Introduction to Optimization

NoteLearning objectives

• Compute an average rate of change between two points.
• Understand the instantaneous rate of change and the intuitive notion of the derivative.
• Differentiate affine and degree-2 polynomial functions.
• Use the derivative to find a maximum or a minimum.

📊 Lecture slides: View the slides for this module

1 Diagnostic Activity

Answer these questions before reading further. They review the tools you will need throughout this module.

1. Compute \(f'(x)\) for \(f(x) = 3x^2 - 4x\).
2. Find \(x\) where \(f'(x) = 0\) for \(f(x) = -x^2 + 6x\).
3. For \(f(x) = -x^2 + 6x\), is \(x = 3\) a maximum or a minimum? Justify your answer.

NoteCheck your answers

1. Using the power rule: \(f'(x) = 6x - 4\).
2. \(f'(x) = -2x + 6 = 0 \;\Rightarrow\; x = 3\).
3. Maximum, because \(a = -1 < 0\) so the parabola opens downward.

Use your results to identify topics needing attention.

2 Opening Problem

A manufacturer produces \(q\) units of a product. The revenue function is \(R(q) = 60q - 0.2q^2\) (euros) and the total cost function is \(C(q) = 500 + 20q\) (euros).

What quantity \(q\) maximises the manufacturer’s profit? Write down your estimate before continuing.

3 Core Concepts

What Does “Optimise” Mean?

In mathematics, to optimize means to find the maximum or the minimum of a function. In management there are many situations that call for an optimal decision:

• maximisation: firm profit, production, yield, consumer utility;
• minimisation: average cost, production or delivery time, inventory, the risk of a portfolio.

To find a maximum or a minimum we study how the function varies: over which intervals is it increasing, and over which is it decreasing? The derivative is the tool that answers this question precisely.

Average Rate of Change

NoteDefinition

The average rate of change of a function \(f\) between \(x_1\) and \(x_2\) is \[ \tau = \frac{f(x_2) - f(x_1)}{x_2 - x_1}. \]

Geometrically, it is the slope of the secant line joining the two points \((x_1, f(x_1))\) and \((x_2, f(x_2))\) on the graph. Economically, it measures the average change of \(f\) per unit of \(x\) — for instance the average change of cost per additional unit produced.

Worked example — average rate of change

For \(f(x) = x^{2}\) between \(x_1 = 1\) and \(x_2 = 3\):

\[ \tau = \frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4. \]

Instantaneous Rate of Change and the Derivative

If we let the second point get closer and closer to the first (\(x_2 \to x_1\)), the secant line tends towards the tangent line at the point. Its slope is the instantaneous rate of change, called the derivative of \(f\) at \(x_1\), written \(f'(x_1)\).

TipIntuition

• \(f'(x) > 0\) on an interval \(\;\Rightarrow\; f\) is increasing there.
• \(f'(x) < 0\) on an interval \(\;\Rightarrow\; f\) is decreasing there.
• \(f'(x) = 0\) at a point where \(f\) changes direction \(\;\Rightarrow\;\) a candidate maximum or minimum (the tangent is horizontal).

CautionDerivation — the derivative of \(f(x) = x^{2}\) (click to expand)

Take a point \(x\) and a small increment \(h\). The average rate of change between \(x\) and \(x + h\) is

\[ \frac{f(x + h) - f(x)}{h} = \frac{(x + h)^{2} - x^{2}}{h} = \frac{x^{2} + 2xh + h^{2} - x^{2}}{h} = \frac{2xh + h^{2}}{h} = 2x + h. \]

As \(h\) becomes infinitely small (\(h \to 0\)), this tends to \(2x\). Hence the instantaneous rate of change is

\[ f'(x) = 2x. \]

TipDerivatives of the reference functions

• Constant \(f(x) = k\): \(\;f'(x) = 0\).
• Affine \(f(x) = a x + b\): \(\;f'(x) = a\) (constant slope).
• Power \(f(x) = x^{n}\): \(\;f'(x) = n\,x^{n-1}\).
• Quadratic \(f(x) = a x^{2} + b x + c\): \(\;f'(x) = 2a x + b\).

