Session 03-03 - Quadratic Functions & Basic Optimization

Section 03: Functions as Business Models

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz - 10 Minutes

Review from Session 03-02

Work individually, then we discuss together as group

  1. Find the market equilibrium for:

    • Demand: \(Q_d = 200 - 2p\)
    • Supply: \(Q_s = 50 + 3p\)

  2. Write the equation of a line passing through points (2, 8) and (5, 20).

  3. For the cost function \(C(x) = 500 + 12x\) and revenue \(R(x) = 25x\), find the profit when \(x = 100\).

Homework Discussion - 15 Minutes

Discussing Tasks 03-02

Let’s discuss the most difficult tasks from last lecture

  • Problem 5: Market competition analysis
    • How did you determine the break-even data usage?
  • Problem 6: Production planning with constraints
    • Challenges with multiple constraints?
  • Problem 7: Dynamic pricing (if attempted)
    • What price seemed optimal in your testing?

Today we’ll learn the exact method to find that optimal price!

Learning Objectives

Learning Objectives

By the end of this session, you can:

  • Interpret the parameters \(a\), \(b\), \(c\) of a quadratic function
  • Switch between standard, vertex, and factored form - and know when each helps
  • Find the vertex with \(x_v = -\frac{b}{2a}\) and use it to optimize
  • Distinguish revenue maximization from profit maximization
  • Respect real-world constraints that may override the mathematical optimum

Introduction to Quadratic Functions

From Linear to Quadratic

Quadratic functions model accelerating change

Linear vs. Quadratic:

  • Linear: \(f(x) = mx + b\) → Constant rate of change
  • Quadratic: \(f(x) = ax^2 + bx + c\) → Changing rate of change
  • Graph shape: Quadratic → Parabola (U-shaped or ∩-shaped)
  • Business meaning:
    • Linear → Fixed relationships
    • Quadratic → Optimization opportunities!

Standard Form

The foundation: f(x) = ax² + bx + c

Key components:

  • a: Direction and width
    • \(a > 0\): Opens upward (has minimum)
    • \(a < 0\): Opens downward (has maximum)
    • \(|a|\) larger → Narrower parabola
  • b: Affects position of vertex
  • c: y-intercept (value when \(x = 0\))

The Role of a

Larger \(|a|\) → narrower parabola; negative \(a\) flips it upside down.

Anatomy of a Parabola

Example: \(f(x) = x^2 - 4x + 3\) - every feature has a business meaning when \(f\) models profit or cost.

Factored Form

Third representation: f(x) = a(x - r₁)(x - r₂)

  • Roots: \(r_1, r_2\) - directly visible!
  • Direction: \(a\) (same as always)
  • Vertex: halfway between the roots, \(x_v = \frac{r_1 + r_2}{2}\)
  • Example: \(f(x) = 2(x - 1)(x - 3)\)
    • Roots at \(x = 1\) and \(x = 3\), vertex at \(x = 2\)
    • Expanded: \(f(x) = 2x^2 - 8x + 6\) (same function, standard form)

For a profit function, the roots are the break-even points!

Example: Profit Function

Quadratic Profit Function: \(P(x) = -2x^2 + 100x - 800\)

Quick Practice - 10 Minutes

Work individually, then we discuss

  • Determine: Does it open upward (U) or downward (∩)?
  • Determine: Does it have a maximum or minimum?
  • Determine: What is the y-intercept?

  1. \(R(x) = -3x^2 + 120x - 500\)

  2. \(C(x) = 2x^2 + 40x + 1000\)

  3. \(P(x) = -x^2 + 50x - 300\)

Challenge: For c. find the break-even points.

Break - 10 Minutes

Finding the Vertex

The Vertex Formula

The key: x = -b/2a

For \(f(x) = ax^2 + bx + c\):

  • Vertex x-coordinate: \(x_v = -\frac{b}{2a}\)
  • Vertex y-coordinate: \(f(x_v) = f(-\frac{b}{2a})\)
  • Vertex represents:
    • Maximum if \(a < 0\) (parabola opens down)
    • Minimum if \(a > 0\) (parabola opens up)
  • Axis of symmetry: Vertical line \(x = x_v\)

Vertex Example: Revenue Optimization

A company’s revenue depends on price:

\[R(p) = -50p^2 + 2000p\]

  • Find optimal price: \(p_v = -\frac{2000}{2(-50)} = -\frac{2000}{-100} = 20\) euros
  • Maximum revenue: \(R(20) = 20000\) euros
  • Interpretation: Charging €20 maximizes revenue at €20,000

The axis of symmetry divides the parabola into mirror images. Points equidistant from it have equal revenue!

