7️⃣ Introduction to Optimization — Slides

Mathematical Fundamentals · MS01-001-G

Learning Objectives

Compute an average rate of change between two points
Understand the instantaneous rate of change and the intuitive notion of the derivative
Apply derivative rules to affine and degree-2 polynomial functions
Find and classify maxima and minima to support decisions

Formula

Between points \(a\) and \(b\):

\[\text{ARC} = \frac{f(b) - f(a)}{b - a}\]

Geometric meaning: slope of the secant line through \((a,\, f(a))\) and \((b,\, f(b))\)
Economic meaning: average variation of the quantity per unit change in the input

Example: revenue goes from €800 to €1,040 as sales go from 40 to 80 units.

\[\text{ARC} = \frac{1040 - 800}{80 - 40} = \frac{240}{40} = 6 \text{ €/unit}\]

The derivative \(f'(x)\) is the instantaneous rate of change — the limit of the ARC as the interval shrinks to a point.

Geometric meaning: slope of the tangent to the curve at \(x\).

Key Derivative Rules

Function	Derivative
\(f(x) = ax + b\)	\(f'(x) = a\)
\(f(x) = ax^2 + bx + c\)	\(f'(x) = 2ax + b\)

Procedure

Compute \(f'(x)\)
Solve \(f'(x) = 0\) — the critical point is \(x^* = -\dfrac{b}{2a}\)
Build a sign table for \(f'\)
Read the nature of the critical point:
- \(f'\) changes from \(+\) to \(-\) → maximum
- \(f'\) changes from \(-\) to \(+\) → minimum

Profit function: \(\pi(q) = -2q^2 + 120q - 800\)

Step 1 — Compute the derivative: \[\pi'(q) = -4q + 120\]

Step 2 — Solve \(\pi'(q) = 0\): \[q^* = 30 \text{ units}\]

Step 3 — Sign table: \(\pi' > 0\) for \(q < 30\), \(\quad \pi' < 0\) for \(q > 30\)

Step 4 — \(q^* = 30\) is a maximum.

\[\pi(30) = -2(900) + 120(30) - 800 = -1800 + 3600 - 800 = €1{,}000\]

The derivative measures the instantaneous rate of change — it is the slope of the tangent at any point.
Set \(f'(x) = 0\) to find critical points, then use a sign table to classify them as maxima or minima.
For \(f(x) = ax^2 + bx + c\): \(a < 0\) → maximum at \(x^*\); \(a > 0\) → minimum at \(x^*\).