousands): \(r_e(10) = \frac{8}{100 + 0.01(10)} = \frac{8}{100 + 0.1} = \frac{8}{100.1} \approx 0.0799 = 7.99\%\) Return: €10,000 × 0.0799 = €799
    • Total return: €800 + €2,360 + €799 = €3,959
  2. Domain for positive stock returns:
    • Need \(r_s(x) > 0\)
    • \(0.12 - 0.0001x > 0\)
    • \(0.12 > 0.0001x\)
    • \(x < \frac{0.12}{0.0001} = 1200\)
    • Domain: \([0, 1200)\) (in thousands of euros)
    • So investments up to €1,200,000 maintain positive returns
  3. Diminishing returns explanation:
    • As \(x\) increases, the denominator \((100 + 0.01x)\) increases
    • This makes the fraction \(\frac{8}{100 + 0.01x}\) smaller
    • The rate decreases as investment increases
    • This models market saturation or decreased efficiency with scale
    • Example: \(r_e(0) = 8\%\), \(r_e(100) \approx 4\%\), \(r_e(1000) \approx 0.73\%\)
  4. Strategy comparison:
    • Split strategy (€25,000 each):
      • Bonds: €25,000 × 0.04 = €1,000
      • Stocks: \(r_s(25) = 0.12 - 0.0001(25) = 0.1175 = 11.75\%\) Return: €25,000 × 0.1175 = €2,937.50
      • Total: €1,000 + €2,937.50 = €3,937.50
    • All bonds strategy:
      • €50,000 × 0.04 = €2,000
    • Conclusion: The split strategy (€3,937.50) yields significantly higher returns than all bonds (€2,000)
    • Difference: €1,937.50 more per year with the split strategy