Applying the Derivative to Optimization

ImportantMethod to optimise a quadratic

1. Compute the derivative \(f'(x)\).
2. Solve \(f'(x) = 0\) to find the candidate point \(x^{*}\).
3. For \(f(x) = a x^{2} + b x + c\), the extremum is at \(x^{*} = -\dfrac{b}{2a}\): it is a maximum if \(a < 0\) and a minimum if \(a > 0\).

Worked example — maximising profit

A firm’s profit is \(P(x) = -2x^{2} + 40x - 50\) (in thousands of euros), where \(x\) is the quantity produced. Then \(P'(x) = -4x + 40\). Setting \(P'(x) = 0\) gives \(x = 10\). Since \(a = -2 < 0\), the parabola opens downward, so \(x = 10\) is a maximum. The maximum profit is \(P(10) = -2(100) + 400 - 50 = 150\) (thousand euros).

CautionPractice exercises (click for solutions)

1. Average rate of change of \(f(x) = 3x + 2\) between \(x = 0\) and \(x = 4\). \(\tau = \dfrac{14 - 2}{4} = 3\) (equal to the slope, as expected for an affine function).
2. Differentiate \(f(x) = 5x^{2} - 3x + 7\). \(\;f'(x) = 10x - 3\).
3. Minimise the average cost \(C(x) = x^{2} - 12x + 40\). \(C'(x) = 2x - 12 = 0 \Rightarrow x = 6\); since \(a > 0\) it is a minimum, \(C(6) = 4\).

4 Worked Example

Worked example — maximising the manufacturer’s profit (Opening Problem)

We return to the opening problem. Given:

\[ R(q) = 60q - 0.2q^2, \qquad C(q) = 500 + 20q. \]

Step 1 — Write the profit function.

\[ P(q) = R(q) - C(q) = 60q - 0.2q^2 - 500 - 20q = -0.2q^2 + 40q - 500. \]

Step 2 — Compute the derivative.

\[ P'(q) = -0.4q + 40. \]

Step 3 — Solve \(P'(q) = 0\).

\[ -0.4q + 40 = 0 \;\Rightarrow\; q^* = 100 \text{ units.} \]

Step 4 — Confirm the nature of the critical point.

Since \(a = -0.2 < 0\), the parabola opens downward, so \(q^* = 100\) is a maximum.

Step 5 — Compute the maximum profit.

\[ P(100) = -0.2(10{,}000) + 40(100) - 500 = -2{,}000 + 4{,}000 - 500 = 1{,}500 \text{ €.} \]

The manufacturer maximises profit at 100 units, earning €1,500.

5 Guided Practice

A firm’s average cost function is \(C(x) = x^2 - 12x + 40\), where \(x\) is the production level (hundreds of units).

1. Find the production level that minimises average cost.
2. What is the minimum average cost?
3. On which intervals is average cost decreasing? Increasing?

TipHint

Compute \(C'(x)\) and set it equal to zero. Since \(a > 0\), the critical point is a minimum.

CautionShow the solution

• \(C'(x) = 2x - 12 = 0 \;\Rightarrow\; x = 6\) (hundreds of units).
• Minimum average cost: \(C(6) = 36 - 72 + 40 = 4\) (per unit).
• Decreasing on \(]-\infty, 6[\,\); increasing on \(\,]6, +\infty[\).

6 Interactive Exploration

Use the slider to choose a production quantity \(q\) and observe how revenue, cost, and profit change. The red dashed line marks the analytically derived optimum \(q^* = 100\).

viewof quantity = Inputs.range([0, 200], {
  value: 50,
  step: 1,
  label: "Quantity q (units)"
})

qRange = Array.from({length: 201}, (_, i) => i)
curveData = qRange.flatMap(q => [
  {q, value: 60*q - 0.2*q*q, curve: "Revenue R(q)"},
  {q, value: 500 + 20*q, curve: "Cost C(q)"},
  {q, value: 60*q - 0.2*q*q - 500 - 20*q, curve: "Profit P(q)"}
])

qStar = 100
revenue_q = 60*quantity - 0.2*quantity*quantity
cost_q = 500 + 20*quantity
profit_q = revenue_q - cost_q
revenue_opt = 60*qStar - 0.2*qStar*qStar
profit_opt = revenue_opt - (500 + 20*qStar)

