Phase 2 - Lesson 2.5

Optimization in One Variable

Finding maxima and minima with the first- and second-derivative tests, and applying them to profit maximization and utility.

⏱ 55 min● Intermediate🔗 Prereqs: 2.4

↖ Phase 2 hub

Builds on: 2.2-2.3 gave derivatives; optimization is their first big payoff.
Leads to: These ideas generalize to gradients and the KKT conditions of convex optimization (Phase 10).

Learning Objectives

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Locate critical points and classify them with the first- and second-derivative tests.
Distinguish local from global extrema and handle closed-interval endpoints.
State the roles of Fermat’s theorem and the Extreme Value Theorem in optimization.
Solve a profit-maximization problem using marginal analysis.

Key Vocabulary

Critical point: A point where f′=0 or f′ is undefined; the only candidates for interior extrema.

Fermat’s theorem: If f has a local extremum at an interior point where it is differentiable, then f′=0 there.

First-derivative test: f′ changing + to − marks a local max; − to + a local min.

Second-derivative test: At a critical point, f″<0 ⇒ local max, f″>0 ⇒ local min, f″=0 inconclusive.

Global (absolute) extremum: The largest/smallest value over the whole domain, possibly at an endpoint.

Concavity: f″>0 means convex (cup up); f″<0 means concave (cap down).

Where extrema can hide

Fermat’s theorem: at an interior local max or min of a differentiable function, the tangent is horizontal, so $f'(x)=0$. Thus interior extrema occur only at critical points. On a closed interval $[a,b]$ the Extreme Value Theorem guarantees a global max and min exist, and they sit either at a critical point or an endpoint.

Key Idea

Global-extremum recipe on [a,b]: (1) find all critical points inside, (2) evaluate f at them and at both endpoints, (3) the largest value is the global max, the smallest the global min. No calculus theorem beyond this is needed.

Classifying a critical point

Two tests decide whether a critical point is a max, a min, or neither. The first-derivative test reads the sign change of $f'$; the second-derivative test reads curvature: $f''\gt 0$ (convex) means a min, $f''\lt 0$ (concave) means a max.

Worked Example - Classify the critical points of f(x)=x³−3x

$f'(x)=3x^2-3=3(x-1)(x+1)$, so critical points at $x=\pm1$.

$f''(x)=6x$. At $x=1$, $f''=6\gt 0$ → local min $f(1)=-2$.

At $x=-1$, $f''=-6\lt 0$ → local max $f(-1)=2$.

Neither is global: $x^3-3x\to\pm\infty$, so on all of $\R$ there is no absolute extremum.

Finance link: profit maximization

A firm with revenue $R(q)$ and cost $C(q)$ maximizes profit $\pi(q)=R(q)-C(q)$. Setting $\pi'(q)=0$ gives the classic condition $R'(q)=C'(q)$ - marginal revenue equals marginal cost. The second-order condition $\pi''\lt 0$ (i.e. $R''\lt C''$) confirms it is a maximum, not a minimum.

Interactive: hunt the optimum

Common Mistakes to Avoid

Assuming every critical point is an extremum - f′=0 can be a saddle/inflection, as x³ at 0.
Reporting a local max as the global max without checking endpoints and end-behavior.
Forgetting critical points where f′ is undefined (corners), not just where f′=0.
Trusting the second-derivative test when f″=0 - it is inconclusive; fall back to the first-derivative test.

Quant Practitioner Tips

On a closed interval, always tabulate f at critical points AND endpoints before declaring a global winner.
Marginal-equals-marginal (MR=MC) is Fermat’s theorem wearing an economics hat.
Concavity (sign of f″) doubles as a max/min classifier and a convexity check for optimization theory.
If the second-derivative test stalls (f″=0), examine the sign change of f′ directly.

Knowledge Check

Q1 Easy

At an interior local maximum of a differentiable function, Fermat’s theorem guarantees:

f″=0

f′=0

f=0

f is increasing

Q2 Medium

If f′(c)=0 and f″(c)>0, then c is a:

local maximum

local minimum

saddle point

global maximum

Q3 Medium

A firm maximizes profit π=R−C. The first-order condition is:

R=C

R′=C′ (marginal revenue = marginal cost)

R′=0

C′=0

Practical Exercise

A manufacturer’s profit is $\pi(q)=-q^3+9q^2-15q-10$ (thousands of dollars) for $q\in[0,7]$ units (thousands). Find the production level that maximizes profit and the maximum profit.

▶ Show full solution

Differentiate: $\pi'(q)=-3q^2+18q-15=-3(q^2-6q+5)=-3(q-1)(q-5)$. Critical points $q=1,5$.

Second derivative $\pi''(q)=-6q+18$: at $q=5$, $\pi''=-12\lt 0$ (local max); at $q=1$, $\pi''=12\gt 0$ (local min).

Evaluate candidates and endpoints: $\pi(0)=-10,\ \pi(1)=-17,\ \pi(5)=15,\ \pi(7)=-17$. The global maximum is at $q=5$ with profit $\$15{,}000$.

After the reveal, answer for yourself: Notice the interior local min at q=1 was the WORST candidate. Always compare every critical point and endpoint numerically before crowning the maximum.

Lesson Summary

Interior extrema of a differentiable function occur only at critical points (Fermat), and on a closed interval the global extrema (guaranteed by the Extreme Value Theorem) lie among critical points and endpoints. The first- and second-derivative tests classify each candidate. In economics this is marginal analysis: profit peaks where marginal revenue equals marginal cost.

Formula Sheet Additions

Profit optimum

\[\pi'(q)=R'(q)-C'(q)=0\ \Rightarrow\ R'(q)=C'(q)\]

Marginal revenue equals marginal cost - Fermat’s theorem in economics.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: Give the recipe for the global max/min of a continuous f on [a,b].
A: Find interior critical points (f′=0 or undefined), evaluate f there and at both endpoints, and pick the largest and smallest values; the Extreme Value Theorem guarantees they exist.

Q: State the second-derivative test and when it fails.
A: At a critical point c: f″(c)<0 gives a local max, f″(c)>0 a local min; if f″(c)=0 it is inconclusive and you use the first-derivative sign-change test instead.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

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