Visualization

\(R(p) = -50p^2 + 2000p\)

Vertex Form

Alternative representation: f(x) = a(x - h)² + k

  • Vertex: \((h, k)\) - directly visible!
  • Direction: \(a\) (same as standard form)
  • Advantage: Vertex immediately apparent
  • Transformation from vertex:
    • Horizontal shift by \(h\)
    • Vertical shift by \(k\)
  • Example: \(f(x) = 2(x - 3)^2 + 5\) → Vertex at \((3, 5)\), minimum
  • Example: \(g(x) = -(x + 4)^2 + 10\) → Vertex at \((-4, 10)\), maximum

Recap: Completing the Square

From standard form to vertex form - you know this from Session 02-02

Example: \(f(x) = 2x^2 - 12x + 10\)

  1. Factor out 2: \(f(x) = 2(x^2 - 6x) + 10\)
  2. Add and subtract \((\frac{-6}{2})^2 = 9\): \(f(x) = 2((x - 3)^2 - 9) + 10\)
  3. Simplify: \(f(x) = 2(x - 3)^2 - 8\) → Vertex \((3, -8)\), minimum

The full technique was covered in Session 02-02 (Quadratic, Biquadratic & Cubic Equations) - here we only use it to read off the vertex.

Practice: From Graph to Formula

FSP-style: determine the function from its properties

A parabola has its vertex at \((2, -8)\) and passes through the point \((0, 4)\).

  1. Write \(f(x)\) in vertex form
  2. Expand to standard form and give the y-intercept

Think-Pair-Share: Which Form When?

3 minutes alone, 2 minutes with your neighbour, then we discuss

Your colleague claims: “Vertex form is always best - I would convert every quadratic immediately.”

  • What does each form show at a glance: standard, vertex, factored?
  • For which business question would you pick which form?
  • When is converting a waste of time?

Standard → y-intercept (fixed costs), vertex → optimum, factored → break-even points. Pick the form that answers your question directly!

Business Applications

Price-Dependent Demand

When price affects quantity: Revenue becomes quadratic!

Basic Scenario:

  • Demand function: \(Q = a - bp\) (quantity depends on price)
  • Revenue: \(R = p \times Q = p(a - bp)\)
  • Expanded: \(R(p) = ap - bp^2 = -bp^2 + ap\)
  • This is quadratic in \(p\)!

Remember, we have seen this in the past!

Example: Concert Venue

A venue (capacity: 1000) has ticket demand: \(Q = 1000 - 20p\)

  • Revenue function: \(R(p) = p(1000 - 20p) = 1000p - 20p^2\)
  • Optimal price: \(p^* = -\frac{1000}{2(-20)} = \frac{1000}{40} = 25\) euros
  • Tickets sold: \(Q = 1000 - 20(25) = 500\)
  • Maximum revenue: \(R(25) = 25 \times 500 = 12,500\)
  • At €0: Demand = 1000 (full capacity if free)
  • At €50: Demand = 0 (too expensive, no one buys)

Note: This maximizes revenue, not necessarily profit!

Example: Minimizing Unit Costs

Not every optimum is a maximum - upward parabolas have minima!

A workshop’s cost per unit depends on the batch size \(x\):

\[c(x) = 0.5x^2 - 20x + 350\]

  • \(a = 0.5 > 0\) → opens upward → the vertex is a minimum
  • Cost-minimizing batch size: \(x_v = -\frac{-20}{2(0.5)} = 20\) units
  • Minimum unit cost: \(c(20) = 200 - 400 + 350 = 150\) euros
  • Too-small batches waste setup time, too-large batches cause overtime - the parabola captures both

Guided Practice - 20 Minutes

Spot the Error

Three statements - each contains a mistake. Find them!

  1. “For \(f(x) = x^2 - 6x + 5\), the vertex is at \(x_v = \frac{b}{2a} = \frac{-6}{2} = -3\).”

  2. “Since \(a = 1 > 0\), the vertex of \(f(x) = x^2 - 6x + 5\) is its maximum.”