profitChart = Plot.plot({
  marks: [
    Plot.line(curveData, {x: "q", y: "value", stroke: "curve"}),
    Plot.ruleY([0], {stroke: "#aaa", strokeDash: [2,2]}),
    Plot.ruleX([quantity], {stroke: "white", strokeDash: [3,3]}),
    Plot.ruleX([qStar], {stroke: "red", strokeDash: [4,2]})
  ],
  x: {label: "Quantity q"},
  y: {label: "Value (€)"},
  color: {legend: true},
  width: 550,
  height: 320
})

html`
<div style="margin-bottom:1rem">${profitChart}</div>
<p style="font-size:0.85rem;opacity:0.7">White dashed line: your slider
position q = ${quantity}. Red dashed line: optimal q* = 100.</p>
<div class="metric-grid">
  <div class="metric-card">
    <span class="metric-label">Revenue R(q)</span>
    <strong>${revenue_q.toFixed(0)} €</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Cost C(q)</span>
    <strong>${cost_q.toFixed(0)} €</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Profit P(q)</span>
    <strong>${profit_q.toFixed(0)} €</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Maximum profit at q* = 100</span>
    <strong>${profit_opt.toFixed(0)} €</strong>
  </div>
</div>
`

Investigation

1. Move the slider to \(q = 100\). Does the displayed profit match the worked example result (€1,500)?
2. Move the slider to \(q = 50\) and \(q = 150\). Are both profits the same? Why? What does this say about the shape of the profit function?
3. At what quantity does revenue equal cost (the break-even point)? Use the slider to find it approximately, then verify algebraically.
4. Does producing more always increase profit? Find the quantity beyond which profit starts to decline.

7 Applied Problem

A bakery sells artisan bread at €4 per loaf. The daily cost function is \(C(q) = 0.05q^2 + 0.5q + 80\) (euros), where \(q\) is the number of loaves baked.

1. Write the daily profit function \(P(q)\).
2. Find the number of loaves that maximises daily profit.
3. What is the maximum daily profit?
4. Find the break-even quantities (where profit = 0).

8 Common Errors

WarningCommon error — not verifying whether the critical point is a maximum or minimum

A student finds \(f'(x) = 0\) at \(x = 5\) and immediately declares “the function has a maximum at \(x = 5\).”

Finding \(f'(x^*) = 0\) identifies a candidate extremum — it could be a maximum, a minimum, or neither (an inflection point). For a quadratic \(f(x) = ax^2 + bx + c\):

• If \(a < 0\): the critical point is a maximum (parabola opens downward).
• If \(a > 0\): the critical point is a minimum (parabola opens upward).

Always check the sign of \(a\) before concluding.

9 Check Your Understanding

1. Compute the average rate of change of \(f(x) = 2x^2 - 3\) between \(x = 1\) and \(x = 4\).
2. Differentiate \(g(x) = -3x^2 + 12x - 5\).
3. A cost function is \(C(q) = 2q^2 - 40q + 300\). Find the production level that minimises cost.
4. A student sets the derivative of a profit function equal to zero and gets \(q = 80\). She concludes profit is maximised. What additional check should she perform?
5. Revenue is \(R(q) = 50q\) and cost is \(C(q) = q^2 + 100q + 200\). Is there a quantity that maximises profit? What does a negative optimal \(q\) mean in this context?

10 Summary

1. The derivative \(f'(x)\) measures the instantaneous rate of change of \(f\) at \(x\): it is positive when \(f\) is increasing, negative when decreasing, and zero at a candidate extremum (where the tangent is horizontal).
2. To optimise a quadratic function, compute \(f'(x)\), set it to zero to find \(x^*\), then confirm whether \(x^*\) is a maximum (if \(a < 0\)) or minimum (if \(a > 0\)) by checking the sign of the leading coefficient.
3. Calculus provides the exact optimal decision — the exact quantity, price, or production level — rather than a graphical approximation; combining the derivative with the economic context (does a maximum make sense?) is what turns mathematics into decision support.

11 Independent Work

• Compute derivatives for functions 1–8 in the exercise bank.
• Solve optimisation problems A–F.
• Compare the derivative-based solution with the slider-based estimate from the Interactive Exploration.
• Record errors in your error log.
• Review glossary entries for: derivative, rate of change, maximum, minimum, optimisation.

12 Continue

Continue to Module 8 →