  3. “The y-intercept of \(f(x) = 2(x - 1)(x - 3)\) is \(-3\), because the constant term is \(-3\).”

Individual Exercise Block

Work alone for 15 minutes, then we compare solutions

  1. For \(f(x) = x^2 - 8x + 12\):
    1. Find the vertex using the formula
    2. Determine if it’s a maximum or minimum and find the y-intercept
  2. A profit function is \(P(x) = -3x^2 + 240x - 3600\):
    1. Find the number of units that maximizes profit
    2. Calculate the maximum profit and the break-even points
  3. Convert \(f(x) = 2x^2 - 12x + 14\) to vertex form by completing the square, then identify the vertex.

Practice: Setting Up a Profit Function

FSP-style: from words to optimum

A food stall sells \(x\) portions per day. The price per portion falls with quantity: \(p(x) = 20 - 0.5x\) euros. Daily costs are \(C(x) = 2x + 50\) euros.

  1. Set up the revenue function \(R(x)\)
  2. Set up the profit function \(P(x)\)
  3. How many portions maximize profit, and what is the maximum daily profit?

Coffee Break - 15 Minutes

Projectile Motion

Product Launch Campaign

Marketing models new product awareness like projectile motion

\[A(t) = -2t^2 + 24t\] where \(A\) is awareness score and \(t\) is weeks after launch.

  • Peak awareness time: \(t = -\frac{24}{2(-2)} = 6\) weeks
  • Maximum awareness: \(A(6) = -72 + 144 = 72\) points
  • Campaign ends when \(A(t) = 0\): at \(t = 0\) and \(t = 12\) weeks

Campaign follows symmetric pattern: builds to peak at 6 weeks, then decays at same rate.

Campaign Awareness

Product Launch Campaign: \(A(t) = -2t^2 + 24t\)

Area Optimization

Maximizing Area with Constraints

Classic problem: Maximum area with fixed perimeter

Rectangular Storage Area with 200 meters of fencing available. One side against a building (no fence) and we want to maximize storage area.

  • Let \(x\) = width, \(y\) = length parallel to building
  • Constraint: \(2x + y = 200\) (fencing)
  • So: \(y = 200 - 2x\)
  • Area: \(A = xy = x(200 - 2x) = 200x - 2x^2\)
  • Maximum at: \(x = -\frac{200}{2(-2)} = 50\) meters
  • Dimensions: 50m × 100m, Area = 5000 m²

Visualization

Too narrow or too wide both shrink the area - the optimum sits in the middle!

Collaborative Problem-Solving - 30 Minutes

Comprehensive Business Optimization

The Scenario: Smart Tech Product Launch

Smart Tech is launching a new tablet. Market research indicates:

  • At €200: would sell 8,000 units per month
  • At €400: would sell 4,000 units per month
  • At €600: would sell 0 units (too expensive)
  • Production cost: €150 per tablet
  • Fixed monthly costs: €200,000

Assume linear demand relationship.

Your Tasks:

Work in groups of 3-4 students

  1. Derive the demand function \(Q(p)\) where \(p\) is price

  2. Express revenue \(R(p)\) as a function of price (this will be quadratic!)

  3. Find the price that maximizes revenue

  4. Express profit \(\Pi(p)\) as a function of price

  5. Find the price that maximizes profit (different from revenue-maximizing price!)

  6. If the company can only produce 5,000 tablets per month, should they use the profit-maximizing price? Explain.

Wrap-Up & Key Takeaways

Key Takeaways

  • The vertex formula \(x = -\frac{b}{2a}\) is your optimization tool
  • Quadratic functions model scenarios with changing rates
  • Maximum/minimum depends on sign of \(a\)
  • Revenue maximization ≠ Profit maximization
  • Completing the square reveals the vertex form
  • Real constraints may override mathematical optima

Remember

Every parabola has a minimum or a maximum point!

Final Assessment

5 minutes - Individual work

A small bakery’s daily profit for chocolate cakes is modeled by: \[P(x) = -x^2 + 14x - 33\] where \(x\) is the price in euros.

  1. Find the price that maximizes profit
  2. Calculate the maximum daily profit
  3. Find the break-even prices

Next Session Preview

Session 03-04: Transformations, Composition & Inverses

  • Shifting functions horizontally and vertically
  • Stretching and reflecting graphs
  • Combining functions through composition
  • Inverse functions and when they exist

Homework Assignment: Complete Tasks 03-03!